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University Physics Volume 2

# 14.2Self-Inductance and Inductors

University Physics Volume 214.2 Self-Inductance and Inductors

### Learning Objectives

By the end of this section, you will be able to:

• Correlate the rate of change of current to the induced emf created by that current in the same circuit
• Derive the self-inductance for a cylindrical solenoid
• Derive the self-inductance for a rectangular toroid

Mutual inductance arises when a current in one circuit produces a changing magnetic field that induces an emf in another circuit. But can the magnetic field affect the current in the original circuit that produced the field? The answer is yes, and this is the phenomenon called self-inductance.

### Inductors

Figure 14.5 shows some of the magnetic field lines due to the current in a circular loop of wire. If the current is constant, the magnetic flux through the loop is also constant. However, if the current I were to vary with time—say, immediately after switch S is closed—then the magnetic flux $ΦmΦm$ would correspondingly change. Then Faraday’s law tells us that an emf $εε$ would be induced in the circuit, where

$ε=−dΦmdt.ε=−dΦmdt.$
14.6

Since the magnetic field due to a current-carrying wire is directly proportional to the current, the flux due to this field is also proportional to the current; that is,

$Φm∝I.Φm∝I.$
14.7
Figure 14.5 A magnetic field is produced by the current I in the loop. If I were to vary with time, the magnetic flux through the loop would also vary and an emf would be induced in the loop.

This can also be written as

$Φm=LIΦm=LI$
14.8

where the constant of proportionality L is known as the self-inductance of the wire loop. If the loop has N turns, this equation becomes

$NΦm=LI.NΦm=LI.$
14.9

By convention, the positive sense of the normal to the loop is related to the current by the right-hand rule, so in Figure 14.5, the normal points downward. With this convention, $ΦmΦm$ is positive in Equation 14.9, so L always has a positive value.

For a loop with N turns, $ε=−NdΦm/dt,ε=−NdΦm/dt,$ so the induced emf may be written in terms of the self-inductance as

$ε=−LdIdt.ε=−LdIdt.$
14.10

When using this equation to determine L, it is easiest to ignore the signs of $εanddI/dt,εanddI/dt,$ and calculate L as

$L=|ε||dI/dt|.L=|ε||dI/dt|.$

Since self-inductance is associated with the magnetic field produced by a current, any configuration of conductors possesses self-inductance. For example, besides the wire loop, a long, straight wire has self-inductance, as does a coaxial cable. A coaxial cable is most commonly used by the cable television industry and may also be found connecting to your cable modem. Coaxial cables are used due to their ability to transmit electrical signals with minimal distortions. Coaxial cables have two long cylindrical conductors that possess current and a self-inductance that may have undesirable effects.

A circuit element used to provide self-inductance is known as an inductor. It is represented by the symbol shown in Figure 14.6, which resembles a coil of wire, the basic form of the inductor. Figure 14.7 shows several types of inductors commonly used in circuits.

Figure 14.6 Symbol used to represent an inductor in a circuit.
Figure 14.7 A variety of inductors. Whether they are encapsulated like the top three shown or wound around in a coil like the bottom-most one, each is simply a relatively long coil of wire. (credit: Windell Oskay)

In accordance with Lenz’s law, the negative sign in Equation 14.10 indicates that the induced emf across an inductor always has a polarity that opposes the change in the current. For example, if the current flowing from A to B in Figure 14.8(a) were increasing, the induced emf (represented by the imaginary battery) would have the polarity shown in order to oppose the increase. If the current from A to B were decreasing, then the induced emf would have the opposite polarity, again to oppose the change in current (Figure 14.8(b)). Finally, if the current through the inductor were constant, no emf would be induced in the coil.

Figure 14.8 The induced emf across an inductor always acts to oppose the change in the current. This can be visualized as an imaginary battery causing current to flow to oppose the change in (a) and reinforce the change in (b).

One common application of inductance is to allow traffic signals to sense when vehicles are waiting at a street intersection. An electrical circuit with an inductor is placed in the road underneath the location where a waiting car will stop. The body of the car increases the inductance and the circuit changes, sending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and a receiver. The pulsed signal from the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path (Figure 14.9). Metal detectors can be adjusted for sensitivity and can also sense the presence of metal on a person.

