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# Science:Math Exam Resources/Courses/MATH102/December 2011/Question 09

MATH102 December 2011
Other MATH102 Exams

### Question 09

Consider the isosceles triangle given in the following figure, where a and b indicate the side lengths. Assume that the triangle has a circumference of 2. Find the lengths a and b for which the area A of the triangle is maximized. You must also check that you found a maximum and your solution must include that check. Hint: Heron's formula may be useful which states that

${\displaystyle A^{2}=s(s-x)(s-y)(s-z)}$

where A is the area of the triangle, x, y, z are its side lengths and s = (x+y+z)/2.

 Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

 Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.


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# Money Metric Utility Function: Budget constraints become Utility Functions?

Im currently reading up on the "money metric utility function" (also known as the minimum income function or direct compensation function).

By definition it is defined as:

$$m(\text{p},\text{x})\equiv e(\text{p},u(\text{x}))$$

Hal Varian writes in Microeconomic analysis (page 109):

it is easy to see that for a fixed $$\text{x}$$, $$u(\text{x})$$ is fixed, so $$m(\text{p},\text{x})$$ behaves exactly like an expenditure function: its monotonic, homogenous, concave in $$\text{p}$$, and so on. What is not so obvious is that when when $$\text{p}$$ is fixed, $$m(\text{p},\text{x})$$ is in fact a utility function.

The proof is simple: for fixed prices the expenditure function is increasing in the level of utility: if you want a higher utility level, you have to spend more money. In fact, the expenditure function is strictly increasing in $$u$$ for continuous, local non-satiated preferences.

Hence for fixed $$\text{p}$$, $$m(\text{p},\text{x})$$ is a monotonic transform of the utility function therefore itself a utility.

Does this (the bolded statement) mean that we essentially convert our consumer's budget constraint/budget line into his indifference curves when prices are fixed?

$m(\mathbf p,\mathbf x)$ specifies the minimum amount of money required for a consumer to attain the same utility as consuming bundle $\mathbf x$, taking prices $\mathbf p$ as given. In other words, since all bundles on the same indifference curve as $\mathbf x$ have the same utility value, and to achieve this utility, one needs at least an income of $m(\mathbf p,\mathbf x)$, we can thus establish a one-to-one mapping between utility levels and income levels. As utility is ordinal, we could as well use income as our utility measure.
This does not, however, suggest that the consumer's budget line is converted to her indifference curve; the two remain distinct objects. To see this, consider a two-goods case. With fixed prices, the slope of the budget line is $p_2/p_1$, which is constant for all $\mathbf x$'s. In contrast, the slope of the indifference curves is generally a non-constant function over $\mathbf x$: $$\frac{\partial x_2}{\partial x_1}=\frac{\partial e(\mathbf p,u(\mathbf x))/\partial x_1}{\partial e(\mathbf p,u(\mathbf x))/\partial x_2}=\frac{MU_1(\mathbf x)}{MU_2(\mathbf x)}\,.$$


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# Homework Help: Charge distribution in an electric field

1. Aug 3, 2015

### teme92

1. The problem statement, all variables and given/known data
An infinitely long cylinder of radius R is placed above an infinite grounded plane. The centre of the cylinder is a distance (d) above the plane, and the cylinder has a surface charge density of σ.

(a) Initially ignoring the grounded plane, what is the electric field due to the charged cylinder, assuming a cylindrical coordinate system that is centred on the cylinder?

(b) Use your result to provide and equation for the electric field in the xyz coordinate system, where the grounded plane is at z=0, and the centre of the cylinder is along the y-direction at z=d?

(c) What is the Electric field due to the charged sphere and the grounded plane for z>0?

(d) What is the distribution of the charge that is generated on the infinite grounded plane?

(e) Calculate the charge per unit length that is generated on the grounded plane in the y-direction, and use this to show that charge is conserved.

2. Relevant equations
Gauss's Law: $\int E{\cdot}da = \frac{Q_{enc}}{\epsilon_0}$

3. The attempt at a solution

(a) I used Gauss' Law where $da = 2{\pi}RL$ and $Q_{enc} = \sigma$, simplifying down to get:

$E = \frac{{\sigma}R}{2\epsilon_0}$

(b) I simply said $R = \sqrt{x^2+d^2}$ and subbed into (a) and got:

$E = \frac{{\sigma}(\sqrt{x^2+d^2})}{2\epsilon_0}$

(c) I got the $R_+$ and $R_-$ and then their respective $E_+$ and $E_-$. Then added them together for total $E$. This is a long equation that I won't type out unless its relevant for later parts.

(d),(e) These parts are confusing me and I don't understand how to answer them. Any help in the right direction would be really appreciated. Thanks in advance.

2. Aug 3, 2015

### tommyxu3

There may be a trivial error here that $da$ must have to match a also little amount like $2\pi RdL.$

A hint based on my opinion: The method of image. Have you learned it?

3. Aug 4, 2015

### teme92

It says for part (a) to ignore the grounded plane so would that not therefore mean $d$ is irrelevant?

4. Aug 4, 2015

### tommyxu3

I'm sorry that I didn't explain what I typed clearly. I used $da$ and $dL$ to present little amount. These just can make the both sides of the equation equal.

5. Aug 5, 2015

### vela

Staff Emeritus
The equation $da = 2{\pi}RL$ isn't correct. The lefthand side is infinitesimal but the righthand side is finite. Writing $Q = \sigma$ doesn't make sense either. The units don't match.

Don't mix up the radius of the cylinder $R$ with the radial coordinate $r$. Also, to be complete, you should consider the regions inside the cylinder and outside the cylinder.

Why would the radius of the cylinder $R$ depend on $x$ and the distance $d$ from the plane?

6. Aug 5, 2015

### teme92

Hey thanks for the replies guys, I got help of a class mate today so everything is good.


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## Algebra 1

The initial amount can be found by plugging in 0 for x: g(x) = (14)(2$^{x}$) g(0) = (14)(2$^{0}$) g(0) = (14)(1) g(0) = 14 The growth factor is 2 because this is the number attached to x. In other words, this is the growth factor because this is the number that allows x to grow as x changes.


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# Chemistry - Freundlich Equation's dimensions

Technically speaking, one typically wouldn't write the left hand side unitless, as it is the ratio of the mass of two different components (usually mg of some chemical per gram of the surface).

But, otherwise your intuition is correct and $$k$$ just needs units that will cancel with $$p^{\frac{1}{n}}$$ to give the units on the left hand side. So $$k$$ could have units of $$\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{bar^{\frac{-1}{n}}}\right)$$.

We could also reformulate Freundlich's equation in terms of concentration (since we are dealing with an isotherm) as: $$q=KC^{\frac{1}{n}}$$ where $$q=m/x$$, $$C$$ is concentration of the chemical and $$K$$ is a new constant with units of $$\left(\dfrac{\text{mg chemical}}{\text{g surface}}\right)\left(\mathrm{{\dfrac{L}{\text{mg chemical}}}}\right)^{\frac{1}{n}}$$.

Source


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# What would happen if the Earth struck a tiny but immovable object

We are living on this big lump of rock called the Earth, drifting ever so continuously through space.

