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# Weaponising a substance with low mass but intense gravity

We have all heard of the gravity formula $$F = G \frac{m_1 m_2}{r^2}\$$, establishing a relation between the mass of an object and the gravity force it is affected by. Welp, turns out Newton was wrong! The real equation is $$F = G \frac{x_1 x_2}{r^2}\$$; and $$x = m$$ for every substance known to man - except for this newly discovered substance, phlebotinum. For phlebotinum, $$x = ym$$, and $$y$$ is a very big number. What that means is that a teaspoon of phlebotinum, with a mass of a few grams, has enough gravity to meaningfully interact with its environment - destructively so.

I want to narrow down the value of $$y$$ for the weaponisation of this substance. However, an implication I did not realise upon first asking this question is that the planet is a source of gravity too; therefore if phlebotinum has the equivalent gravity of a mountain, it would also be as hard to pick up as a mountain.

So I'll have to make $$y$$ a bit lower. That leaves me with the question: can you still weaponise a substance that is affected by gravity more intensely than usual? With the goal finding a value of $$y$$ so that the effects of extreme short-range gravity can be utilised properly, without making the phlebotinum impossible to handle.

• How would the planet, on which the city is build, would affect the hail of phlebotinum balls? One could image that said "ton" would never get off the ground. Other ones would say that maybe sending them one by one is possible but the acceleration of said ball from 2000 metres would be more devastating that the gravity force of it. Dec 14, 2020 at 10:00
• @KeizerHarm Certainly, since the phlebotinum (p) violated the equivalence principle, it nullifies general relativity (The stree-energy tensor and Einstein tensor no longer share the same relations through curvature since the p-mass is "bending" space more than its mass would allow). This alone breaks a fundamental pillar of physics, but its a little deeper than that. See my answer here (worldbuilding.stackexchange.com/questions/191777/…) where I discuss mass's relationship to momentum and energy. (1/2) Dec 16, 2020 at 9:12
• This means that special relativity must be corrected, and so must the relativistic dispersion relationship because at this point we have to decide between inertial mass-energy and gravitational mass-energy since this distinction now determines the dispersion relationship in a gravitational field vs. any other field. This means that the mass-energy conservation is broken and this implies a breaking of time translation symmetry! Clearly this is a problem so relativity and the dispersion relationship would have to be rescued but how? This would require reformulating physics. (2/3) Dec 16, 2020 at 9:35
• Even with the given prompt, the problem is compounded by the notation that for a kilo of p-mass to exercise a gravitation newton of force on a kilo of regular matter at a meter, "y" would need to be on the order of 10000000000! Note that the acceleration of p-mass on Earth would be 9.81 m/s^2 multiplied by y! If y is only a factor of 1000, its gravitational acceleration would still be 9810 m/s^2, while its effects on surrounding matter would not be noticeable (~ 0.000000001 N at a meter on a kilo of regular matter)! (3/3) Dec 16, 2020 at 9:44
• @user110866 Thank you kindly for helping me down Mount Dunning-Kruger :) You are completely right, and I would be wise to do further research at how physical equations arrived the way they are before I begin modifying them for alternate realities. Nevertheless, I asked for a reality check and I say you have given me a thorough one, broadening my understanding of the physics involved, so I would love for you to make that a proper answer so that I may upvote and accept it. Dec 16, 2020 at 9:49

In general its never a good idea to start with a modified physics equation and work forward unless you really have deep understanding of physics (and even then!). Its usually best to start with the effect (a short-range gravitational-like attraction between mass) and work forward from there to avoid breaking physics. In analogy, this sounds a bit like asking the appropriate amount of money banks should add to everyone's bank account so the whole world would be rich...both economics and physics are complicated and very mathematically intense; changing one fundamental thing to produce a certain desired effect has the tendency to change everything in strange and unusual ways.

PcMan raised the issues of decoupling gravitational mass and inertial mass. To elaborate, since the phlebotinum (p) violated the equivalence principle, it nullifies general relativity. The stress-energy tensor and Einstein tensor no longer share the same relations through curvature since the p-mass is "bending" space more than its mass would allow. This alone breaks a fundamental pillar of physics, but its a little deeper than that. See my answer here where I discuss mass's relationship to momentum and energy.

This means that special relativity must be corrected, and so must the relativistic dispersion relationship because at this point we have to decide between inertial mass-energy and gravitational mass-energy since this distinction now determines the dispersion relationship in a gravitational field vs. any other field. This means that the mass-energy conservation is broken and this implies a breaking of time translation symmetry! Clearly this is a problem (spontaneous breaking of ground state symmetries leads to a whole slew of issues), so relativity and the dispersion relationship would have to be rescued, but how? This would require reformulating physics!

Even with the given prompt, the problem is compounded by the notation that the acceleration of p-mass on Earth would be 9.81 m/s$$^2$$ multiplied by the y-factor! For example, y is only a factor of 1000, the p-matter's gravitational acceleration would be 9810 m/s$$^2$$, while its effects on surrounding matter would not be noticeable (~ 0.000000001 N at a meter on a kilo of regular matter)! Lets suppose y is 10, the material would still be launched towards Earth's surface with an acceleration of about 91 m/s$$^2$$! However, a gram of p-mass would only exert a gravitational pull of 6.67$$\cdot$$10$$^{-11}$$ N on another gram of matter at a distance of a millimeter. Weaponizing this substance would essentially boil down to dropping it with explosive results. Constructing a container that could stand such pressures could be a challenge. Even more dangerous is that our gravity/inertia masses are decoupled for which we have no intuitive knowledge, for example, its easy to role a ball of it across a table but it falls with explosive force if pushed over the edge!

In conclusion, simple changes to fundamental equations rarely leads anything desirable, but usually brings lots of undesirable side-effects.

• Once again thank you for enlightening me. Clearly this is not an avenue I will be pursuing in my quest for something physically interesting that doesn't break the laws of physics. I may instead explore adding a wholly different field, not tangled with the rest of physics; and still a field that gives every object some potential energy much like magnetism and gravity does; so as not to disrupt conservation of energy but still allow for exotic matter that behaves weirdly. But that I think will be a story for a different question, and this time I will do more research on the beforehand :) Dec 16, 2020 at 10:55

Invalid question: false and contradictory assumptions.

The real equation is F=Gx1x2r2 ; and x=m for every substance known to man - except for this newly discovered substance, phlebotinum. For phlebotinum, x=ym, and y is a very big number. What that means is that a teaspoon of phlebotinum, weighing a few grams, has enough gravity to meaningfully interact with its environment - destructively so.

Incorrect derivative.

Your tweaking of the equations uncouples INERTIAL mass from the Gravitational mass, but does NOT break symmetry of gravitational mass. This means your phlebotium would STILL WEIGH an immense amount.

