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Lecture Comments (6)

3 answers

Last reply by: Professor Dan Fullerton
Sat Jul 9, 2016 6:15 AM

Post by Peter Ke on July 7 at 06:28:39 PM

For problem 45, how does rotating the spring about a diameter causes a change in flux?

1 answer

Last reply by: Professor Dan Fullerton
Sun May 5, 2013 5:40 AM

Post by Saki Amagai on May 4, 2013

For #65, shouldn't the area of the wire also increase, if the radius increases? cuz a=pi r^2

AP Practice Exam: Multiple Choice, Part 2

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Problem 36 0:18
  • Problem 37 0:42
  • Problem 38 2:13
  • Problem 39 4:10
  • Problem 40 4:47
  • Problem 41 5:52
  • Problem 42 7:22
  • Problem 43 8:16
  • Problem 44 9:11
  • Problem 45 9:42
  • Problem 46 10:56
  • Problem 47 12:03
  • Problem 48 13:58
  • Problem 49 14:49
  • Problem 50 15:36
  • Problem 51 15:51
  • Problem 52 17:18
  • Problem 53 17:59
  • Problem 54 19:10
  • Problem 55 21:27
  • Problem 56 22:40
  • Problem 57 23:19
  • Problem 58 23:50
  • Problem 59 25:35
  • Problem 60 26:45
  • Problem 61 27:57
  • Problem 62 28:32
  • Problem 63 29:52
  • Problem 64 30:27
  • Problem 65 31:27
  • Problem 66 32:22
  • Problem 67 33:18
  • Problem 68 35:21
  • Problem 69 36:27
  • Problem 70 36:46

Transcription: AP Practice Exam: Multiple Choice, Part 2

Hi everyone! I am Dan Fullerton and I would like to welcome you back to

In this lesson we are going to continue our work through the 1998 AP Physics B examination for practice.0004

Starting at multiple choice question Number 36, we will continue through the end of the multiple choice section.0011

Let us dive right in.0017

As I look at Number 36 here, we are talking about number of un-decayed atoms as a function of time, and what we can see here is if we are looking for the half-life -- well that is the point where the number of un-decayed atoms has cut in half...0019 on the graph, that should be pretty obvious to see that that happens right at point A, which is choice A.0033

Moving on to 37, we are talking about photons of light that have a frequency and momentum, but what about the momentum if they have a frequency of 2 times their initial frequency?0042

Well to do that, let us start by looking at some of the relationships involved. 0055

We will start with our wave equation, V = F(λ), but we also know that wavelength = H/momentum.0058

I could write this as V = FH/P, and now what I am going to do is I am going to try and get all the constants on one side.0068

Since it is a photon, (V) is going to be constant, and that is a speed and we also have this H constant, so I am going to write that V/H = F/P.0077

Knowing that this is constant, we know that F/P must remain the same, so if V/H = F/P -- well let us take a look here -- V/H which is C/H (the speed of light) must equal F/P.0090

You can ignore this part now, C/H = F/P.0109

What happens then if we have a frequency of 2F?0112

Well if we put a frequency of 2 here and we want this to remain constant, we must have 2 down there.0117

So what do we do? We doubled the momentum -- answer choice A.0123

All right. Number 38 -- This problem is a little more involved, there is a little more to it than it looks like at first glance.0133

So we have the block that has a 3 kg mass and we know what its stretch is when it is at equilibrium.0139

First thing, let us use that to find its spring constant.0145

For the 3 kg mass that is on there, we know that the spring constant is the force divided by the displacement, so that is going to be mg/displacement, or 3 kg × 10 m/s2 over its displacement which is...0149

...let us leave it in centimeters to make life easy -- 12 cm, so that will be 30/12 N/cm.0166

Now we are going to replace that by a 4 kg block and when we do that, we release it from its highest position, and it is going to fall and they want to know how far it will fall before it reverses, so it is going to fall to the equilibrium position and then another same distance past that until it stops.0175

To figure that out, let us first figure out that equilibrium position -- X = F/K, so that is going to be our force (4 kg) × 10 m/s2/K (30/12), so I could re-write that as 4 × 10 = 40/30...0195