Now imagine a tiny but immovable object in space, a single atom of 'unexplanium', locked in space-time (relative to our universe or whatever you need). It will not move, bulge or transform; it's just there having indestructible mass. Assume it has the properties of a hydrogen atom (1 proton + 1 electron) for all physical and chemical purposes, except that it is endowed with infinite inertia.

What would happen if our Earth hit that one immovable atom? For simplicity, we may assume that the Earth's movement is linear and that the atom will hit the center of the Earth. I imagine something like a truck hitting a stationary object. If it's super narrow it'll slice the half, if it's broad it'll smash into it.

Would we even know it happened? Or would the amount of energy rip a giant tunnel through the planet? If a single atom is too small for something significant to happen, what would happen if it's the size of a marble or something bigger?

I've found this XKCD, which is somewhat related and might be an interesting read. It also described super fast and tiny particles:

"If a meteor made out of diamond and 100 feet in diameter was traveling at the speed of light and hit the earth, what would happen to it?”

• Sorry, this is the archetype of opinion based question: a totally fictional object with totally vague properties, and how are you going to pick an answer? – L.Dutch - Reinstate Monica May 23 '18 at 13:20
• You just need to specify the interactions that your unexplanium has with regular matter (making it less of an unexplanium I guess). – agaitaarino May 23 '18 at 13:22
• This may have already happened: telegraph.co.uk/news/science/science-news/3294473/… – Thucydides May 23 '18 at 13:24
• tiny but immovable object in space That makes no sense based on science - so unless you provide us with the physics for your fictional universe, we cannot answer. In the end it is completely up to you. – Renan May 23 '18 at 13:59
• There is no immovable in space; nothing is 'at rest'. – user3106 May 23 '18 at 14:14

## 11 Answers

As @Thucydides mentioned, researchers posited a similar magnitude of impact and thought they had found evidence for it, though they later retracted their findings they did calculate the potential effects.

The team, from the Southern Methodist University in Texas, analysed more than a million earthquake reports, looking for the tell-tale signal of strangelets hitting Earth.

While their very high speed gives strangelets a huge amount of energy their tiny size suggests that any effects might be extremely localised, and there is unlikely to be a blast big enough to have widespread effects on the surface.

The scientists looked for events producing two sharp signals, one as it entered Earth, the other as it emerged again. They found two such events, both in 1993. The first was on the morning of October 22. Seismometers in Turkey and Bolivia recorded a violent event in Antarctica that packed the punch of several thousand tons of TNT. The disturbance then ripped through Earth on a route that ended with it exiting through the floor of the Indian Ocean off Sri Lanka just 26 seconds later - implying a speed of 900,000 mph.

The speed of 900,000mph is important here. Since you're determining that the particles are absolutely stationary we need to know how fast the Earth is moving relative to them.

What is the speed of earth relative to absolute space?

In practice we approximate the comoving frame as the frame at rest with respect to the microwave background radiation. Thus we measure the speed of Earth relative to the comoving coordinates assuming, that the background radiation has no "natural" dipole anisotropy.

The speed of Earth wrt the local comoving frame measured this way in of the order of 500 km/s. And, of course, it varies as the Earth orbits the Sun. If we subtract the velocity of Earth in orbit around the sun, velocity of the sun relative to galactic center and velocity of the Milky Way with respect to the centre of the local group of galaxies, we find, that the local group is moving and slightly above 600 km/s relative to the comoving frame.

900,000mph is ~400km/s, you're looking for an impact appropriate to 600km/s, but the orders of magnitude have the same approximation as the incident in the Telegraph article

several thousand tons of TNT

• And what would happen if it whould be the size of a marble? – Martijn May 23 '18 at 14:36
• It seems unlikely this has already happened as the researchers later retracted their claim, apparently. en.wikipedia.org/wiki/Strangelet#cite_note-12 – goPlayerJuggler May 23 '18 at 16:22
• @goPlayerJuggler none the less the answer shows this is hypothetically possible, and what the signs were to look like if this were the case. – whn May 23 '18 at 16:57
• -1 This doesn't actually answer the right question. You answered what would happen if Earth hit a stationary object. Not an "immovable" object. That's not at all the same. I don't think that question is even answerable. – WakeDemons3 May 24 '18 at 17:23
• @WakeDemons3 An "immovable" object differs only in that it does not at all change it trajectory/velocity/acceleration relative to anything it collides with, which I think should be mathematically equivalent to a stationary object upon which additional force is magically applied in perfect sync and opposition to the force it experiences from the Earth colliding with it. Fudging for simplicity, this should have roughly the same effect as "stationary object being collided with that much harder". – mtraceur May 25 '18 at 23:51

Conclusion:

In the case of the atom, you wouldn't see anything without special instruments. This is because the object is so small that very few atoms will interact with it, so it imparts almost no energy to any matter that it encounters.

In the case of the marble, there would be a bright flash streaking through the sky, comparable to a large meteor, but much faster. Upon impact, the top layers of soil would be vaporized, resulting in a blast comparable in intensity to a very large conventional explosion (tens of tons of TNT). This would coincide with a small localized earthquake and a ground shockwave comparable to an underground explosion, due to the object passing through the rock at greater depths.

At the exit site, almost identical effects would occur, except there would be a large explosion followed by a flash streaking upward into the sky.

The exit event would happen $34$ seconds after the original impact, assuming the object comes down vertically and is moving at $370 \text{km} / \text{s}$ (see below about these assumptions).

Detailed explanation and calculations:

I. Hydrogen Atom

Let's first do a quick back-of-the-envelope estimate for how much energy would be imparted to the atmosphere.

1. Object's speed relative to the Earth.

First of all, what does "at rest" mean? Let's suppose the object is at rest with respect to the cosmic microwave background (CMB). The rationale is that if it originated in the primordial Universe, then it would have originally been at rest with respect to the CMB, and if it has infinite inertia, then nothing can change its motion, so it will remain at rest with respect to the CMB. The Earth moves at about $370 \text{km} / \text{s}$ with respect to the CMB, so that's how fast the object would hit.

1. Interaction with the atmosphere.

When the object passes through the atmosphere, then any atoms it comes into contact with will bounce off at approximately this speed, on average. Let's first calculate how much mass of air it will encounter. A column of Earth's atmosphere (from the surface to space) has a mass of about $10\mathpunct{,}000$ kilograms per square meter of surface. The object is about the size of a hydrogen atom, with a radius of $10^{-10}$ meters, so if it's a sphere, it will take out a cylinder of atmosphere with an area of $\pi R^2 \approx 3\mathrm{x}10^{-20} \text{m}^2$. Multiplying this by $10\mathpunct{,}000$ kilograms gives us $3\mathrm{x}10^{-16} \text{kg}$ of air that directly interacts with the object.

After this object passes, these air molecules will be moving around at about $370 \text{km} / \text{s}$. Using the expression $E = \frac{MV^2}{2}$ for kinetic energy, we get $$E = \frac{(10^{-16} \text{kg})(370\mathpunct{,}000 \text{m} / \text{s})^2}{2} \approx 7 \mathrm{x}10^{-6} \text{J}$$ energy imparted, in Joules. This is an absolutely minuscule amount of energy; for comparison, a normal household 10W halogen bulb emits 10 Joules of light per second.

1. Effects on rock would also be negligible, so there is no need to calculate them.

The amount of energy deposited when the object strikes the surface will be similarly negligible.