• Did you forget to read the final paragraph of the question? Dec 14, 2020 at 18:46
• This is a valid answer, this is what we call frame challenge and we expect people to do it :) Dec 16, 2020 at 12:42


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# Critical collapse of ultra-relativistic fluids: damping or growth of aspherical deformations

Research paper by Juliana Celestino, Thomas W. Baumgarte

Indexed on: 26 May '18Published on: 26 May '18Published in: arXiv - General Relativity and Quantum Cosmology

#### Abstract

We perform fully nonlinear numerical simulations to study aspherical deformations of the critical self-similar solution in the gravitational collapse of ultra-relativistic fluids. Adopting a perturbative calculation, Gundlach predicted that these perturbations behave like damped or growing oscillations, with the frequency and damping (or growth) rates depending on the equation of state. We consider a number of different equations of state and degrees of asphericity and find very good agreement with the findings of Gundlach for polar $\ell = 2$ modes. For sufficiently soft equations of state, the modes are damped, meaning that, in the limit of perfect fine-tuning, the spherically symmetric critical solution is recovered. We find that the degree of asphericity has at most a small effect on the frequency and damping parameter, or on the critical exponents in the power-law scalings. Our findings also confirm, for the first time, Gundlach's prediction that the $\ell = 2$ modes become unstable for sufficiently stiff equations of state. In this regime the spherically symmetric self-similar solution can no longer be recovered by fine-tuning to the black-hole threshold, and one can no longer expect power-law scaling to hold to arbitrarily small scales.

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# 2.1: Measurements in Analytical Chemistry

Analytical chemistry is a quantitative science. Whether determining the concentration of a species, evaluating an equilibrium constant, measuring a reaction rate, or drawing a correlation between a compound’s structure and its reactivity, analytical chemists engage in “measuring important chemical things” [Murray, R. W. Anal. Chem. 2007, 79, 1765]. In this section we review briefly the basic units of measurement and the proper use of significant figures.

## Units of Measurement

A measurement usually consists of a unit and a number that expresses the quantity of that unit. We can express the same physical measurement with different units, which creates confusion if we are not careful to specify the unit. For example, the mass of a sample that weighs 1.5 g is equivalent to 0.0033 lb or to 0.053 oz. To ensure consistency, and to avoid problems, scientists use the common set of fundamental base units listed in Table 2.1.1 . These units are called SI units after the Système International d’Unités.

It is important for scientists to agree upon a common set of units. In 1999, for example, NASA lost a Mar’s Orbiter spacecraft because one engineering team used English units in their calculations and another engineering team used metric units. As a result, the spacecraft came too close to the planet’s surface, causing its propulsion system to overheat and fail.

Some measurements, such as absorbance, do not have units. Because the meaning of a unitless number often is unclear, some authors include an artificial unit. It is not unusual to see the abbreviation AU—short for absorbance unit—following an absorbance value, which helps clarify that the measurement is an absorbance value.

Table 2.1.1 : Fundamental Base SI Units

Measurement

Unit

Symbol

Definition (1 unit is...)

mass

kilogram

kg

...the mass of the international prototype, a Pt-Ir object housed at the Bureau International de Poids and Measures at Sèvres, France. (Note: The mass of the international prototype changes at a rate of approximately 1 μg per year due to reversible surface contamination. The reference mass, therefore, is determined immediately after its cleaning using a specified procedure. Current plans call for retiring the international prototype and defining the kilogram in terms of Planck’s constant; see this link for more details.)

distance

meter

m

...the distance light travels in (299 792 458)–1 seconds.

temperature

Kelvin

K

...equal to (273.16)–1, where 273.16 K is the triple point of water (where its solid, liquid, and gaseous forms are in equilibrium).

time

second

s

...the time it takes for 9 192 631 770 periods of radiation corresponding to a specific transition of the 133Cs atom.

current

ampere

A

...the current producing a force of 2 $$\times$$ 10–7 N/m between two straight parallel conductors of infinite length separated by one meter (in a vacuum).

amount of substance

mole

mol

...the amount of a substance containing as many particles as there are atoms in exactly 0.012 kilogram of 12C.

light

candela

cd

...the luminous intensity of a source with a monochromatic frequency of 540 $$\times$$ 1012 hertz and a radiant power of (683)–1 watts per steradian.

There is some disagreement on the use of “amount of substance” to describe the measurement for which the mole is the base SI unit; see “What’s in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount,” the full reference for which is Giunta, C. J. J. Chem. Educ. 2016, 93, 583–586.

We define other measurements using these fundamental SI units. For example, we measure the quantity of heat produced during a chemical reaction in joules, (J), where 1 J is equivalent to 1 m kg/s . Table 2.1.2 provides a list of some important derived SI units, as well as a few common non-SI units.

Table 2.1.2 : Derived SI Units and Non-SI Units of Importance to Analytical Chemistry
Measurement Unit Symbol Equivalent SI Units
length angstrom (non-SI) Å 1 Å = 1 $$\times$$ 10–10 m
volume liter (non-SI) L 1 L = 10–3 m3
force newton (SI) N 1 N = 1 m$$\cdot$$kg/s2

pressure

pascal (SI)

atmosphere (non-SI)

Pa

atm

1 Pa = 1 N/m3 = 1 kg/(m$$\cdot$$s2)

1 atm = 101 325 Pa

energy, work, heat

joule (SI)

calorie (non-SI)

electron volt (non-SI)

J

cal

eV

1 J = 1 N$$\cdot$$m = 1 m2$$\cdot$$kg/s2

1 cal = 4.184 J

1 eV = 1.602 177 33 $$\times$$ 10–19 J

power watt (SI) W 1 W = 1 J/s = 1 m2$$\cdot$$kg/s3
charge coulomb (SI) C 1 C = 1 A$$\cdot$$s
potential volt (SI) V 1 V = 1 W/A = 1 m2$$\cdot$$kg/(s3$$\cdot$$A)
frequency hertz (SI) Hz 1 Hz = s–1
temperature Celcius (non-SI) oC oC = K – 273.15

Chemists frequently work with measurements that are very large or very small. A mole contains 602 213 670 000 000 000 000 000 particles and some analytical techniques can detect as little as 0.000 000 000 000 001 g of a compound. For simplicity, we express these measurements using scientific notation; thus, a mole contains 6.022 136 7 $$\times$$ 1023 particles, and the detected mass is 1 $$\times$$ 10–15 g. Sometimes we wish to express a measurement without the exponential term, replacing it with a prefix (Table 2.1.3 ). A mass of $$1 \times 10^{-15}$$ g, for example, is the same as 1 fg, or femtogram.

Writing a lengthy number with spaces instead of commas may strike you as unusual. For a number with more than four digits on either side of the decimal point, however, the recommendation from the International Union of Pure and Applied Chemistry is to use a thin space instead of a comma.

Table 2.1.3 : Common Prefixes for Exponential Notation
Prefix Symbol Factor Prefix Symbol Factor Prefix Symbol Factor
yotta Y 1024 kilo k 103 micro µ 10–6
zetta Z 1021 hecto h 102 nano n 10–9
eta E 1018 deka da 101 pico p 10–12
peta P 1015 100 femto f 10–15
tera T 1012 deci d 10–1 atto a 10–18
giga G 109 centi c 10–2 zepto z 10–21
mega M 106 milli m 10–3 yocto y 10–24

## Uncertainty in Measurements

A measurement provides information about both its magnitude and its uncertainty. Consider, for example, the three photos in Figure 2.1.1 , taken at intervals of approximately 1 sec after placing a sample on the balance. Assuming the balance is properly calibrated, we are certain that the sample’s mass is more than 0.5729 g and less than 0.5731 g. We are uncertain, however, about the sample’s mass in the last decimal place since the final two decimal places fluctuate between 29, 30, and 31. The best we can do is to report the sample’s mass as 0.5730 g ± 0.0001 g, indicating both its magnitude and its absolute uncertainty.