Figure 14.9 The familiar security gate at an airport not only detects metals, but can also indicate their approximate height above the floor. (credit: “Alexbuirds”/Wikimedia Commons)

Large induced voltages are found in camera flashes. Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. Recall from Oscillations on oscillations that “oscillation” is defined as the fluctuation of a quantity, or repeated regular fluctuations of a quantity, between two extreme values around an average value. Also recall (from Electromagnetic Induction on electromagnetic induction) that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil. The oscillator system does this many times as the battery voltage is boosted to over 1000 volts. (You may hear the high-pitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash.

### Example 14.2

#### Self-Inductance of a Coil

An induced emf of 20 mV is measured across a coil of 50 closely wound turns while the current through it increases uniformly from 0.0 to 5.0 A in 0.10 s. (a) What is the self-inductance of the coil? (b) With the current at 5.0 A, what is the flux through each turn of the coil?

#### Strategy

Both parts of this problem give all the information needed to solve for the self-inductance in part (a) or the flux through each turn of the coil in part (b). The equations needed are Equation 14.10 for part (a) and Equation 14.9 for part (b).

#### Solution

1. Ignoring the negative sign and using magnitudes, we have, from Equation 14.10,
$L=εdI/dt=20mV5.0A/0.10s=4.0×10−4H.L=εdI/dt=20mV5.0A/0.10s=4.0×10−4H.$
2. From Equation 14.9, the flux is given in terms of the current by $Φm=LI/N,Φm=LI/N,$ so
$Φm=(4.0×10−2H)(5.0A)50turns=4.0×10−5Wb.Φm=(4.0×10−2H)(5.0A)50turns=4.0×10−5Wb.$

#### Significance

The self-inductance and flux calculated in parts (a) and (b) are typical values for coils found in contemporary devices. If the current is not changing over time, the flux is not changing in time, so no emf is induced.

Check Your Understanding Current flows through the inductor in Figure 14.8 from B to A instead of from A to B as shown. Is the current increasing or decreasing in order to produce the emf given in diagram (a)? In diagram (b)?

Check Your Understanding A changing current induces an emf of 10 V across a 0.25-H inductor. What is the rate at which the current is changing?

A good approach for calculating the self-inductance of an inductor consists of the following steps:

### Problem-Solving Strategy

#### Self-Inductance

1. Assume a current I is flowing through the inductor.
2. Determine the magnetic field $B→B→$ produced by the current. If there is appropriate symmetry, you may be able to do this with Ampère’s law.
3. Obtain the magnetic flux, $Φm.Φm.$
4. With the flux known, the self-inductance can be found from Equation 14.9, $L=NΦm/IL=NΦm/I$.

To demonstrate this procedure, we now calculate the self-inductances of two inductors.

### Cylindrical Solenoid

Consider a long, cylindrical solenoid with length l, cross-sectional area A, and N turns of wire. We assume that the length of the solenoid is so much larger than its diameter that we can take the magnetic field to be $B=μ0nIB=μ0nI$ throughout the interior of the solenoid, that is, we ignore end effects in the solenoid. With a current I flowing through the coils, the magnetic field produced within the solenoid is

$B=μ0(Nl)I,B=μ0(Nl)I,$
14.11

so the magnetic flux through one turn is

$Φm=BA=μ0NAlI.Φm=BA=μ0NAlI.$
14.12

Using Equation 14.9, we find for the self-inductance of the solenoid,

$Lsolenoid=NΦmI=μ0N2Al.Lsolenoid=NΦmI=μ0N2Al.$
14.13

If $n=N/ln=N/l$ is the number of turns per unit length of the solenoid, we may write Equation 14.13 as

$L=μ0(Nl)2Al=μ0n2Al=μ0n2(V),L=μ0(Nl)2Al=μ0n2Al=μ0n2(V),$
14.14

where $V=AlV=Al$ is the volume of the solenoid. Notice that the self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current. This is true for inductors in general.

### Rectangular Toroid

A toroid with a rectangular cross-section is shown in Figure 14.10. The inner and outer radii of the toroid are $R1andR2,andhR1andR2,andh$ is the height of the toroid. Applying Ampère’s law in the same manner as we did in Example 13.8 for a toroid with a circular cross-section, we find the magnetic field inside a rectangular toroid is also given by

$B=μ0NI2πr,B=μ0NI2πr,$
14.15

where r is the distance from the central axis of the toroid. Because the field changes within the toroid, we must calculate the flux by integrating over the toroid’s cross-section. Using the infinitesimal cross-sectional area element $da=hdrda=hdr$ shown in Figure 14.10, we obtain

$Φm=∫Bda=∫R1R2(μ0NI2πr)(hdr)=μ0NhI2πlnR2R1.Φm=∫Bda=∫R1R2(μ0NI2πr)(hdr)=μ0NhI2πlnR2R1.$
14.16
Figure 14.10 Calculating the self-inductance of a rectangular toroid.