In the above calculation, the energy imparted to an object is proportional to its column mass (mass per surface area).

Consider the first three meters of soil that the object passes. A layer of rock 3 meters deep has about the same column mass as the atmosphere. Therefore, the object would also deposit about $7 \mathrm{x}10^{-6} \text{J}$ in the first three meters of rock that it penetrates. Again, this is practically undetectable. It will, of course, continue depositing these tiny amounts of energy as it passes through the Earth.

II. Marble

In the case of the marble, the calculation is almost the same as above, except the object's radius is now more like $0.01$ meters instead of $10^{-10}$.

So its radius is a factor of $10^{8}$ bigger, the area of the column of atmosphere it takes out is $10^{16}$ times bigger (as it's proportional to the radius squared) and the amount of energy deposited is also $10^{16}$ times bigger.

The marble would therefore deposit $7 \mathrm{x}10^{-6} \text{J} * 10^{16} = 7 \mathrm{x}10^{10} \text{J}$ of energy in the atmosphere, and a similar amount in the first three meters of rock.

Since we're looking at explosive-like effects, let's convert this to tons of TNT. One ton of TNT releases $4.2 \mathrm{x}10^{9} \text{J}$. So, our object would deposit $$(7 \mathrm{x}10^{10} \text{J})(\frac{1 \text{ ton}}{4.2 \mathrm{x}10^{9} \text{J}}) \approx 17 \text{ tons}$$ of TNT in the atmosphere and in the first three meters of soil.

The effects from the atmosphere would be comparable to a sizable meteor fireball. On the ground, the energy from the first few meters will reach the surface, producing an explosion comparable to a few tens of tons of TNT (a very large bomb).

Anything deeper than a few tens of meters would produce a relatively short-range surface rumble, like a small earthquake, but relatively little energy will reach the surface.

The same effects would happen at the exit site but in reverse order. The diameter of the earth is about $12700 \text{km}$, and the object is moving at $370 \text{km} / \text{s}$, so if the object comes down vertically, the exit event would happen $12700 / 370 \approx 34$ seconds after the first impact.

• Haha, so it would be a case of "I'm telling you, I really saw a faint lightbulk just plow into the grond!" and nobody to believe you – Martijn May 24 '18 at 14:14
• @Martijn: In the case of the marble, however, it would be far more impressive. Edited the answer to include that case. – Alexey Vlasenko May 24 '18 at 14:39
• 10 Joules of light per second. Light bulbs are not 100% efficient :P Maybe you meant to say that's how much light + heat a 10W bulb emits. – Peter Cordes May 25 '18 at 8:30
• @Wilf: There shouldn't be any magma leaking, because by the time the object gets deep enough for magma, the pressure in the rock will be so high that the cavity left behind by the object will collapse almost immediately. No magma will reach the surface. Of course, if the marble-sized object hits something important (like a city block), it can cause some serious destruction. Tens of tons of TNT is no joke. – Alexey Vlasenko Jun 4 '18 at 6:41
• I would expect the effects of the marble to be much bigger due to the (very) supersonic speed of the marble. The molecules in front of it don't have time to "move out of the way", and accumulate in front of it, slowly bleeding around. but as they bleed around, they have the same effect on molecules further in front, in effect making the frontal area larger. The most obvious result would be the mother of all supersonic shockwaves. – Eth Jun 4 '18 at 16:23

With an impactor the size of a hydrogen atom it's probable that we wouldn't even notice overly much. There might be a bright flash as it hits the atmosphere, it's even more probable there'd be on as it hit the ground, hard enough to break some molecular bonds. That flash is going to go all the way through too but we're not going to see it below ground level. It's almost impossible, certainly at such low relative velocities, to get an individual atom to actually hit such an object, even one with no charge, but because of it's exotic nature there might be some atomic fission events. Taken altogether I would expect very little energy to be imparted to the Earth, and very little if any noticeable damage.

• @JanDoggen Yeah but no it would hit the atmosphere, and then the ground, and then go the whole way through the planet and out the other side, it doesn't move or transform in any way so it's going to hit things all the way along. – Ash May 23 '18 at 14:26
• @Martijn Yeah no I'll do your marble and then you're on your own, these are not simple equations. – Ash May 23 '18 at 15:58
• @Ash: What is "the epicenter of the big bang"? The Big Bang happened everywhere. – Beta May 24 '18 at 1:26
• @Beta: Everywhere was very small back then. – Joe Bloggs May 24 '18 at 6:38
• @somebody: As far as we know, space itself doesn't have a velocity, i.e. there's no such thing as absolute velocity, and no special frame of reference. Fortunately, for the purposes of this question we can just consider small objects moving relative to the earth at whatever velocity we think is interesting. – Peter Cordes May 25 '18 at 8:08

At first I thought this would be a planet-ending event, but upon further consideration, it wouldn't be that big of a deal.

First off, nothing can really be immovable, and nothing is really stationary in an absolute sense as others have mentioned. However, if we assume that the earth is going around a few hundred thousand miles per hour relative to the cosmic background, that's a number we can work with: let's call it 400,000 mph. That's not a relativistic number, so nuclear reactions are unlikely.

Neutrinos have a small cross section and routinely pass through the earth unimpeded. However, a hydrogen atom has a much larger area; this is the reason hydrogen can be stored in a pressure vessel, because it bounces off the walls of the container.

Now hydrogen has a diameter of about 50 picometers, so at the very least the particle would scrape out a 50 picometer-wide tunnel all the way through the earth. The mass of that tunnel would be about $6\times10^{-10}\text{ kg}$.  If all of that mass were converted to pure energy, which is a worst-case scenario ($\text{E}=\text{mc}^2$), it would be about $5.6\times10^{7}\text{ Joules}$, which is about the amount of energy released by burning a couple liters of gasoline.

So the good news is we'd all survive.

• Earth's orbital velocity in a Sun-centric reference frame is about 30 km/s, and the Sun's orbital velocity in a reference frame fixed to the centre of the Milky Way is on the order of 200 km/s if I recall correctly. 200 km/s is 720,000 km/h or about 450,000 miles per hour. Or, alternatively, 0.07% of the speed of light. For our purposes, relativistic effects at such velocities should be completely negligible. – a CVn May 23 '18 at 21:04
• The biggest difference between a neutrino and an H atom (or H2 molecule) is that neutrinos don't interact with matter via the electromagnetic force. Or via the en.wikipedia.org/wiki/Pauli_exclusion_principle, which applies between the electron shell in an H atom and the electron shells of any atoms / molecules it approaches. This is why neutral atoms repel if pushed close together, and why you don't fall through your chair even though at an atomic level even "solid" matter is mostly empty space. Neutrinos can go right through atomic nuclei with a low chance of interacting. – Peter Cordes May 25 '18 at 8:16

As we now know - everything is relative.

A fixed particle in space might as well be a moving particle in space, relative to us, when we are moving.

This being the case, the atom in question would simply be like many other atoms already bombarding the Earth - individually the effect would likely be negligible, collectively a different story.

The only difference in your case is your comment 'infinite inertia' - I presume this atom doesn't interact like normal atoms do then when in proximity or colliding with another atom. Still, one atom is only one atom, it may not directly interact with others at all on its brief journey through the Earth, or if so only imperceptibly.