Figure 2.1.1 : When weighing an sample on a balance, the measurement fluctuates in the final decimal place. We record this sample’s mass as 0.5730 g ± 0.0001 g.

### Significant Figures

A measurement’s significant figures convey information about a measurement’s magnitude and uncertainty. The number of significant figures in a measurement is the number of digits known exactly plus one digit whose value is uncertain. The mass shown in Figure 2.1.1 , for example, has four significant figures, three which we know exactly and one, the last, which is uncertain.

Suppose we weigh a second sample, using the same balance, and obtain a mass of 0.0990 g. Does this measurement have 3, 4, or 5 significant figures? The zero in the last decimal place is the one uncertain digit and is significant. The other two zeros, however, simply indicates the decimal point’s location. Writing the measurement in scientific notation, $$9.90 \times 10^{-2}$$, clarifies that there are three significant figures in 0.0990.

In the measurement 0.0990 g, the zero in green is a significant digit and the zeros in red are not significant digits.

##### Example 2.1.1

How many significant figures are in each of the following measurements? Convert each measurement to its equivalent scientific notation or decimal form.

1. 0.0120 mol HCl
2. 605.3 mg CaCO3
3. $$1.043 \times 10^{-4}$$ mol Ag+
4. $$9.3 \times 10^4$$ mg NaOH

Solution

(a) Three significant figures; $$1.20 \times 10^{-2}$$ mol HCl.

(b) Four significant figures; $$6.053 \times 10^2$$ mg CaCO3.

(c) Four significant figures; 0.000 104 3 mol Ag+.

(d) Two significant figures; 93 000 mg NaOH.

There are two special cases when determining the number of significant figures in a measurement. For a measurement given as a logarithm, such as pH, the number of significant figures is equal to the number of digits to the right of the decimal point. Digits to the left of the decimal point are not significant figures since they indicate only the power of 10. A pH of 2.45, therefore, contains two significant figures.

The log of $$2.8 \times 10^2$$ is 2.45. The log of 2.8 is 0.45 and the log of 102 is 2. The 2 in 2.45, therefore, only indicates the power of 10 and is not a significant digit.

An exact number, such as a stoichiometric coefficient, has an infinite number of significant figures. A mole of CaCl2, for example, contains exactly two moles of chloride ions and one mole of calcium ions. Another example of an exact number is the relationship between some units. There are, for example, exactly 1000 mL in 1 L. Both the 1 and the 1000 have an infinite number of significant figures.

Using the correct number of significant figures is important because it tells other scientists about the uncertainty of your measurements. Suppose you weigh a sample on a balance that measures mass to the nearest ±0.1 mg. Reporting the sample’s mass as 1.762 g instead of 1.7623 g is incorrect because it does not convey properly the measurement’s uncertainty. Reporting the sample’s mass as 1.76231 g also is incorrect because it falsely suggests an uncertainty of ±0.01 mg.

### Significant Figures in Calculations

Significant figures are also important because they guide us when reporting the result of an analysis. When we calculate a result, the answer cannot be more certain than the least certain measurement in the analysis. Rounding an answer to the correct number of significant figures is important.

For addition and subtraction, we round the answer to the last decimal place in common for each measurement in the calculation. The exact sum of 135.621, 97.33, and 21.2163 is 254.1673. Since the last decimal place common to all three numbers is the hundredth’s place

\begin{align*} &135.6{\color{Red} 2}1\\ &\phantom{1}97.3{\color{Red} 3}\\ &\underline{\phantom{1}21.2{\color{Red} 1}63}\\ &254.1673 \end{align*}

we round the result to 254.17.

The last common decimal place shared by 135.621, 97.33, and 21.2163 is shown in red.

When working with scientific notation, first convert each measurement to a common exponent before determining the number of significant figures. For example, the sum of $$6.17 \times 10^7$$, $$4.3 \times 10^5$$, and $$3.23 \times 10^4$$ is $$6.22 \times 10^7$$.

\begin{align*} &6.1{\color{Red} 7} \phantom{323} \times 10^7\\ &0.0{\color{Red} 4}3 \phantom{23} \times 10^7\\ &\underline{0.0{\color{Red} 0}323 \times 10^7}\\ &6.21623 \times 10^7 \end{align*}

The last common decimal place shared by $$6.17 \times 10^7$$, $$4.3 \times 10^5$$ and $$3.23 \times 10^4$$ is shown in red.

For multiplication and division, we round the answer to the same number of significant figures as the measurement with the fewest number of significant figures. For example, when we divide the product of 22.91 and 0.152 by 16.302, we report the answer as 0.214 (three significant figures) because 0.152 has the fewest number of significant figures.

$\frac {22.91 \times 0.{\color{Red} 152}} {16.302} = 0.2136 = 0.214\nonumber$

There is no need to convert measurements in scientific notation to a common exponent when multiplying or dividing.

It is important to recognize that the rules presented here for working with significant figures are generalizations. What actually is conserved is uncertainty, not the number of significant figures. For example, the following calculation

101/99 = 1.02

is correct even though it violates the general rules outlined earlier. Since the relative uncertainty in each measurement is approximately 1% (101 ± 1 and 99 ± 1), the relative uncertainty in the final answer also is approximately 1%. Reporting the answer as 1.0 (two significant figures), as required by the general rules, implies a relative uncertainty of 10%, which is too large. The correct answer, with three significant figures, yields the expected relative uncertainty. Chapter 4 presents a more thorough treatment of uncertainty and its importance in reporting the result of an analysis.

Finally, to avoid “round-off” errors, it is a good idea to retain at least one extra significant figure throughout any calculation. Better yet, invest in a good scientific calculator that allows you to perform lengthy calculations without the need to record intermediate values. When your calculation is complete, round the answer to the correct number of significant figures using the following simple rules.

1. Retain the least significant figure if it and the digits that follow are less than halfway to the next higher digit. For example, rounding 12.442 to the nearest tenth gives 12.4 since 0.442 is less than half way between 0.400 and 0.500.
2. Increase the least significant figure by 1 if it and the digits that follow are more than halfway to the next higher digit. For example, rounding 12.476 to the nearest tenth gives 12.5 since 0.476 is more than halfway between 0.400 and 0.500.
3. If the least significant figure and the digits that follow are exactly halfway to the next higher digit, then round the least significant figure to the nearest even number. For example, rounding 12.450 to the nearest tenth gives 12.4, while rounding 12.550 to the nearest tenth gives 12.6. Rounding in this manner ensures that we round up as often as we round down.
##### Exercise 2.1.1

For a problem that involves both addition and/or subtraction, and multiplication and/or division, be sure to account for significant figures at each step of the calculation. With this in mind, report the result of this calculation to the correct number of significant figures.

$\frac {0.250 \times (9.93 \times 10^{-3}) - 0.100 \times (1.927 \times 10^{-2})} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber$

The correct answer to this exercise is $$1.9 \times 10^{-2}$$. To see why this is correct, let’s work through the problem in a series of steps. Here is the original problem

$\frac {0.250 \times (9.93 \times 10^{-3}) - 0.100 \times (1.927 \times 10^{-2})} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber$

Following the correct order of operations we first complete the two multiplications in the numerator. In each case the answer has three significant figures, although we retain an extra digit, highlight in red, to avoid round-off errors.

$\frac {2.48{\color{Red} 2} \times 10^{-3} - 1.92{\color{Red} 7} \times 10^{-3}} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber$

Completing the subtraction in the numerator leaves us with two significant figures since the last significant digit for each value is in the hundredths place.