Now from Equation 14.16, we obtain for the self-inductance of a rectangular toroid

$L=NΦmI=μ0N2h2πlnR2R1.L=NΦmI=μ0N2h2πlnR2R1.$
14.17

As expected, the self-inductance is a constant determined by only the physical properties of the toroid.

Check Your Understanding (a) Calculate the self-inductance of a solenoid that is tightly wound with wire of diameter 0.10 cm, has a cross-sectional area of $0.90cm20.90cm2$, and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid?

Check Your Understanding (a) What is the magnetic flux through one turn of a solenoid of self-inductance $8.0×10−5H8.0×10−5H$ when a current of 3.0 A flows through it? Assume that the solenoid has 1000 turns and is wound from wire of diameter 1.0 mm. (b) What is the cross-sectional area of the solenoid?

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# Parsec

This article is about the unit of length. In computer programming, Parsec [1] (http://research.microsoft.com/~emeijer/Papers/Parsec.pdf) is an XML syntax analyzer, like Lark, or a parsing library for the Haskell programming language [2] (http://www.cs.uu.nl/~daan/parsec.html). There is also a FLOSS computer game named Parsec [3] (http://openparsec.sourceforge.net/).

The parsec (symbol pc) is a unit of length used in astronomy. It stands for "parallax of one arc second", and is approximately 19,131,554,073,600 (19 trillion) miles.

It is based on the method of trigonometric parallax, the most ancient and standard method of determining stellar distances. The angle subtended (http://www.answers.com/subtended) at a star by the mean radius of the Earth's orbit, around the Sun, is called the parallax. The parsec is defined to be the distance from the Earth of a star that has a parallax of 1 arcsecond. Alternatively, the parsec is the distance at which two objects, separated by 1 astronomical unit, appear to be separated by an angle of 1 arcsecond. It is, therefore, approximately:

[itex]\frac{360\cdot60\cdot60}{2\pi}[itex] AU = 206,265 AU = 3.08568×1016 m = 30.8568 Pm (petametres) = 3.2616 ly (light-years).

See 1 E16 m for a list of comparable lengths and scientific notation for an explanation of the notation.

Astronomers usually express distances to astronomical objects in units of parsecs, instead of light-years. This is both for historical reasons, and because it avoids including conversion factors such as the length of an astronomical unit, which was not known to a high level of precision until the latter part of the 20th century. The first direct measurements of an object at interstellar distances (of the star 61 Cygni, by Friedrich Wilhelm Bessel in 1838) were done using trigonometry using the width of the Earth's orbit as a baseline. The parsec follows naturally from this method, since the distance (in parsecs) is simply the reciprocal of the parallax angle (in arcseconds).

Though it had probably been used before, the term parsec was first mentioned in an astronomical publication in 1913, when Frank Watson Dyson expressed his concern for the need of a name for that unit of distance. He himself proposed the name astron, while Carl Charlier had suggested siriometer. However, Herbert Hall Turner's suggestion, parsec, was eventually adopted.

There is no star whose parallax is 1 arcsecond. The greater the parallax of the star the closer it is to the Earth, and the smaller its distance in parsecs. Therefore the closest star to the Earth will have the largest measured parallax. This belongs to the star Proxima Centauri, with a parallax of 0.772 arcseconds, and thus lying approximately 1.29 parsecs, or 4.22 light-years, away from us.

The measurement of distances of celestial bodies from the Earth in parsecs is a key aspect of astrometry, the science of making positional measurements of celestial bodies.

Because of the extremely small scale of parallactic shifts, ground-based parallax methods provide reliable measurements of stellar distances of no more than 325 light-years, or about 100 parsecs, corresponding to parallaxes of no less than 1/100 of 1 arcsecond, or 10 mas (1 mas or milliarcsecond = 1/1000 arcsecond).