• This is not a bad answer, but not an answer to the question that has been asked. It might not have any influence on the fate of mankind, but what would actually happen during the event? – Raditz_35 May 23 '18 at 13:55

## Nothing will happen.

Remember, there are (comparatively, when you think about the ratio of macroscopic distances to the sizes of macroscopic objects), huge gaps between atoms, which is why hydrogen will gradually leak out of any container. In much the same way, the fixed hydrogen atom will mostly push other atoms out of the way. In the atmosphere, we might see a parabolic arc of lightning as the the atoms gets knocked away at the speed of the earth's rotation. When it hits the ground, the fixed atom will simply force its way through, displacing atoms where it comes in contact with the nucleus, until it comes out on the other side. Then we might see another parabolic lightning arc going the other way.

• Thabnks for your answer. Because I sort of expected this outcome, I've added the "what if it would be the size of a marble" :) – Martijn May 24 '18 at 12:18
• @Martijn: It will be crushed. Immovable isn't indestructible. – nzaman May 24 '18 at 12:20
• It might be worth noting that the Earth isn't a rigid body -- it's spherical precisely because gravity was able to deform it into a (more or less) evenly-distributed shape, and much of its center is liquid. It seems plausible to me that an indestructible marble with infinite inertia (assuming such a thing could exist) might just plow right through the core of the Earth and come out the other side again, like a bullet flying through a cube of butter. As for how much that would damage the areas near the entry and exit areas, I can't guess. – Jeremy Friesner May 24 '18 at 14:08
• @JeremyFriesner: Nobody said it was indestructible. Simply that it was moving with sufficient inertia so as not to be deflected by the mass of the earth (which is what immovable actually means in an expanding universe). We know nothing of its density, only its volume and a rough idea of its speed – nzaman May 24 '18 at 14:30
• @nzaman "immovable" rather implies "indestructible" to me, in that if the object is destroyed, it will no longer be found at the location where it originally was at, and therefore will have been effectively -- moved. – Jeremy Friesner May 24 '18 at 15:06

As the asker of the question has already learned, it was asked using some phraseology which is difficult to mesh with contemporary physics. As such, I am attempting to answer as if “immovable” is describing:

• an elastic collision in which the relative motion between a third body which does not collide and one participant in the collision is unchanged.

While the Earth is moving in relation to this “immovable atom” — IA — which you propose, each atom of the Earth has a velocity vector with respect to the IA. The distance between the IA center of mass and the Earth's center of mass is decreasing. When any atom considered to be a part of the Earth collides with this IA, the result of that collision does not cause any change of the IA's relative velocity vector with respect to the velocity vector of any other atom.
In simpler language, the incident atom which is not the IA will bounce back with exactly the same speed but with a deflected direction. Any other atom will continue to see the IA as having the same velocity vector as it did prior to the collision.

Ergo, the IA will continue to pass through the Earth with each successive collision — however, the relative velocity vector between the Earth's center of mass and the center of mass of the IA will not be unchanged, but will be apparently slowed, albeit almost immeasurably. Such will occur because the velocity of the Earth's center of mass is composited from the velocity vectors of each atom which is factored in the arbitrarily defined center of mass …

Anyway. So, we have not yet defined exactly what fields are contributing to these aforementioned collisions. If the IA is uncharged, i.e. not ionic, then the collision is simply a result of the repulsion between electrons — which is, in turn, a consequence of the Pauli Exclusion property which electrons have — when its electron orbitals approach the similar orbitals of other atoms. It is very improbable that nuclear orbitals will ever interact, but at higher velocities such interactions become more probable — i.e. like the velocities you'd see in an atom smasher. And, even then, not very likely.

Of course, all that lends problems to considering exactly how the IA is indeed ‘immovable’, and how that pertains to its constituent particles: its electrons, quarks, and all the other stuff that perhaps fluxes about between in unperceived spaces inside. Gluons and photons and whatnot.

What if the IA were larger — like a pea, or a ball-bearing, or a marble? Well, all that really does — aside from making its ‘immovability’ more difficult to define — is give the IA a larger cross–section for its collisions: i.e., collisions are more likely.
Because it is immovable, and any collision with it is perfectly elastic, its own mass is irrevelant. Only its size is relevant.

What would be the state of that big mass of atoms which we call the Earth when the IA has passed through them and produce no further interactions?
Well, it really depends on what the initial relative velocity was between the two centers of mass. Many other answers here seem to confuse ‘immovable’ with ‘motionless’.
Whatever relative velocity you finally decide to use in your world, know that a larger magnitude will cause more deflection and thus more damage.

All in all, the more details you want require a more rigid definition understanding the concept of ‘immovable’. Others have more or less explained why immovability is almost like a physical limit: you can become more or less immovable, but never perfectly immovable.
Unless, of course, you exist outside the laws of physics which are known to contemporary science.

If that is the case, then this answer is really only useful to you if your physical laws operate in any way which allows my explanation to seem reasonable. Obviously.

• The OP changed the question from an atom to a marble (see comments on my answer) – nzaman May 25 '18 at 5:16
• I have never chaged it from atom to marble, it has been marble from the first version of the question :) I've had both, in case a single atom would be boring :) – Martijn May 25 '18 at 7:43
• +1 finally a mention of en.wikipedia.org/wiki/Pauli_exclusion_principle and the en.wikipedia.org/wiki/Exchange_interaction "force" being how (the electron shells of) neutral atoms repel each other when very close together. – Peter Cordes May 25 '18 at 8:36
• Size of the ‘immovable whatever’ simply gives it a larger cross–section, but the concept is the same — hang on, I'll add that to this answer – can-ned_food May 26 '18 at 6:39

Immovable (relative to the Earth), and potentially indestructible, marble sized object? "Immovable" could only mean it exists at the exact center of the universe, around which everything else is in relative motion.

It would be surrounded by a cloud of dust and asteroid fragments from previous impacts, would it not? Unless it sprang into being shortly before impact with Earth. Dust zone size depending on how long this marble has existed.

We would enter that zone of space dust, causing a massive and unexpected meteor shower. Then a flare of light as the marble passed through the atmosphere, followed by minimal impact, and a shudder as it plowed through Earth.

This would be a "bullet" that did not flatten or break on impact. The collision would be no big deal, but pressure would surely increase as it bulled its way through. Resulting in confusing seismic readings on the opposite side, then a small eruption as the marble exited. Or, if it entered and exited through oceans, no discernable effect on the surface of the ocean.

• When you say “immovable relative to the Earth”, it seems like you are saying that it doesn't move with respect to the Earth, but simply floats at a constant distance and position. Would that be geosynchronous, or how exactly does the rotation of the Earth in the solar frame of reference contribute here? – can-ned_food Jun 4 '18 at 21:20
• This marble of unobtanium would have to be at the "center" of the known universe, around which every other object is in relative motion, right? My aim in posting this answer was to bring up the cloud of debris. – Mary C Jun 6 '18 at 4:36
• Oh, i gotcha. So it could actually appear to move with respect to one thing or another, but if you were able to compute the exact positions and movements of everything, then you'd see that the immovable unobtanite was like the stationary center. I think i disagree with the stuff about the collision, but i like the definition of ‘immovable’. – can-ned_food Jun 6 '18 at 6:39

The answer is known, and it’s that nothing happens at all. If something is immovable, it means that nothing can move it, which can only mean that nothing we know of will interact with it. You’re describing a particle of dark matter, which the Earth undoubtedly hits all the time, but since it doesn’t interact with the matter of the Earth (except gravitationally, which is too weak and too long-range to be noticeable) nothing happens.