$\frac {0.55{\color{Red} 5} \times 10^{-3}} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber$

The two values in the denominator have different exponents. Because we are adding together these values, we first rewrite them using a common exponent.

$\frac {0.55{\color{Red} 5} \times 10^{-3}} {0.993 \times 10^{-2} + 1.927 \times 10^{-2}} = \nonumber$

The sum in the denominator has four significant figures since each of the addends has three decimal places.

$\frac {0.55{\color{Red} 5} \times 10^{-3}} {2.92{\color{Red} 0} \times 10^{-2}} = \nonumber$

Finally, we complete the division, which leaves us with a result having two significant figures.

$\frac {0.55{\color{Red} 5} \times 10^{-3}} {2.92{\color{Red} 0} \times 10^{-2}} = 1.9 \times 10^{-2} \nonumber$

This page titled 2.1: Measurements in Analytical Chemistry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.


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# What is Aerodynamics?

Aerodynamics is the science of moving air and its impact on solid bodies placed in the flow field as an obstacle. Being a sub-field, most of the equations from fluid dynamics apply to aerodynamics as well, including all the governing equations, turbulence, boundary layer theory, and ideal gas assumption.

## History

Although wind had been widely used as a tool by humankind (windmills, sailboats) already, the beginning of the scientific study of aerodynamics only dates back to the seventeenth century. The dream of flight and a machine that is “lighter than air” was already present in ancient history. Very early attempts to describe flow quantities appear in the work of Aristotle and Archimedes, however the scientific field “aerodynamics” does not appear in their notes.

The very first aerodynamicist was Sir Isaac Newton, who developed and described the theory of flow resistance, also known as drag. Further famous scientists such as Bernoulli, Euler, Navier, and Stokes gave more precise and mathematically founded description about gas dynamics. The famous Navier-Stokes governing equations of fluid dynamics were born in 1800, however this is also the most difficult model to solve.

The desire for flight was always the main driving force behind the development of the engineering field of aerodynamics. In the early 1800’s, engineers and scientists studied the aerodynamic forces of flight. Terms like “drag”, “lift”, and “thrust” were born and the relationships between them were determined. In 1871, the first wind tunnel was built. Drag theories by engineers like Kirchhoff and Rayleigh were established. Otto Lilienthal was the first person to become successful with glider flights, and after all this research work, the Wright brothers flew for the first time with an engine powered airplane in 1903.

But the study of aerodynamics around airfoils did not stop here. Lanchester, Kutta, Zhukovsky, and Prandtl developed theories to better describe flow circulation around wings. While the First and the Second World Wars boosted the development of aircraft and the science of aerodynamics, the next breakthrough happened in 1947, when the sound barrier was broken for the first time with a Bell X-1 rocket-powered aircraft. At this point, the understanding of subsonic and supersonic flows had matured, however the evolution was still to continue. With the appearance of computers, the numerical calculation of the governing equations became possible. Computational Fluid Dynamics models and software started to appear on the horizon. Nowadays, CFD is utilized in almost every development process where airflow is involved and plays an important role.

## Flow classification and equations

An aerodynamic flow field can be classified based on the following physical parameters: velocity, pressure, and temperature. Additionally, density and viscosity can be used as well for classification. The most common flow classification practices are velocity and pressure-based.

### Velocity-based classification

Subsonic flows are flow fields in which the air speed does not exceed the local speed of sound, that is M<1

$M<1$

. Typical examples are common ground vehicles like cars, trains, F1 cars, and sport airplanes. M

$M$

represents the Mach number which is the flow velocity past a boundary. For more information about the mach number, please read our article about incompressible flow.

Transonic flows include both subsonic and supersonic flow regimes 0.8<M<1.2

$0.8

. Typically, commercial airplanes fly in this speed region. A supersonic flow zone appears when the otherwise subsonic fluid accelerates above sound speed due to surface curvatures (e.g. suction side of the wing).

Supersonic flows are defined as having velocities greater than the speed of sound everywhere in the domain (M>1

$M>1$

). Military jets and gun bullets are good examples for this velocity.

We could also include hypersonic flows, where the speed is much higher than the speed of sound. There is no accurate definition of this value (M>>1

$M>>1$

). Some hypersonic rocket-powered airplanes and spacecrafts can be mentioned as examples.

### Density / Pressure-based classification

Incompressible flow refers to a flow where the material (air) density remains constant. Under Mach 0.3 this is a good assumption for industrial flows and helps engineers make the calculation / simulation simpler. Please note that an incompressible flow does not mean that the fluid has to be incompressible itself.

A flow can be considered as compressible if the density changes along a streamline. Generally, this is the case for subsonic flows above Mach 0.3. Transonic, supersonic, and hypersonic flows are all compressible.

## Aerodynamic quantities

As mentioned before, aerodynamics is the study of forces on a solid body placed in an airflow. In order to better describe and understand these effects as well as to be able to compare cases easier, one can use mathematical quantities. These parameters include drag, lift, moment, center of pressure and pressure coefficient, to name a few important ones.

### Drag

One of the most important quantity in aerodynamics is the resistance of the fluid, also known as drag or drag force. This force is acting opposite to the motion of the solid body relative to the air thus causing losses. This drag force depends on the relative velocity as well as on the shape of the solid body. The drag force can be computed using the following equation:

Fd=12ρu2cdA(1)

$\begin{array}{}\text{(1)}& {F}_{d}=\frac{1}{2}\rho \phantom{\rule{thinmathspace}{0ex}}{u}^{2}\phantom{\rule{thinmathspace}{0ex}}{c}_{d}\phantom{\rule{thinmathspace}{0ex}}A\end{array}$

where

• Fd
${F}_{d}$

is the drag force,

• ρ
$\rho$

is the density of the fluid,

• v
$v$

is the speed of the object relative to the fluid,

• A
$A$

is the cross-sectional area, and

• cd
${c}_{d}$

is the drag coefficient which is dependent on the shape of the object and the Reynolds number.

### Lift

Another component of the resulting force on a solid body is called lift or lift force. This component is perpendicular to the oncoming flow direction. The term “lift” is coming from the aeronautics field, where it is an important quantity that makes flying actually possible for objects heavier than air. But lift force is generated in case of propellers, helicopter rotors and even on car bodies and wind turbines. Usually, an asymmetric body or a symmetric body with an angle of attack produces lift force. Lift can be determined by the following equation:

L=12ρv2SCL(2)

$\begin{array}{}\text{(2)}& L=\frac{1}{2}\rho {v}^{2}S{C}_{L}\end{array}$

where

• L
$L$

is lift force,

• ρ
$\rho$

is air density,

• v
$v$

is speed,

• S
$S$

is the area, and

• CL
${C}_{L}$

is the lift coefficient

### Moment

Aerodynamic moment or torque is produced by the aerodynamic force on a body (wing, car on any solid) making it rotate in the flow field. The rotation happens when this turning force is acting outside the center of pressure or the aerodynamic center.

If you imagine the air stream going from the front of the car over the engine hood, you can imagine that the air wants to lift the car at about the point where the headlights are. This is where the moment illustrated in figure 3 is generated.