Between 1989 and 1993 the Hipparcos satellite, launched by the European Space Agency (ESA) in 1989, measured parallaxes for about 100,000 stars, with a precision of about 0.97 mas, and obtained accurate measurements for stellar distances of around 1000 pc.

NASA's FAME satellite was due to be launched in 2004, to measure parallaxes for about 40 million stars with sufficient precision to measure stellar distances of up to 2000 pc. However, the mission's funding was withdrawn by NASA in January 2002.

The ESA's GAIA satellite, due to be launched in mid-2012, will be of sufficiently high astrometric precision to measure stellar distances to within 10% accuracy as far as the galactic centre about 8 kpc away in the Sagittarius constellation.

 Contents

## Distances in parsecs

One kiloparsec, abbreviated kpc, is one thousand parsecs. One megaparsec, abbreviated Mpc, is one million parsecs.

## How to calculate the value of a parsec

Missing image
Parsec.png
Image:Parsec.png

In the diagram above (not to scale), S represents the Sun, and E the Earth at one point in its orbit. D is an object at a distance of one parsec from the Sun. By definition, the angle D is one arcsecond and the distance ES is one astronomical unit. By trigonometry, the distance SD is

[itex]SD = {{ES} \over {\tan 1''}} = 206 265 \mbox{ au}[itex]

One astronomical unit is equal to approximately 1.49598×108 km, so

[itex] 1 \mbox{ pc} = 206 265 \times 1.49598 \times 10^{11} \mbox{ m} = 3.08568 \times 10^{16} \mbox{ m} \,[itex]

The attoparsec (atto being the prefix indicating 10-18) is a unit humorously used by some hackers. It is approximately 3.1 centimetres - just a little over an inch. See also microfortnight.

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# Tag Info

17

Method 1: Foucault Pendulum As user Rob asks, what is wrong with a short Foucault pendulum? There is a problem, but it can be overcome inexpensively, but with some DIY effort at home. The problem is that, by dint of imperfections in the suspending fibre and bob, no pendulum will swing in a plane even if the Earth were not rotating. Instead, the bob will ...