• Not necessarily dark matter — indeed, not at all, I would say — but the premise is sound, albeit a tad pedantic. I like how this answers the question as if it wants something to warrant the adjective “immovable” at the expense of any other properties which it also stipulates. – can-ned_food May 29 '18 at 1:15

This scenario already exists. Neutrinos can travel through dense matter such as the Earth without interacting with a single atom, leaving no trace of their passage. To observe even just a few of the extremely rare interactions of neutrinos with matter, physicists build detectors with massive amounts of target material and operate them for many years. The detectors record the tracks of the particles that emerge from the rare collisions of neutrinos with atoms of the target material.

I remember hearing that if you wanted to block a neutrino, you would need a sheet of lead as thick as the distance the Earth is from the Sun.

• The premise of the question is an immovable object which interacts like an atom of normal matter. Neutrinos are not immovable, and do not interact like an atom of normal matter. – Sneftel May 24 '18 at 11:14

For every action there is an equal and opposite reaction. If the supposed object is immovable, that tells us that the reaction is zero, and therefore we know that the action is zero; if there is no effect on one of the objects that are colliding, then there can be no effect on the other one either.

• This seems like a philosophical dodge, and not an attempt to engage the question. – Misha R May 28 '18 at 0:17
• I'm sorry you see it that way. It's a genuine attempt to see what would happen under Newtonian mechanics if there was such a thing as an "immovable object". My conclusion was that it can't be immovable unless it somehow is incapable of reacting with the other body in a collision, and if it doesn't react with the other body then the collision has no effect on the other body either. (Essentially @MikeScott came to the same conclusions.) – Michael Kay May 28 '18 at 8:25
• Your conclusion ignores the question, which asks us to consider a scenario that does have an immovable object. If you believe that the objects will have no effect on each other, however, then you are welcome to describe it in practical terms, rather than as an abstract statement. If you cannot do that, then your answer is - like I said - a dodge, and nothing more. – Misha R May 28 '18 at 9:24
• I absolutely did not ignore the question. I specifically focused on the part of the question regarding the behaviour of an "immovable object", and considered what Newtonian mechanics would have to say about immovable objects. – Michael Kay May 28 '18 at 18:40
• What the classical approach would say about the impossibility of immovable objects is irrelevant, since an immovable object is present. The hypothetical asks you to suspend that part of Newtonian physics, and your answer, in essence, is "no, I won't." Consider that this is a process scientists are more than familiar with - thinking creatively every time some new piece of information comes in that defies a current accepted model. In this case, the piece of information is that we have an immovable object. To point at the classical model and say "no" is to refuse to engage. – Misha R May 28 '18 at 20:26


text:
## applications of definite integrals

QUESTION1) A dome is made by cutting a sphere of radius 5 meters at 2.5 meters from its center. For having a better idea, look at the picture of problem 30, page 475. Find the area of the surface of the dome. Hint: express the curve being revolved using parametric equations, and then revolve about the y-axis. ..QUESTION 2) Consider the curve x = sqr of (y + 1) for 0?y?1. Find the area of the surface obtained by revolving that curve about the y-axis...QUESTION 3) Solve the following differential equation, where y > 0 and -?/4? x? ?/4. y cos^2(x)dy/dx =(y^2 +1)tan(x)


text:
# Difficult question on pressure in liquids

1. Jun 30, 2009

### jontyjashan

1. The problem statement, all variables and given/known data
There is a container of height 2 m
it is filled with water upto the brim
there are 2 holes made in it
one hole of area 5cm^2 is made at a height of 1.5 m
another hole of area 10cm^2 is made at a height of 0.5m
through which hole will the water rush out with more pressure

2. Relevant equations

P=hdg

3. The attempt at a solution
I cannot make out through which hole the water will rush out with more pressure.
i dont know whether pressure in liquids depend upon area

2. Jun 30, 2009

### minger

This question is worded quite poorly if you ask me. When people say that a liquid is flowing with a lot of pressure, they often mean that it exerts a lot of force upon a stationary object; or more plainly, it has a high velocity.

If we are to look at pressure alone, then there are two pressures to consider: the pressure just before the fluid leaves the container, and the pressure afterwards. The pressure just before it leaves the container is simply the hydrostatic pressure that you have listed $$p = \rho g h$$. As the fluid then leaves the container, this pressure or energy is converted into kinetic energy, which can be illustrated in Bernoulli's equation which basically says that along a steamline the pressure energy + the velocity energy + the height equals a constant. Now, the final pressure is simply the pressure of the air, ambient.

So, whatever case has the higher initial pressure has the higher velocity. However, this is not pressure, as both fluids will leave at atmospheric pressure.

3. Jun 30, 2009

### tiny-tim

Hi jontyjashan!
Well, pretend there's only one hole, and that the density of water is 1 …

what speed will the water leave the hole?

4. Jun 30, 2009

### jontyjashan

First of all there is only 1 container and 1 fluid not 2 fluids.
how can water from both holes leave with the same pressure.It is simple logic.What wordings inthe question you did not understand by the way.I just want to know from which hole water will come out with more velocity

5. Jun 30, 2009

### minger

I understood the wordings [sic] better than you explained them. Pressure != velocity. Remember that. Any jet discharging into the atmosphere will have an internal pressure of 0 gauge. However, I was trying to explain to you the difference between the two.

Now, onto velocities. Before anyone simply throws you the answer, why don't you give it a try. What do you know, what do you need to find? What are you looking for?

6. Jun 30, 2009

### jontyjashan

sorry ,If i said something wrong but
if we do this experiment practically,velocity of water from the upper hole is much high
but if we look at it mathematically,velocity of water from the lower hole should be high

7. Jun 30, 2009

### RoyalCat

Yes, you are incorrect. Take the limit of the 'upper hole', the very top of the container of liquid. The velocity at which it exits the container is 0, it does not leak out at all!

Interestingly enough, the maximum horizontal distance the water will reach isn't achieved from a hole at either extreme, but from a hole at the very middle. :)

8. Jun 30, 2009

### tiny-tim

Hi jontyjashan! Thanks for the PM.

Yes, mathematically the speed depends only on the pressure, and therefore on the depth, not on the size of the hole.

Did you do the experiment yourself?

How did you measure the water coming out of the two holes?

9. Jun 30, 2009

### RoyalCat

Wouldn't the fact that $$P\propto{Area^{-1}}$$ have a say in that?
So that means that the acceleration the pressure provides to the liquid is inversely proportional to the area of the hole it's bursting through, doesn't it? (Sorry for the poor wording, by the way. Haven't dealt with pressure properly yet)

And the acceleration it provides to the liquid, is what determines its exit velocity, and this would be determined by making an equation something like the following, right?

$$a=\frac{P}{mA}$$
$$a=Acceleration, A=Area, P=Pressure$$
$$P=\rho gh$$
$$a(A,h)=\frac{\rho gh}{mA}$$

So the acceleration, and in turn, exit velocity, is directly proportional to the height, and inversely to the area, isn't it?
Sorry if I'm way off on this, this is just my intuition talking here, other than knowing the units of pressure, I'm lost, haha!