...and I will put the 12 in the numerator and with just a little bit of math here, we have 4 × 12 = 48/3, and that is going to be 16 cm, so that is the new equilibrium position.0216

If it starts at the highest point, it falls 16 cm to the equilibrium position and then another 16 cm before it stops and will pass back up, its total must be...0228

...How far does it fall? 32 cm and that answer would be choice D.0239

Number 39 -- We know an object has a weight (W) when it is on the surface of the planet at a radius (r), so it starts with weight (W).0250

What happens to the gravitational force when you quadruple that distance, when it goes from r to 4r?0258

Well remember, gravity is an inverse square law. 0265

If you increase the distance, the force must be going down, and if you quadruple the distance, multiply it by 4 -- well it has squared the effect so that is 16, so you will get 1/16 the initial weight -- Answer E.0268

Number 40 -- We are trying to find the kinetic energy of a satellite that orbits the earth in a circular orbit of radius (r).0286

Well if it is moving in a circular orbit, that must be caused by a centripetal force which is mv2/R and what is causing that centripetal force? 0296

Well the force of gravity, which is G times the mass of our satellite and the mass of our planet divided by r2.0307

I am going to solve to get mv2 by itself, multiply both sides by R and I get that mv2 is going to be equal to GmM/R.0316

Why did I solve for mv2? Well I know kinetic energy is just 1/2 mv2.0328

So it is going to be 1/2 × mv2 (GmM/R), so our correct answer there must be choice B.0336

Number 41 -- We have a couple of objects that are moving parallel to the x-axis and then they are going to collide and we are trying to find out the y-component of the velocity of one of these objects.0354

Well first thing I think of when I hear collision is conservation of momentum, and in this case we only have to worry about the conservation of momentum in the y-direction.0366

We do not have to worry about the (x) at all because all we are asked about is the y-component of the velocity. 0375

We know that the initial momentum in the y-direction must equal the final momentum in the y-direction, and since they are both traveling purely horizontally to start the problem, the initial momentum in the y-direction must be 0.0380

Therefore 0 must be equal to -- well the final momentum in the y-direction, we have the mass of the .2 kg object and it is traveling with a velocity of 1 up, so that must be added to the momentum of the second object, which has a mass of 0.1 kg and some unknown velocity.0393

That is a pretty easy equation to solve: 0 = .2 + .1 V, which implies then that .1 V = -.2 or therefore V = -2 m/s, where all that negative means is it is downward, so the correct answer must be A, 2 m/s downward.0416

Number 42 -- We have a beam of white light incident on a prism and it talks about what is going on here, producing a spectrum and as that is filled with air, you get one effect. 0442

Then what happens -- if you fill it in with water, which has an index of refraction of 1.3 -- well you are still going to get a spectrum. 0454

You are still going from a lower index to a higher index into the prism and a higher to a lower index as you leave the prism.0462

But as you have that smaller change in the index of refraction, you are going to get less refraction; you are going to get less dispersion, and therefore you are going to get a compressed spectrum.0469

It is going to take a lot less space than it did before in order to show all of those colors. 0478

You are going to get less separation between the red end of the spectrum and the violet end of the spectrum, so our correct answer must be E.0484

Numbers 43 and 44 have to do with these distance-time graphs and as I look at these distance-time graphs, the first thing I notice is the first graph has a constant slope.0496

That means it must have a constant velocity, so constant velocity for graph 1.0507

Graph 2 has 0 slope, therefore that must have a 0 velocity.0514

Graph 3 has an increasing slope, therefore it must have an increasing velocity.0520

In this case you have some acceleration and acceleration does not equal 0.0526

Now let us look at the questions.0531

Talking about the magnitude of the momentum is increasing in which case?0533

Well the mass is not increasing so the only way momentum could increase is if velocity is increasing, so that rules out 1 and 2 and the answer must be 3 only, so for answer 43, the correct answer is B.0537

If we go to 44, the sum of the forces is 0, well when the sum of the forces is 0, that means the acceleration must be 0 by Newton's Second Law.0551