15

Ok, here is my (hopefully rigorous) demonstration of the origin of these forces here, from first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy! Angular velocity vector Let us start with the principal equation defining angular velocity in three dimensions, \dot{\vec{r}} = \vec{\omega} \times ... 14 The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random. If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials. The Coriolis force is ... 10 The answer: the ball appears to be deflected ~10 cm. The calculation: For simplicity, say we tee off at the north pole. The effects are a bit weaker at more typical locations, you multiply by sin(latitude) = 0.64 for a 40 degree (central california or washington DC) latitude. The Coriolis effect exists because the Earth rotates while the ball is in ... 8 The Coriolis acceleration goes like -2\omega \times v, which for the sake of an order of magnitude estimate we can take to be a\sim \omega v. But in order to get an observable effect, we don't just need an acceleration, we need a difference in acceleration between the two ends of the tub, which are separated by some distance L\sim 1 m. The ... 5 Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) ... 4 Firstly, is that correct? Yes your intuitive understanding for this part of the Coriolis effect is correct. The second part, that is, why wind in the East direction is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation: F = \frac{mv^2}{r} If we re-arrange the above equation, we can find ... 4 An alternate derivation for a due-north ball, ignoring the diminishing effect of latitude, that confirms Kevin's order of magnitude: Acceleration due to the Coriolis effect: a_C = -2 \, {\Omega \times v} \Omega = 2 \pi/day v = 45 m/s a_C = -0.00654498469 m/s^2 Horizontal displacement d is given by d = 1/2a_C\,t^2  Using earlier estimate of ... 4 Coriolis force is not an actual force, but rather an effect observed in rotating frame of reference. The light path is not actually bent, so it doesn't matter that the photon has no mass, the Earth's rotation will have an affect on the photon's apparent path. This does not contradict your calculation of F_{Coriolis}=0, because you have to put this force ... 4 Indirectly, yes. It is called a gyro compass. When you constrain a gyroscope to spin about an axis in the horizontal plane it will experience a Coriolis like force unless it is pointing due North-South - if it isn't it will slowly align, then stay there, so you can use this as a compass - in fact it is widely used as it doesn't suffer from magnetic ... 3 Note that Fitzpatrick states towards the beginning, The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, \Omega can be treated as a small parameter Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: ... 3 The Coriolis force \vec F_{\text{coriolis}} = -2m \, \vec \omega \times \vec v only depends on velocity. The centrifugal force \vec F_{\text{centrifugal}} = -m \, \vec \omega \times (\vec \omega \times \vec r) only depends on position. Finally, if the object is not rotating uniformly (\dot {\vec \omega} \ne 0), then yet another fictitious force comes ... 3 could the Coriolis effect on snowing be so dramatic...? No. The Coriolis effect is only noticeable for objects traveling long distances with respect to Earth's surface for significant periods of time. For example, a ballistic missile fired hundreds of miles or a hurricaine that is hundreds of miles in diameter and lasts for days. Across the street is ... 3 The Coriolis force on the equator indeed does point outwards, if you are moving west to east. This is not the same as the centrifugal force, because the centrifugal force is present always - even if you are not moving. But when you move (west to east), there is an additional force on top of the centrifugal force - the Coriolis force. If you travel east to ... 3 Well, there is a partial yes that is a direct result of the Coriolis force: If you go up in a hot air balloon, you will be subject to various winds which will move you. And these winds are a result of the Earth spinning. In principle you should be able to navigate to most places on the globe by choosing height etc. in reality it is much too complex to do ... 3 The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface. Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even ... 2 I am not particularly an expert either, but my understanding is that shuttle flight is a very active process compared to ballistic motion, so any effects the Coriolis effect might have can just as well be considered as additional errors in the trajectory, which is being adjusted. There's an active feedback loop at work: "observe flightpath -> identify ... 2 You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the ... 2 Rockets lean as they climb on purpose, in order to obtain the high orbital velocities needed to stay in space once they get there. For a nice explanation, see Orbital Speed at xkcd what-if, but the gist is the following. Being in orbit means going so fast that the Earth begins to curve away from you as you fall down towards it. The classic image to keep in ... 2 The Coriolis effect cannot possibly account for this. Other factors (e.g the shape of the basin and initial conditions for the water flow) should have a comparatively huger effect. This is intuitively obvious---if you are not convinced you should perform an experiment. Could be fun! You could also estimate characteristic scales (via dimensional analysis) ... 2 Well, first we know that the Coriolis force acting on the scale of a toilet is going to be a pretty small force. So the question can generally be tackled in the following way: Estimate the magnitude of the Coriolis force on the toilet water. Use this to estimate the magnitude of effect of the Coriolis force on the toilet water spin. Enumerate the ... 2 At any given moment, the pendulum is swinging in a certain plane. The driving force should be within this plane. If the earth weren't rotating, then such a force could never cause the plane of swing to rotate about a vertical axis, since by symmetry there would be no preferred direction for the rotation. 2 The rotation of your coordinate system causes the Coriolis effect. Things move in straight lines, but if your coordinate system is rotating, then the straight lines look curved from the perspective of your coordinate system. If you want to insist that objects move in straight lines in your coordinate system, then you must invent a fictitious reason why ... 2 Surprisingly, the answer is that yes you do, though the effect is very small. To see this consider the following (highly exaggerated) diagram of the lift shaft: The Earth rotates at a constant angular velocity of one rotation every 24 hours (\omega = 7.27 \times 10^{-5} radians/sec). The tangential velocity of a part of the lift shaft at a distance r ... 2 Look in Wikipedia http://en.wikipedia.org/wiki/Coriolis_effect. For understanding intuitively the Coriolis force effect, assume an object moving according to a static (inertial) frame of reference, in the plane perpendicular to the rotation axis, and along the radius, In the rotating frame, see the animation in Wikipedia, the Coriolis force imposes an ... 2 Ahh, Richard Fitzpatrick. Great guy. Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" 2\Omega \sin\lambda t is small (note that the t is not within the sin function!), you get the first expressions, e.g.\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$and then ... 1 Disclaimer: borrowed from wikipedia with minor modifications. The acceleration is the second time derivative of position,$$ \mathbf{a}_{\mathrm{i}} \ \stackrel{\mathrm{ }}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}}\right)_{\mathrm{i}} = \left( \frac{d\mathbf{v}}{dt} \right)_{\mathrm{i}} = \left[ \left( \frac{d}{dt} \right)_{\mathrm{r}} + ...