10. Jun 30, 2009

### minger

The velocity is inversely proportional compared to the area in the case of pipe flow or when the fluid is forced through the opening. In this case of a free jet, where the inlet can be assume to be an infinite resevoir, this is not the case.

The fact of any disparities you may see in a small experiment would be in any difficulties keeping the level of the fluid high enough such that constant pressure is maintained in both holes.

11. Jun 30, 2009

### tiny-tim

Sorry, RoyalCat, but you're just burbling.

12. Jun 30, 2009

### RoyalCat

:shy:
Guess I should give this a try once I actually study hydrodynamics, heh.

13. Jul 1, 2009

### jontyjashan

i have tried this experiment practically,and i found that from the upper hole water just trickles out but from the second all the water comes out very fast

14. Jul 1, 2009

### tiny-tim

But that is mathematically correct!

(I'd thought you were saying it was the other way round )

At the lower hole, there is more pressure , so the water should come out faster.

15. Jul 1, 2009

### minger

Just so we don't confuse him tiny...

At the lower hole, just before the hole there is more pressure, which is a function of depth. That pressure then becomes atmospheric as the fluid leaves the hole, with the energy being converted into kinetic energy, the difference in pressure being the driving force.

16. Jul 1, 2009

### modulus

As minger mentioned, the experiment is practically impossible, because it is not possible to keep a uniform pressure while the water flows out of the holes.

But, if we take this theoretically and conceptually, the hole from which water comes out with more pressure (considering an infinite reservoir of water), is the one from which water goes a larger horizontal distance.

So, I think we might try to attempt this by figuring out the initial velocity with which the water comes out from each hole, and then, use that knowledge to figure out the horizontal range (considering each water particle is a projectile under horizontal projection) with the formula u*sqrt(2h/g).

But, here, we come at another problem- if we use this formula, we must consider the area around the infinite reservoir to be a vaccum, and, if we consider that, then, the water inside the reservoir would have infinite pressure (being exerted on the walls of the container).

Or, we can use the formulas of projectile motion which also consider drag provided by air... but I don't know those...

17. Jul 2, 2009

### jontyjashan

ya,there is need to find the range of the water from both holes.but is it necessary that water having more range will have more velocity

18. Jul 2, 2009

### ideasrule

You need to find the range? I thought you just needed to find out which hole water will rush out faster from.

Also, no, it's not true that higher velocity = higher range. If the hole is too close to the ground, then the water will get to travel almost no distance at all before hitting the ground. if the hole is too high, the water's exit velocity would be very low. Somewhere between those two extremes is the height that gives the longest range.

Last edited: Jul 2, 2009
19. Jul 2, 2009

### jontyjashan

thats a part of it

20. Jul 2, 2009

### jontyjashan

i think what you said is not true


text:
# How do you solve the equation for y: (2x)/ 5 - (x / 3) = 3?

Mar 23, 2018

You cannot...

#### Explanation:

In this equation you can not solve for $y$, for there is no $y$ in it.

But we can surely solve for $x$, which goes as follows :-

$\left(\frac{2 x}{5}\right) - \left(\frac{x}{3}\right) = 3$

(3×2x -5×x)/15=3

6x-5x=3×15

$x = 75$

$\therefore \textcolor{B r o w n}{x = 75}$


text:
Span

In mathematics Span has many concepts.

Vector span

The span of a list of vectors $\{\boldsymbol{x}_{1},\boldsymbol{x}_{2},\dots,\boldsymbol{x}_{n}\}$ is the space formed by all linear combinations of the vectors,

$Span\{\boldsymbol{x}_{1},\boldsymbol{x}_{2},\dots,\boldsymbol{x}_{n}\}:=\{\boldsymbol{v}\in\mathbb{R}^{n}|\boldsymbol{v}=a_{1}\boldsymbol{x}_{1}+a_{1}\boldsymbol{x}_{1}+\dots+a_{n}\boldsymbol{x}_{n},\; \forall a_{i}\in\mathbb{R}\}.$


text:
# Calculate boolean matrix multiplication (BMM) using transitive closure

Let us say I am given an algorithm that calculates the transitive closure of a given graph $$G = \{ V, E \}$$.
How can I use this algorithm in order to perform the Boolean Matrix Multiplication of two matrices $$X$$ and $$Y$$?

I know that in order to calculate the transitive closure of a matrix $$I$$ need to compute $$I^{(V-1)}$$. But what else?

• How is the graph $G$ given as input? Feb 18, 2020 at 16:52
• G={V,E},directed graph Feb 18, 2020 at 16:52

Let us build the tripartite graph $$G = (S := U\dot\cup V \dot\cup W, E)$$, where $$U := \{u_1, \dots u_n\}$$ and similarly $$V := \{v_1, \dots v_n\}$$ and $$W := \{w_1, \dots w_n\}$$. Define $$E$$ as follows: For $$i, j \in [n]$$, we add $$(u_i, v_j)$$ to $$E$$ for $$u_i \in U$$ and $$v_j \in V$$, if and only if $$X_{ij} = 1$$. Similarly we add $$(v_i, w_j)$$ to $$E$$ for $$v_i \in V$$ and $$w_j \in W$$ if and only if $$Y_{ij} = 1$$.
Let $$G^T := (S, E')$$ be the transitive closure of $$G$$. This means $$(x, y) \in E'$$ if and only if there is a path from $$x$$ to $$y$$ in $$G$$.
Claim. Let $$Z := X \cdot Y$$ be the matrix resulting from the multiplication. We claim that $$Z_{ij} = 1$$ if and only if $$(u_i, w_j) \in E'$$.
Proof. $$Z_{ij} = 1$$ if and only if $$\bigvee\limits_{k=1}^nX_{ik}\land Y_{kj}$$ if and only if there is a $$k\in [n]$$ such that $$X_{ik} = 1$$ and $$Y_{kj} = 1$$ which is the case if and only if there is a $$k\in [n]$$ such that $$(u_i, v_k) \in E$$ and $$(v_k, w_j) \in E$$ which is the case if and only if$$^{(*)}$$ there is a path in $$G$$ from $$u_i$$ to $$w_j$$ and hence $$(u_i, w_j) \in E'$$.
Try to prove $$(*)$$ as an exercise. Hint: for the harder direction use the fact that the graph is directed.


text:
## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.1 - Linear and Nonlinear Systems of Equations - 9.1 Exercises - Page 635: 1

#### Answer

$Solution$

#### Work Step by Step

The definition of a solution is given as: An ordered pair which satisfies each equation in the system of equations.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Minimum steps to sort array

Consider you have a permutation of $$1$$ to $$n$$ in an array $$array$$. Now select three distinct indices $$i$$,$$j$$,$$k$$, there is no need to be sorted. Let $$array_i$$, $$array_j$$ and $$array_k$$ be the values at those indices and now you make a right shift to it, that is $$new$$ $$array_i$$= $$old$$ $$array_j$$ and $$new$$ $$array_j$$= $$old$$ $$array_k$$ and $$new$$ $$array_k$$=$$old$$ $$array_i$$. Find the minimum number of operations required to sort the array or if is impossible how to determine it.