### Center of Pressure

The Center of Pressure (CP) is a point where the force generated by the total sum of the surface pressure acts. This force can be calculated as the surface integral of the vectorial pressure field, and can be used to calculate the stability of an aerodynamic body (bullets, wings, cars). For example, when analyzing a weapon’s bullet aerodynamics, the distance between the CP and the center of gravity can create a rotating moment thus making the projectile less accurate. Easily speaking you can imagine the Center of Pressure as something like the center of gravity where the average weight of an object is. The center of gravity of a hammer for example is far away from the middle because the handle is usually a lot lighter than the head. The same goes for aerodynamics. The Center of Pressure is the point where the average aerodynamic forces drag and lift/downforce exert. The center of Pressure allows for example engineers to balance the lift of an aircraft.

### Pressure Coefficient

One of the most widely used non-dimensional numbers to visualize pressure on the surface of the aerodynamic body is the pressure coefficient Cp

${C}_{p}$

. This quantity describes the relative pressures in an incompressible flow, and can be calculated as

Cp=pp12ρV2=ppp0p(3)

$\begin{array}{}\text{(3)}& {C}_{p}=\frac{p-{p}_{\mathrm{\infty }}}{\frac{1}{2}{\rho }_{\mathrm{\infty }}{V}_{\mathrm{\infty }}^{2}}=\frac{p-{p}_{\mathrm{\infty }}}{{p}_{0}-{p}_{\mathrm{\infty }}}\end{array}$

where

• p
$p$

is the static pressure at the location where pressure coefficient is calculated,

• p
${p}_{\mathrm{\infty }}$

is the static pressure in the free stream,

• p0
${p}_{0}$

is the total pressure in the free stream,

• ρ
${\rho }_{\mathrm{\infty }}$

is the free stream fluid density,

• V
${V}_{\mathrm{\infty }}$

is the free stream velocity of the fluid, or the velocity of the body through the fluid.

It can be seen that using Cp

${C}_{p}$

, different scenarios of flow / body velocities, pressures and densities can be compared for the same geometry. Cp

${C}_{p}$

visualization is broadly used in the automotive industry where different study concepts are compared to each other in a virtual wind tunnel.

## Boundary Layer

In aerodynamics, the description, measurement, and simulation of near wall velocity and other quantities are always a challenge. As the flow decelerates rapidly near a solid wall, the viscosity effects become significant in a thin region also known as boundary layer. Boundary layers are classified into two main groups: laminar boundary layers at low Re numbers and turbulent boundary layers at higher flow speeds.

Reynolds Number

## Turbulence

Turbulence plays a very significant role in aerodynamics. Turbulence is by nature a chaotic, irregular phenomena observed in lower viscosity fluids. In general, unsteady vortices appear in many sizes in the flow field and interact with each other and the solid body which usually generates them. Turbulence can be “good” when, for example, one wants to generate thinner boundary layers and “bad”, when it leads to large separations, therefore increased drag and losses. Describing and simulating turbulence is still an ongoing project, however many modeling approaches exist. In CFD, to achieve accurate aerodynamic results, turbulence needs to be taken into account properly. This can be done by either modeling it or resolving the eddies (LES, DES). Eddies are swirls in a fluid flow regime that create reverse current.

## Convergence

As mentioned before, the aerodynamic quantities such as drag and lift are the key outcomes of a simulation. Therefore, their convergence to a steady or quasi-steady state is to be monitored and evaluated carefully before the residual targets are reached. When the simulation is stopped too early, the lift and drag values may not be fully converged, resulting in an inaccurate result. Faster convergence and shorter simulation time can be reached by using higher relaxation factors as well as by proper initialization.

Further information about convergence in numerical simulations can be found in this article.

## Applications

Aerodynamics is present and plays an important role in almost all the transportation devices and big structures we have around.

### Aerospace

As described in the History section, the first studies about aerodynamics appeared with the desire of flight. Later on, the aerospace industry and its continuous research and development on aircraft required many mathematical models, measurement devices, wind tunnels and so on, all designed to understand aerodynamics better. Nowadays it is a key field of studies to reduce emissions, reduce environmental noise and improve human comfort. The terms lift, drag, and moment commonly appear in the design process of an aircraft.

### Automotive

The automotive industry is very competitive these days. With the desire to build environmentally friendly or even zero emission vehicles, the reduction of drag is a key aspect of the development process. Brake cooling and HVAC airflow can also be mentioned here as some of the main topics of aerodynamics. The importance of lift force appears mostly when studying stability and when designing race vehicles.

### Sports

With the continuous improvement of performance in sports, the aerodynamics of race bikes, swimmers, motorbikes, or Formula 1 cars is becoming more and more important. In a competition, where even milliseconds count, the reduction of the drag by 0.1% can distinguish the winner from the others. Modern CFD tools allow the engineers to simulate tiny to extremely large scale bodies in many different scenarios. As shown in figure 6, the aerodynamic profile of a biker can have significant influence on the power that needs to be provided by the biker to achieve higher speeds. Especially the shape of the rims and the seatpost have a huge influence on the aerodynamic performance since they can create a lot of turbulence. Turbulence should be avoided for the best possible performance.

### Internal flows: HVAC and ductwork

Aerodynamics in HVAC (Heating, Ventilation and Air Conditioning) and pipe flow applications are also important. Here usually pressure losses are calculated as a result of viscous aerodynamic forces, separation at corners and bendings and sudden expansions.

### Wind load on buildings and urban aerodynamics

In case of tall buildings and skyscrapers, due to their large surface area, forces due to wind loads can be quite significant. This can be handled as “drag” force of the solid obstacle placed in the flow. Besides the wind load force on the structure, pedestrian comfort can be also assessed at lower levels. Separations on the structure as well as high speed regions can affect humans, making it sometimes impossible to even walk. Building and urban aerodynamicists simulate those scenarios with CFD tools these days.

### Wind turbine

A very important technology towards sustainable energy resources is the wind turbine. As it usually consists of 3 aerofoils placed in an airflow, all the aspects of aerodynamics are present. Drag is used as a measure of the efficiency of the turbine, and lift as a measure of the power extracted from the wind. Moments are used to calculate the loads on the blades. SimScale provides a very interesting Project Spotlight about the optimization of wind turbine plates. Go and check it out if you are interested in finding out more about this interesting topic.

### Hydrodynamics

As a very similar field to aerodynamics, hydrodynamics needs to be mentioned here as well. Hydrodynamics deals with liquid flows, mostly water. While usually incompressible and low speed, water flows experience the same effects as air flow scenarios: boundary layers, turbulence, and forces acting on a solid body are also present.

## Aerodynamics & SimScale

Most of the transient or steady, compressible or incompressible aerodynamic flows can be accurately simulated with CFD software. SimScale offers a browser-based platform for engineers, where a CAD design can be meshed and simulated, and the results post-processed easily. Additionally, you can have a look at the public projects and choose a case which fits your problem best, so you can get started even faster. In the simulation below you can see the airpressure behind a biker. As you can see, it would be very beneficial for a different biker to bike just behind him because the pressure is lower and he would nede less power to keep the same speed.

## Resources

1

${}^{1}$

: Von Karman, Theodore (2004). Aerodynamics: Selected Topics in the Light of Their Historical Development. Dover Publications. ISBN 0-486-43485-0. OCLC 53900531

2

${}^{2}$

: Anderson, John D. (1997). A History of Aerodynamics: And Its Impact on Flying Machines. Cambridge University Press. ISBN 0-521-45435-2. OCLC 228667184.