1

You are right: the coriolis effect in the equator is converted to a kind of centrifugal force. But in many situations only the horizontal Coriolis forces are accountes, because in the vertical one the gravity is so much greater. The vertical coriolis and the centrifugal force may point parallel, but they are still different forces. Let's think of a ...

1

If you project the velocity of a water parcel traveling up the channel into radial coordinates in both the inertial frame (space frame) and the rotating frame of reference (earth frame) then I think the necessary effects will become apparent. In the earth's frame, the velocity is (use a ' to symbolize the rotating frame): \vec{v} = 0\cdot\hat{r}' + ...

1

Ok, let's see if I get this right. The circumference of the equator is 24902 miles. The Earth rotates 360° in 86,164.098 seconds. This means the surface of the Earth moves 1525.96 ft/sec at the equator. Adding 1000 ft to the height give a circumference of 24903 miles. This means the lateral speed is 1526.03 ft/sec; a difference of about 0.87 inches/sec. It ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Sigma Function of Non-Square Semiprime/Examples

## Examples of $\sigma$ of Non-Square Semiprime

Let $\sigma: \Z_{>0} \to \Z_{>0}$ denote the $\sigma$ function: the sum of all the positive integer divisors of $n$.

### $\sigma$ of $14$

$\sigma \left({14}\right) = 24$

### $\sigma$ of $15$

$\sigma \left({15}\right) = 24$

### $\sigma$ of $22$

$\map \sigma {22} = 36$

### $\sigma$ of $26$

$\sigma \left({26}\right) = 42$

### $\sigma$ of $33$

$\sigma \left({33}\right) = 48$

### $\sigma$ of $35$

$\sigma \left({35}\right) = 48$

### $\sigma$ of $38$

$\sigma \left({38}\right) = 60$

### $\sigma$ of $58$

$\sigma \left({58}\right) = 90$

### $\sigma$ of $62$

$\sigma \left({62}\right) = 96$

### $\sigma$ of $65$

$\sigma \left({65}\right) = 84$

### $\sigma$ of $87$

$\sigma \left({87}\right) = 120$

### $\sigma$ of $94$

$\map \sigma {94} = 144$

### $\sigma$ of $115$

$\map \sigma {115} = 144$

### $\sigma$ of $206$

$\map \sigma {206} = 312$

### $\sigma$ of $362$

$\sigma \left({362}\right) = 546$

### $\sigma$ of $1257$

$\sigma \left({1257}\right) = 1680$


text:
sayang naman ! kla ko pa man din gets ko na .. kailangan ko pa pala mag-review mabuti ! by the way, thanks for this chance to have a quiz on Biology .. This one informed me to review more ..

Hopefully, I’ll pass the UPCAT . :)

by akuhxiehluvie08 on UPCAT Sample Questions: Biology

# UPCAT Schedule for AY 2012-2013 admissions

Newsflash for students currently in third year high school. We now have word on the next  UPCAT application and exam schedule. Call it UPCAT 2011 (for it would be administered this year) or call it UPCAT 2012 – call it what you will, but we’re referring to the UPCAT Application for Freshman Admission for the First Semester of the 2012-2013 academic year.

• June 17, 2011 (Friday) – if you’re from a MM school
• June 24, 2011 (Friday)- if you’re from a non-MM school

The UPCAT  for the first semester of the  2012-2013 freshmen intake year will be administered on the following dates:

• August 6, 2011 (Saturday)
• August 7, 2011 (Sunday)

As usual, the UPCAT will be held in various testing centers nationwide. There will be two testing schedules each day: morning and afternoon. Every UPCAT taker will be assigned to only one UPCAT testing schedule and will be assigned only to one UPCAT testing date. In other words, it’s a one-shot deal so take your UPCAT in your testing date and testing schedule.

Important: You don’t have to be in the top 10 percent (or top-whatever-percent) of your class to take the UPCAT. The UPCAT is open to everyone (as long as specific non-grade eligibility requirements are satisfied, of course).

Source: UP Academic Calendar for AY 2011-2012

# UPCAT 2010: SY2011-2012 Schedule Updates

It’s probably around May that parents and schools alike begin preparations for the start of the upcoming school year in June.  The University of the Philippines, however, is also interested in preparing for the next school year, so much so that has its variables already set up—all that remains is to execute its plan.

The kids may still be on vacation mode, yet it’s not too early to brief them on the steps leading up to UPCAT 2011.  There are some formalities they’ll need to deal with before they take the entrance test, as detailed below.