Example : Consider $$array= [3,1,2]$$; consider indices $$(1,3,2)$$ in the given order after applying one operation it is $$s =[1,3,2]$$.


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# Calculus

I have the equation:
cos x - x = f(x)
I am told to find the relative extrema. I am also told to use the Second Derivative Test where applicable.
My question is how do I solve this problem, and how do I know when to use the Second Deriv test?
Thanks!

1. 👍
2. 👎
3. 👁
4. ℹ
5. 🚩
1. Points where the first derivative is zero are relative maxima if the second derivatve is negative and relative minima if the second derivative is positive.

In your case, df/dx = 0 when f'(x) = -sin x -1 = 0; sin x = -1
That happens when x = 3 pi/2.
At that point, f''(x) = -cos x = 0

So the second derivative fails to show if is a maximum or minimum. 3 pi/2 could be an inflection point, where the curve flattens out and then resumes its downward trend

1. 👍
2. 👎
3. ℹ
4. 🚩
2. Can't i figure out if its increasing or decreasing using the first derivative test?
What is the point of the second derivative test?

1. 👍
2. 👎
3. ℹ
4. 🚩

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# Math Help - undetermined coeffecients...

1. ## undetermined coeffecients...

Im doing my DE and were going over complimentery and undetermined coeffeceints. Complimentery equations I could spout out all day long , but the undetermined coeffeceints is giving me some problems.

One of my main problems is im not quiet sure on what to subsitute for on the coeffeceints when the right side of the equations is a repeated root, or a trig function. Also i cant really ask the teacher as it is an online course. Some examples of my problems are:

PROBLEM 1)

y(double dervitave) + 3y = -48(x^2)(e^3x)

i got the complimentery as yc= e^x(c1cos((squareroot(3))x)+C2sin(sqareroot 3))

but i go to build the coeffecients i get stuck. i have come up with A(e^4x)+ B(x(e^4x))+(c(x^2)(e^4x))

its suppose to come out :

y=(c1cos((squareroot(3))x)+C2sin(sqareroot 3))+((-4x^2)+4x-(4/3))(e^3x)

PROBLEM 2)

y(double der.) +4y = 3sin2x

im not sure what to expand this out to anymore...as when i set up the coeffecients i get (Acos(2x))+(bsin(2x)) which once i take the first and second derviate and subsitute i get all my coeffecients =0 which means i need to expand, but i am not sure what i would add to expand it....

I know this post is lengty and ugly but i would greatly appreciate any help or tips to get with the coeffecent building thanks!

2. #1:

$y''+3y=-48x^{2}e^{3x}$

From $m^{2}+3=0$ we find that $m_{1}=\sqrt{3}i$ and $m_{2}=-\sqrt{3}i$.

Then $y_{c}=C_{1}cos(\sqrt{3}x)+C_{2}sin(\sqrt{3}x)$

We can assume $y_{p}=(Ax^{2}+Bx+C)e^{3x}$

Sub into the DE we get:

$2A+6B+12C=0, \;\ 12A+12B=0, \;\ 12A=-48$

Solve the system and get:

Then, $A=-4, \;\ B=4, \;\ C=\frac{-4}{3}$

$y_{p}=(-4x^{2}+4x-\frac{4}{3})e^{3x}$

and therefore, hence, and whence:

$y=C_{1}cos(\sqrt{3}x)+C_{2}sin(\sqrt{3}x)+\left(-4x^{2}+4x-\frac{4}{3}\right)e^{3x}$

3. I might be asking a completly stupid question here:

The yp= (ax^2+bx+c)e^3x, the yp prime and yp double prime would be a long fun product rule?

4. Yes, for y'', you sub in a take the second derivative of $(Ax^{2}+Bx+C)e^{3x}$

which gives $(9Ax^{2}+12Ax+9Bx+2A+6B+9C)e^{3x}$

Then

$(9Ax^{2}+12Ax+9Bx+2A+6B+9C)e^{3x}+3(Ax^{2}+Bx+C)e^ {3x}$

= $(12Ax^{2}+12Ax+12Bx+2A+6B+12C)e^{3x}$

Now, equate coefficients and you get the system form my first post.

It's not that bad.

5. yeh, its not that bad...except i ussally make my errors on stupid mistakes i learned years ago


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How do you find the derivative of 1/sqrt(x)?

Jun 5, 2016

This function can be written as a composition of two functions, therefore we use the chain rule.

Explanation:

Let $f \left(x\right) = \frac{1}{\sqrt{x}}$, then $y = \frac{1}{u} \mathmr{and} u = {x}^{\frac{1}{2}}$, since $\sqrt{x} = {x}^{\frac{1}{2}}$.

Simplifying further, we have that $y = u \mathmr{and} u = {x}^{- \frac{1}{2}}$

The chain rule states $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

This means we have to differentiate both functions and multiply them. Let's start with $y$.

By the power rule $y ' = 1 \times {u}^{0} = 1$.

Now for $u$:

Once again by the power rule we get:

$u ' = - \frac{1}{2} \times {x}^{- \frac{1}{2} - 1}$

$u ' = - \frac{1}{2} {x}^{- \frac{3}{2}}$

$u ' = - \frac{1}{2 \sqrt{{x}^{3}}}$

$f ' \left(x\right) = y ' \times u '$

$f ' \left(x\right) = 1 \times - \frac{1}{2 \sqrt{{x}^{3}}}$

$f ' \left(x\right) = - \frac{1}{2 \sqrt{{x}^{3}}}$

Hopefully this helps!


text:
5

# Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2tx(t) =-2cost y(t) = sin t...

## Question

###### Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2tx(t) =-2cost y(t) = sin t

Find the area bounded curve defined below on the Questlon #10 interval 0 <t <2t x(t) =-2cost y(t) = sin t

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text:
# Fractions

Sort fractions
${z}_{1}=\frac{6}{11};{z}_{2}=\frac{10}{21};{z}_{3}=\frac{19}{22}$
by its size. Result write as three serial numbers 1,2,3.

n(z1):  2
n(z2):  1
n(z3):  3

### Step-by-step explanation:

Did you find an error or inaccuracy? Feel free to write us. Thank you!

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text:
# What is the complete electron configuration of nitrogen?