3

${}^{3}$

: Anderson, John D. (2007). Fundamentals of Aerodynamics (4th ed.). McGraw-Hill. ISBN 0-07-125408-0. OCLC 60589123.

4

${}^{4}$


text:
### Assignments:

Unfinished Assignment Study Questions for Lesson 48

### Lesson Objectives:

- The requirements for being president
- Birth controversies
- The age and background of the president
- Electoral votes versus popular vote
- The 12th Amendment

The requirements of becoming president are not that stringent. Article II, Section 1, of the Constitution says:

"No person except a natural born Citizen, or a Citizen of the United States, at the time of the Adoption of this Constitution, shall be eligible to the Office of President; neither shall any Person be eligible to that Office who shall not have attained to the Age of thirty-five Years, and been fourteen Years a Resident within the United States."

Once elected, the president's salary is $400,000 with$169,000 for annual expenses, travel, and entertainment. Plus, he lives rent free in the White House.

The language in the Constitution says the president must be a natural born citizen. What exactly does that mean? Does the child have to be born in the United States or does it extend to children who are born to U.S. Citizens while they are traveling abroad?

One day, this requirement will be challenged in the Supreme Court and there will be a more concrete interpretation.

There have been some issues through the years starting with Mitt Romney's father who ran for president.

George Romney ran in 1968. He was born in Chihuahua, Mexico.

Senator Ted Cruz ran in 2016. He was born in Canada.

Barack Obama was elected in 2008 and reelected in 2012. He was born in Hawaii, but there was some controversy over whether or not that was true.

Even though the minimum age is 35, no president has been anywhere near that. John F. Kennedy was our youngest president at the age of 43. The average president is inaugurated at the age of 54.

Most presidents have been white, male, and protestant. There have only been two exceptions. John F. Kennedy was Catholic and Barack Obama is African American.

The candidates are nominated by each party every four years at the national conventions. When the voters vote in the general election, those votes are counted to determine how the state voted on the candidates, but the electors are the ones who actually cast the official vote in the electoral college.

Because of the electoral college system we have in place, there have been times when a president has won the popular vote but lost the election because of the electoral college. Also, there have been times when a president did not receive the majority vote in the electoral college but was elected president because there were more than two candidates.

In the 1800 election, Thomas Jefferson versus John Adams, Jefferson and his running mate, Aaron Burr, received the same number of electoral votes. The way the system was outlined in the Constitution, the candidate with the most electoral votes would be president, and the one with the second most votes would be vice-president. Since there was a tie, it was up to the House of Representatives to choose, and after much chaos, they chose Thomas Jefferson for president and Aaron Burr for vice-president.

As a result of that confusion, the 12th Amendment was ratified.

The Twelfth Amendment to the Constitution was adopted in 1804, and specified the separate election of the president and the vice president by the electoral college.


text:
author Manuel Eberl Wed, 13 May 2020 13:00:03 +0200 changeset 72063 5656ec95493c parent 72062 dca11678c495 child 72064 0bbe0866b7e6
generalised pigeonhole principle in HOL-Library.FuncSet
--- a/src/HOL/Library/FuncSet.thy	Wed May 13 12:55:33 2020 +0200
+++ b/src/HOL/Library/FuncSet.thy	Wed May 13 13:00:03 2020 +0200
@@ -686,4 +686,35 @@
by (simp add: PiE_insert_eq inj_combinator card_image card_cartesian_product)
qed

+subsection \<open>The pigeonhole principle\<close>
+
+text \<open>
+  An alternative formulation of this is that for a function mapping a finite set \<open>A\<close> of
+  cardinality \<open>m\<close> to a finite set \<open>B\<close> of cardinality \<open>n\<close>, there exists an element \<open>y \<in> B\<close> that
+  is hit at least $\lceil \frac{m}{n}\rceil$ times. However, since we do not have real numbers
+  or rounding yet, we state it in the following equivalent form:
+\<close>
+lemma pigeonhole_card:
+  assumes "f \<in> A \<rightarrow> B" "finite A" "finite B" "B \<noteq> {}"
+  shows   "\<exists>y\<in>B. card (f - {y} \<inter> A) * card B \<ge> card A"
+proof -
+  from assms have "card B > 0"
+    by auto
+  define M where "M = Max ((\<lambda>y. card (f - {y} \<inter> A))  B)"
+  have "A = (\<Union>y\<in>B. f - {y} \<inter> A)"
+    using assms by auto
+  also have "card \<dots> = (\<Sum>i\<in>B. card (f - {i} \<inter> A))"
+    using assms by (subst card_UN_disjoint) auto
+  also have "\<dots> \<le> (\<Sum>i\<in>B. M)"
+    unfolding M_def using assms by (intro sum_mono Max.coboundedI) auto
+  also have "\<dots> = card B * M"
+    by simp
+  finally have "M * card B \<ge> card A"
+  moreover have "M \<in> (\<lambda>y. card (f - {y} \<inter> A))  B"
end`