UPCAT Forms

Your first order of business as an UPCAT applicant is to download three important forms and familiarize yourself with their content.  Please note that all three documents are free for distribution and photocopying.

The first one is the general information form that is available here.  Like its name suggests, the form is a kind of newbie’s guide on the U.P. admission system, the UPCAT, scholarships and other relevant topics.

The second form is the 2010 edition of the UPCAT application form for incoming freshmen, also known as the Personal Data Sheet.  Look through the PDF file, fill it up completely and keep the reminders in mind.

Finally, the High School Records Form also needs to be filled up, but not by you—your high school principal (or an equivalent school authority) will do the honors.  U.P. officials need to see how well you did quantitatively in secondary school, as this will influence their decision to accept you or not.

Kindly mark the important dates on your calendar as well.  For starters, May 25, 2010 marks the beginning of the application period for next year’s UPCAT.  The deadline for turning everything in to U.P. authorities will depend on whether you’re from a Metro Manila-based school or not.

NCR students have until June 18 2010 to submit all the requirements; non-Metro Manila students have June 25 2010 as their deadliest deadline.  Fortunately, the university has a requirement submission system in place to take some of the stress off of applicants, especially those from far away places.

Besides submitting your documents in person, you may also opt for an authorized school representative to file your application and pick up your test permit on your behalf.  Alternatively, you may also mail your permit to the university and wait for your test permit to be delivered to your school in the same fashion.

The general information sheet should satiate your thirst for U.P. and UPCAT-related information, but if ever you have questions not covered by the form, kindly consult this webpage and you may find the answers you’re looking for.

Sometimes, it’s best to get your updates straight from the horse’s mouth.  In that case, you can request for more information from the folks at the U.P. Office of Admissions.  Just dial (632) 981-8500 and input locals 3827, 3828 or 3831.  They may also be reached via e-mail at oadms@up.edu.ph.

Source:

http://upcat.up.edu.ph/index.html

# UPCAT Schedule and Important Dates

Hello boys and girls!  UPCAT is right around the corner.  We sincerely hope you are prepared to face the grueling (depends on your point of view), five-hour (this is fact) exam.

### UPCAT Schedule:  August 1 to 2, 2009

The UP College Admission Test is set to take place on August 1 to 2, 2009 barring unforeseen issues or problems.  For that matter, I don’t know if UPCAT has ever been postponed.  I will have to look in on that.  Better yet, if you have information you wish to share, please hit me; I’m open.

June 23, 2009 – Deadline for Filing of UPCAT Application if you go to a school in Metro Manila
June 30, 2009 – Deadline for Filing of UPCAT Application if you go to a non-MM school

### UPCAT Application Form

If you have never heard of UPCAT till now (where on Earth are you living, anyway?), you still have a chance to get your application in (as long as you have courier service in your town and your school officials prove to be cooperative).  Simply download the forms and the guidelines listed below (these PDF documents are sourced directly from the UP website), print them, fill them out, and get them over to the Office of Admissions by the above stated dates.

If you need a primer on the UPCAT, donwload the document specified below.  Like the other two, this comes from the UP website.  Unlike the other two, this is an MS Word document.  You’ll be asked to either “Open” or “Save” the document in your computer; the document will not open in your browser.

Now, if you’re wondering how much UPCAT costs, we have just the information you need.  If your annual family income (that’s gross) is P100,000 and above, you have to pay P450.  If you are taking the UPCAT in the Middle East, your UPCAT Fee is $100. Foreigners living abroad have to pay$50.


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# 3.5: The Friction Velocity or Shear Velocity u*

The term friction velocity comes from the fact that $$\ \sqrt{( \boldsymbol{\tau}_{\mathrm{w}} / \boldsymbol{\rho}_{\mathrm{l}}}$$ has the same unit as velocity and it has something to do with the wall friction force. The wall shear stress τw is often represented by friction velocity u*, defined by:

$\ \mathrm{u}_{*}=\sqrt{\frac{\tau_{\mathrm{w}}}{\rho_{\mathrm{l}}}}=\sqrt{\frac{\lambda_{1}}{8}} \cdot \mathrm{v}_{\mathrm{ls}}$

This page titled 3.5: The Friction Velocity or Shear Velocity u* is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Sape A. Miedema (TU Delft Open Textbooks) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.