For the nitrogen atom, $Z = 7$
$1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$....the $p$ electrons occupy each of the $2 p$ orbitals, ${p}_{x} , {p}_{y} , {p}_{z}$ singly....this is the most stable electronic configuration.


text:
×
Get Full Access to C++ For Everyone - 2 Edition - Chapter 6 - Problem R6.1
Get Full Access to C++ For Everyone - 2 Edition - Chapter 6 - Problem R6.1

×

# Write code that fills an array double values[10] with each set of values below.a. 1 2 3

ISBN: 9780470927137 356

## Solution for problem R6.1 Chapter 6

C++ for Everyone | 2nd Edition

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C++ for Everyone | 2nd Edition

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3
Problem R6.1

Write code that fills an array double values[10] with each set of values below.

a)  1 2 3 4 5 6 7 8 9 10

b)  0 2 4 6 8 10 12 14 16 18

c)  1 4 9 16 25 36 49 64 81 100

d)  0 0 0 0 0 0 0 0 0 0

e)  1 4 9 16 9 7 4 9 11

f)  0 1 0 1 0 1 0 1 0 1

g)  0 1 2 3 4 0 1 2 3 4

Step-by-Step Solution:

Step 1 of 7

a) 1 2 3 4 5 6 7 8 9 10 for this loop i is started from 0 and ended when the value of i is smaller to the10 and inside the for loop the value i should be equal to . Here when value of i is 0 it is incremented by 1 so value of i should be equal to 1 then 2 up to 9 index is should be incremented. Because index started from 0 so at index 0 value 1 is present and at the 9 index value 10 is present. So code for this is:

for(int i=0;i<10;i++)

{

value[i]=i+1;

}

Step 2 of 7

Step 3 of 7

##### ISBN: 9780470927137

The full step-by-step solution to problem: R6.1 from chapter: 6 was answered by , our top Engineering and Tech solution expert on 03/08/18, 08:39PM. The answer to “?Write code that fills an array double values[10] with each set of values below. a) 1 2 3 4 5 6 7 8 9 10 b) 0 2 4 6 8 10 12 14 16 18 c) 1 4 9 16 25 36 49 64 81 100 d) 0 0 0 0 0 0 0 0 0 0 e) 1 4 9 16 9 7 4 9 11 f) 0 1 0 1 0 1 0 1 0 1 g) 0 1 2 3 4 0 1 2 3 4” is broken down into a number of easy to follow steps, and 90 words. C++ for Everyone was written by and is associated to the ISBN: 9780470927137. This textbook survival guide was created for the textbook: C++ for Everyone , edition: 2. This full solution covers the following key subjects: . This expansive textbook survival guide covers 10 chapters, and 515 solutions. Since the solution to R6.1 from 6 chapter was answered, more than 343 students have viewed the full step-by-step answer.

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### codingkshatriya's blog

By codingkshatriya, history, 5 weeks ago,

Chronology of events :

1) Solved problem 2, submitted, got "correct answer"

2) Solved problem 3, submitted, got "correct answer"

3) Coded problem 4, got some errors in testing, debugged it, clicked "test" button got Please wait, fetching code

4) Thought it is some server issue, moved to problem 1, coded it and clicked "test" button, got Please wait, fetching code

5) Moved to 5th problem, tried solving it, couldn't get a concrete idea

6) Tried submitting 1 and 4, got the same issue

7) Copy pasted the test link in a fresh window, continued with the contest and tried submitting again, this time got Error: Connection lost.

8) Tried minimizing the window and checking the internet connection, got warning "Go to fullscreen in 10s", checked internet connection worked fine.

9) Tried submitting multiple times but in vain.

I wasted a lot of time and was not able to submit 2 problems. Any help or guidance would be great ?

Did anyone else face this issue ?

Tagging people whom I think would be able to help Ashishgup

• +4

By codingkshatriya, history, 3 months ago,

In the wake of current contests, observation has become very crucial in every problem, thinking from the problem statement and trying to reach the solution is no longer easy, because the crucial observation is no longer easy to arrive from the problem statement alone, I think we have to start from random points and see if we can reach solution from there.

Let's take today's C (Edu round 99), the crucial observation here is Bob can't win more than y matches, and there is a way for him to always win y matches, this was not easy for me, I saw the high number of AC's lowered my thinking scope, saw the sample, observed the pattern then tried to justify it. If there had been no sample, and I had not seen the submissions, I probably would have taken more time.

And unpredictability factor has increased, sometimes you get the observation in a few seconds, and at times you never get it, this definitely has impacted the ratings.

Any suggestions or comments on this are welcome.

EDIT : Some have misinterpreted my post, I am not complaining nor am I urging the problem setters to change the format, my post is more about how to tackle these kinds of problems, before screaming practice, add me as your friend have a look at the number of problems I have solved in the standings. And regarding unpredictability, I think it makes sense, even experienced coders at times have a blind spot, and relatively young ones may get that observation immediately, that's why I said it's more about the thinking style.

• +163

By codingkshatriya, history, 3 months ago,

I would like to hide the problem tags in Mashups, is there any way to do that ?

• +1

By codingkshatriya, history, 4 months ago,

I came across this site (https://kenkoooo.com/atcoder/#/table/ ) that quantifies the difficulty of atcoder problems,but I was wondering how it would relate to CF difficulty, Any inputs would be great

• +3

By codingkshatriya, history, 5 months ago,

How did so many people get D right ? more than 2000 ppl got AC, just curious, my Approach was F[i] be the answer if string starts from i,

if(str[i]==str[i+1])
F[i] = F[j] + 1

where j is the first index diff from i,

else
F[i] = F[i+2] + 1

, this approach though wrong, seemed extremely obvious, so those who sovled it in contest did you land on the right approach initially itself, or did you land on the above mentioned approach then realized it's wrong and then moved on ?

• -25

By codingkshatriya, history, 6 months ago,

MikeMirzayanov Problem 1183H - Subsequences (hard version) has difficulty rating of 1900, which is less than it's corresponding easier version 1183E - Subsequences (easy version) of rating 2000, and the only difference between them is the constraints, can you please look into it?

• +14

By codingkshatriya, history, 7 months ago,

Given a binary array, convert all the elements to zero Array (array with only zeros) in minimum number of steps

In one step we can choose any 2 adjacent elements and flip them together, or you can choose the last or first element and flip that particular element alone.

Input : N ( 1 <= N <= 1e5) Array elements ( 0 <= a[i] <= 1)

Output : Minimum number of steps to convert the binary array to all zeros

Sample : [1,0,1] Anwer is 2 ( one possible way is choose the second and third elements, flip them and then choose the first and second elements and flip them)

Can anyone help me solve this?

• +13

By codingkshatriya, history, 7 months ago,

I understand the entire solution except the part " The probability of reaching any particular green square is the same for all but the green square in the last row", I don't get it why the last row is any different?

My understanding of the solution : We take all the squares which are the extension of the diagonal of the removed rectangle part. We have to pass through one of these squares in order to reach the destination and once we get to these squares the probability of reaching destination is 1. Since there cannot be interesction between these paths, we individually find the probability of reaching each of these squares and add them.

Can anyone please explain how the last row is any different?

My code : https://ideone.com/FIGKWD (got WA)

• -4

By codingkshatriya, history, 7 months ago,

Given an Array, find the maximum difference between the sum of elements in odd indices and even indices. To achieve this you can delete any number of elements from the array,after deleting the elements,the array will be re-numbered.

Example : 1 4 8 2 6

Upon deleting 1,4.

New array 8 2 6

Constraints : 1<=N<=1e5

• 0

By codingkshatriya, history, 14 months ago,

I read the editorial,but the last part,the "second way" is not clear,can anyone please help me out? 962E - Byteland, Berland and Disputed Cities

• 0

By codingkshatriya, history, 14 months ago,

MikeMirzayanov Several of the old contests don't have solutions,and even some of the new ones like Codeforces Round #609 (Div. 2) don't have solutions.

Though the contestant is expected to implement the solution on his own after reading the editorial,but at times,when editorial isn't elaborate enough or if the contestant couldn't get the implementation part,a peek at the solution would be really helpful.

Though we can view other's submissions,but most of the code is not written for reading purpose,and the approach may be completely different from that given in editorial.

It maybe inessential for top coders,but for novices like me,it could be a boon.