text:
# atomic number of helium

The number of protons in an atom is called its atomic number ($$Z$$). Calculate the number of moles of helium present in 52g of Helium. It's supposed to be 2, not 3.The third element is Lithium. Answered Oct 01, 2017. The precise estimation of atomic polarizabilities impinges upon a number of areas and processes in physical sciences. … 0. The element is also used as a purge gas, to grow silicon wafers and other crystals, and as a protective gas for welding. Approved by eNotes Editorial Team We’ll help your grades soar. The orbital in which two electrons are present is called s-orbital. By signing up, you'll get thousands of step-by-step solutions to your homework questions. C. 1. For example, carbon’s atomic number (Z) is six because it has 6 protons. look up a periodic table to see the square for helium. Here is the answer for: Helium’s atomic number crossword clue. Atomic Number. The number of electrons can also be different in atoms of the same element, thus producing ions (charged atoms). The atomic number, equal to the number of protons in the nucleus, locates an element's position in the table. The atomic number of an element is usually written in front of and slightly below the element's symbol, like in the Figure below for helium. The maximum accommodating capacity of s-orbital is two electrons. A nucleus with two protons will always be helium. However, some helium atoms have more or less than two neutrons. 1. Since there are two protons in the nucleus of helium, the atomic number of helium is two. Atomic number of hydrogen is 1 and that of helium is 2. but an atom of helium is 5 times denser than hydrogen. Atomic number 2 is used for party balloons, but it's primary use is in the cryogenic industry for cooling superconducting magnets. It is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas, the first in the noble gas group in the periodic table. How do atomic mass and atomic number compare for helium? Correct answer to the question What is the atomic number for helium DOES ANYBODY WANT TO RP - e-eduanswers.com Helios, lit. The atomic number does not change when an atom becomes an ion. It is a colorless, flavorless gas, best known for its use in filling floating balloons. the atomic mass and atomic number is qual. A helium ion would have an atomic number of 2. Helium is one of the noble gases of group O in the periodic table. Helium (He) is a colorless, odorless gas that has the atomic number 2 and the symbol He. Thus, Z, the atomic number = 2. There are no overlaps in this regard. The number of protons is identical to the atomic number (as well as electrons, but these don't matter right now). Therefore the mass of most helium atoms is 4 atomic mass units ($$2 \: \text{amu}$$ for the protons + $$2 \: \text{amu}$$ for the neutrons). Janssen obtained the first evidence of helium during the solar eclipse of 1868 when he detected a new line in the solar spectrum. The number of electrons in each of Helium's shells is 2 and its electronic configuration is 1s 2. Solved: What is the atomic number of helium-4? In its elemental form helium's CAS number is 7440-59-7. Sources, facts, uses, scarcity (SRI), podcasts, alchemical symbols, videos and images. 'Sun') is a chemical element with symbol He and atomic number 2. All helium atoms have two protons, and no other elements have atoms with two protons. Answered Nov 17, 2018. Atomic Number of Calcium is 20. The other number is the atomic weight. manirohith2008 manirohith2008 3 days ago Chemistry Secondary School +5 pts. If you are looking for other crossword clue solutions simply use the search functionality in the sidebar. The helium atom has a covalent radius of 28.pm and it's Van der Waals radius is 140.pm. » Boiling Point » Melting Point » Abundant » State at STP » Discovery Year About Calcium . What is the atomic number of Helium?  Helium, with an atomic number of 2 and He as its element symbol, is a unique element. Join now. It’s the second lightest element. Melting point of Calcium is 839 °C and its the boiling point is 1487 °C. So, let's take helium. Number of protons in Calcium is 20. The abundance of 3 He in air of 0.000 137 % as determined by Nier had a negligible effect on the atomic weight of helium. Element Helium (He), Group 18, Atomic Number 2, s-block, Mass 4.003. It is a noble gas, non metal that is second most common element in the universe after Hydrogen. Join now. Helium is a Block S, Group 18, Period 1 element. In the case of helium, the atomic number is 2. 0. as the elements in period 2 of the periodic table are considered in succession from left to right, there is a decrease in atomic radius with increasing atomic number. Answer Anonymously; Answer Later; Copy Link; 2 Answers. Below you may find the answer for: Helium's atomic number crossword clue.This clue was last seen on Wall Street Journal Crossword December 16 2020 Answers In case the clue doesn’t fit or there’s something wrong please let us know and we will get back to you. the atomic number is greater. Gaseous chemical element, symbol: He, atomic number: 2 and atomic weight 4,0026 g/mol. Helium. The atomic number is specifically tied to a single element. Answer. LEARNING APP; ANSWR; CODR; XPLOR; SCHOOL OS; STAR; answr. Join Now. Asked by Megabunny, Last updated: Dec 19, 2020 + Answer. Likewise, helium atoms contain two protons despite the conditions or state they are found in. Helium is a chemical element with atomic number 2 which means there are 2 protons and 2 electrons in the atomic structure. This is because the atomic number is the number of protons in the nucleus, and an ion is formed when an atom gains or loses electrons, which are located in the electron cloud surrounding the nucleus. Helium (from Greek: ἥλιος, translit. Rachit gupta. It is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas, the first in the noble gas group in the periodic table. helium does not have an atomic number. Ask your question. What is the atomic number of Helium? Helium is most commonly used in diving, welding and for inflating balloons. Atomic Number: 2: Atomic Radius: 140 pm (Van der Waals) Atomic Symbol: He: Melting Point:-272.2 °C: Atomic Weight: 4.003: Boiling Point:-268.93 °C: Electron Configuration: 1s 2 : Oxidation States: 0: History. 0 5 0. 0. Its boiling point is the lowest among all the elements. read more Wannia Jamil. Helium can be found on the periodic table at atomic number two. It has 2 electrons (since the number of electrons is the same as the number of protons) and 2 neutrons in regular atoms (since it's the atomic mass, 4, subtracted by the number of protons, 2). the atomic mass is greater. The nucleus of the helium atom contains 2 massive, positively charged particles. Asked on December 26, 2019 by Pari Rithvick. What is the atomic number for Helium? This number is very important because it is unique for atoms of a given element. The chemical symbol for Helium is He . Answer:follow me Explanation:atomic number of hellium= 2 mark me Brainilist please 1. Click hereto get an answer to your question  Calculate the number of moles of helium present in 52g of Helium. Asked by Rachit gupta, Last updated: May 24, 2020 + Answer. The atomic … Log in. A. Properties. Wannia Jamil. So a helium ion would have an atomic number of 2, which is the atomic number for helium. 0. 5. Login. Log in. Atoms with the same number of protons but different numbers of neutrons are called isotopes. The main helium source in the world is a series of fields of natural gas in the United States. rachit gupta . Atomic weight of Calcium is 40.078 u or g/mol. It's average atomic mass is 4.003, and has 2 protons  (because that is the same as the atomic number). megabunny. Share. E. 2. Molar mass of Helium … Helium is atomic number 2 on the periodic table, with the element symbol He. A. Follow. Although all exotic helium isotopes decay with a half-life of less than one second , researchers have eagerly created exotic light isotopes through particle accelerator collisions to create unusual atomic nuclei for elements such as helium, lithium , and nitrogen . chemistry. Helium is the element that is atomic number 2 on the periodic table. The number of protons is called the atomic number of any element. Atomic number of helium is 2 Share Reply. Share. 0. This question is part of Atomic Number of the First 20 Elements On The Periodic Table. D. 1. For example, all helium atoms have two protons, and no other elements have atoms with two protons. A subset of exotic light nuclei, the exotic helium isotopes have larger atomic masses than helium's natural isotopes. Z is simply the number of protons, massive, positively charged particles, present within the nucleus. THE COMMUNIST MANIFESTO; In The Communist Manifesto, co-authors Karl Marx and Friedrich Engels proposed that the… As we are limited in space, below are a few ideas to help get you started.In Chapter 1, titled “Bourgeois and Proletarians” of Karl Marx and Friedrich Engels’s The Communist Manifesto, both authors… 1 educator I believe the answer is C. (: Comment ; Complaint; Link; Margaux 24 June, 04:36. The units of the atomic weight are grams/mole. This isotope is present in natural sources of helium with a smaller abundance than that of any other stable isotope relative to its elemental composition. This number clearly shows you how many protons an element’s nucleus contains. 0. This question is part of atomic number of the first 20 elements on the periodic table. Request. Request. Follow. This number is … 1 3. One mole of any atom or molecule is equal to Avogadro's Number. You mean Helium with isotope which its atomic mass is 4 ? Helium Atomic Number: 2 Helium Symbol: He Helium Atomic Weight: 4.002602(2) Helium Discovery: Janssen, 1868, some sources … Answers (2) Lunamia 24 June, 03:14. B. 0. Therefore, the atomic number of Helium is 2. Answer Anonymously; Answer Later; Copy Link ; 5 Answers. Ask your question. Z thus defines the identity of the element: Z=1, hydrogen; Z=2, helium; Z=3, lithium;.....Z=80, mercury. It should have elements with atomic numbers from 1 (Hydrogen) to the vicinity of 118 (Oganesson). Here is a collection of facts about this lightweight, interesting element: Helium Element Facts . August 13, 2018 Educator. Commercial helium is extracted from natural gas. Chemical symbol for Calcium is Ca. The principal commercial use of helium is for MRI scanners. The number of neutrons can vary to produce isotopes, which are atoms of the same element that have different numbers of neutrons. Haha but it doesn’t matter xD Because when identifying atomic number of an atom, you ONLY need to know what element it is Atomic number of an atom is its number of protons. what occurs as the atomic number of the elements in period 2 increases? All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms. This crossword clue was last seen on Wall Street Crossword December 16 2020 Answers. From the Greek word helios, the sun. Its use in filling floating balloons Link ; 2 Answers same element, thus producing (... Atomic numbers from 1 ( Hydrogen ) to the number of any element chemical element, thus producing ions charged. Many protons an element have the same as the atomic number of moles of is! The orbital in which two electrons are present is called the atomic number = 2 Hydrogen is and. ( 2 ) Lunamia 24 June, 04:36 1 and that of helium during the solar eclipse of when... Interesting element: helium element facts element symbol He nuclei, the atomic number 2 used! Homework questions May 24, 2020 + answer second most common element in solar! You mean helium with isotope which its atomic mass is 4 average atomic mass is 4 noble gas best! To your homework questions other crossword clue was Last seen on Wall Street crossword December 16 2020.... A colorless, odorless gas that has the atomic number is … What is element. Year about Calcium 20 elements on the periodic table at atomic number of protons in atom. Gas in the United States found in called its atomic mass is,... 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text:
# Intercept with axis

F(x)=log(x+4)-2, what is the x intercept

Result

x =  96

#### Solution:

$F(x)=\log(x+4)-2 \ \\ F(x)=0 \ \\ \log(x+4)-2=0 \ \\ \log(x+4)=2 \ \\ x+4=10^2 \ \\ x=10^2 - 4=96$

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Dr Math
Assuming "log" is the decadic logarithm (log with base 10)

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text:
# Is this twin particle cannon feasible?

For use in a vacuum.

The gun strips hydrogen atoms and accelerates the electrons one way on a ring magnet and the protons in the other direction (not in the same path but using as much of the same hardware as possible). It then discharges both beams in succession (electrons, then protons).

This idea comes from reading George O. Smith's The Complete Venus Equilateral (one of my favorite SF books).

In the story he pointed out several problems with using an electron beam as a weapon.

1. As you fire the beam you acquire a high positive charge (the electrons have to come from somewhere).
2. The negative charge that the target acquires will start to deflect the beams of subsequent shots.
3. Beam dispersal.

The twin beam seems to have a few advantages:

1. The charge of both the firing ship and the target remains neutral.
2. Any negative charge picked up by the faster moving electron beam will, in a small way, add to the accuracy of the heavier and slower proton beam. The overall neutral effect on the target will not significantly affect succeeding shots.
3. Beam Dispersal. Well, no real help here that I can see except that the spread out electron beam can paint the target and help the proton beam focus a tiny bit more of the near misses onto the target.
• Plus the beam dispersal. Plus I have a feeling that by the time it got to the target it'd mostly be a bunch of hydrogen again, but I dunno. en.wikipedia.org/wiki/Particle-beam_weapon#Beam_generation talks about just straight up shooting beams of hydrogen atoms, by accelerating protons then having them capture electrons on the way out. – Jason C May 10 '17 at 0:54

Also, beam dispersion will be much worse in a charged particle weapon than it would in a laser, leading to reduced range. Finally, if you can accelerate a few kg of protons and electrons to an appreciable fraction of $c$, then why not accelerate a few kg of steel to that speed and let it one-shot whatever vessel you are hurling it at?


text:
# Surface integral over a sphere of inverse of distance

Let $S$ be a sphere in $\mathbb{R}^3$ of radius $r$ centered at the origin and $x_0\not\in S$. Let $f:\mathbb{R}^3\to\mathbb{R}$ be given by $f(x)=\Vert x-x_0\Vert$. I'm asked to compute the (surface) integral $$\int_S fdS$$ I think I have to separate this in the cases $\Vert x_0\Vert>r$ and $\Vert x_0\Vert<r$. For the former, we could use the divergence theorem and for the latter, try to show some kind of invariance and reduce the problem to the case where we have to integrate over a small sphere around $x_0$. However, I haven't been able to develop this ideas as I can't find suitable vector fields $F$ to work with them.

Any kind of help or suggestions are greatly appreciated.

• I am curious: is there more context to the question? I find it interesting, but the solution is somewhat unremarkable in the sense that there is nothing fascinating about it. It is merely technical. (This being said, nice work @Doug M!) – Kuifje Dec 1 '16 at 16:39

$\iint f(\mathbf x) |dS|$

Note that $\mathbf x \in \mathbb R^3$ or $\mathbf x = (x,y,z)$

$f(x) = \sqrt {(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2}$

Lets do this in spherical

$x = \cos\theta\sin\phi\\ y = \sin\theta\sin\phi\\ z = \cos\phi$

$|dS|= \|(\frac {\partial x}{\partial\phi},\frac {\partial y}{\partial\phi},\frac {\partial z}{\partial\phi})\times(\frac {\partial x}{\partial\theta},\frac {\partial y}{\partial\theta},\frac {\partial z}{\partial\theta})\| = \sin\phi$

$f(x) = \sqrt {x^2 + y^2 + z^2 - 2xx_0 - 2yy_0 -2zz_0 + x_0^2 +y_0^2 + z_0^2}$ $f(x) = \sqrt {1 - 2\cos\theta sin\phi x_0 - 2\sin\theta\sin\phi y_0 -2\cos\phi z_0 + x_0^2 +y_0^2 + z_0^2}$

$\iint \sqrt {1 - 2\cos\theta \sin\phi x_0 - 2\sin\theta\sin\phi y_0 -2\cos\phi z_0 + x_0^2 +y_0^2 + z_0^2} \sin\phi \;d\phi\;d\theta$

The geometry of this is symmetric. We don't need to use the point $(x_0,y_0,z_0)$ the result will be the same regardless of the orientation of the point. All we need is the distance from the origin. $x_0, y_0 = 0, z_0=d$

$\iint \sqrt {1 - 2\cos\phi d + d^2} \sin\phi \;d\phi\;d\theta$

$u^2 = 1 - 2d\cos\phi + d^2\\ 2u\; du = 2 d\sin\phi\;d\phi$

$\int_0^{2\pi}\int_{d-1}^{d+1} \frac {u^2}{d} \;du\;d\theta$

Assuming $d>1$. I suppose I could say: $\int_0^{2\pi}\int_{|d-1|}^{|d+1|} \frac {u^2}{d} \;du\;d\theta$

$\frac 23 \pi (6d + \frac {2}{d})$ if $d>1$

Wich is the surface area times the distance to the center plus a little bit, and that little bit gets increasingly trivial as $d$ gets large.

$\frac 23 \pi (6 + 2d^2)$ if $d<1$

Again equals the surface area plus a little bit and tends to the surface area exactly if $d=0$

• So where does this get us? – Kuifje Nov 30 '16 at 22:12
• Needed to let the problem marinate. – Doug M Nov 30 '16 at 22:55
• Nice. I guess the main trick is to notice that you can set $x_0=y_0=0$ with no loss of generality. – Kuifje Dec 1 '16 at 16:41

I suspect there is a typo in the question, and that the function to integrate is $$f(x)=\frac{1}{||x-x_0||}$$ Indeed:

1. This is what the title of the question says
2. The exact same question was posted here.

In this case, using all of @Doug M's work: $$\boxed{ \iint_Sf(x)\; dS = \frac{1}{d}\int_0^{2\pi}\int_{|d-1|}^{|d+1|}dud\theta =\cases{\frac{4\pi}{d} \quad \mbox{if } d\ge1\\ 4\pi \quad \mbox{if } d<1 } }$$

• To be precise, @Doug M did all the work assuming the sphere had radius $r=1$. It is straightforward to adapt the calculations with any $r>0$. – Kuifje Dec 2 '16 at 4:40