The only place where the acceleration is 0, is 1, constant velocity, and 2, no velocity, so the answer there must be 1 and 2, or answer C.0561

Number 45 -- You have a metal spring inside a uniform magnetic field, and we are looking for a condition that will not cause a current to be induced and if you do not want a current, we cannot have any induced EMF.0581

So which of those is going to have some effect where you are not going to have a change in flux because remember the EMF is equal to the opposite of the change in magnetic flux divided by the time interval. 0594

If you were to change the magnitude of the magnetic field, you are going to change the flux -- it cannot be A.0606

Increase the diameter of the circle? -- Well if you do that, you are changing the flux, you have more area.0611

Rotate the spring about the diameter? -- You are going to change the flux as you rotate that.0616

Moving the spring parallel to the magnetic field? -- Just moving the whole thing parallel to the magnetic field is not going to do anything to the flux.0622

I am thinking D must be our best answer, but let us check.0629

E, Moving the spring in and out of the magnetic field? -- No, as you do that you are going to change the flux, so if you want to have no current induced, you cannot have any change in flux, therefore that must be 45 (D).0633

All right. Let us take a look at Number 46 here.0653

We are looking again for 46 and 47 at a magnetic field on a proton beam.0660

It says "Of the following, what is the best estimate of the work done?"0667

Well let us draw a diagram quickly.0671

If there is our magnetic field and we are going to have protons moving in a circle -- let us just draw our circle there and as they do that, the force required for it to move in a circle must be toward the center of the circle.0673

But its velocity is at a right angle to that, and if you recall work is F(δ)R cos(θ), but in this case, θ equals 90 degrees so cos(θ) equals 0, therefore the work done must be 0 -- answer A.0695

A magnetic field cannot do any work on a moving charge.0714

It can change its direction; it can accelerate it but it cannot do work on it.0718

Number 47 -- What is the best estimate of the speed of the proton as it is moving in that circle?0723

Well the centripetal force must be caused by the magnetic force which is QVB sin(θ) and that must be equal to (M) times the acceleration (mv2/R), because it is moving in a circle. 0730

Therefore, we know θ = 90 degrees, so sin(θ) = 1, so we could say that QVB = mv2/R.0746

A little simplification here -- we can divide velocity out of both sides -- solving for V then, I could say that V = QBr/M and substituting in my values, Q (1.6 × 10-19), B (0.1 tesla), radius here (0.1 m) and the mass of our proton (1.67 × 10-27 kg).0760

That looks awfully tough given that you do not have a calculator for the multiple choice part of the question.0794

Let us just estimate the order of magnitude.0799

Right away let us say 1.6/1.67 and that is pretty close to 1.0801

Now we have 10-19 × 10-2, and that is going to be 10-21/10-27, so I would say that that is going to be around 106m/s then which will give us answer C.0807

You do not really have anything else that is really close to that value, so you could figure that one out without having to go to the calculator.0826

Checking out number 48 here -- We have a loop of wire in the plane of a page perpendicular to a magnetic field (B) out of the page.0838

And it looks like this is a Lenz's Law problem -- trying to figure out the direction of the induced current in the wire. 0847

If the magnitude of the magnetic field is decreasing and it is coming out, the induced current wants to create a magnetic field to oppose that, so it is also going out, because that is decreasing; it wants to keep it the same.0855

The right-hand rule, as I then wrap the fingers of my right hand around to the direction of that induced magnetic field, I am going to find that this is counterclockwise around the loop or choice D -- a Lenz's Law right-handed rule problem.0869

Taking a look at Number 49 -- We have an object making ripples or waves of frequency 20 Hz and it tells us the speed and if the source is moving to the right with some speed, at which of those labeled points will you get a highest possible frequency?0889

Well a shift in frequency -- we are talking about the Doppler Effect -- and you get the highest shift in frequency when the source and observer are moving toward each other.0907

That must be point C, which you could also see as you look at the diagram -- you can see that C is going to pick up those wave fronts at a higher frequency than the place behind the point.0914

(A) would be a lower frequency, for example, but C must be our best answer for the highest frequency.0926

And Number 50 -- We have an object in front of a plane mirror.0936

Well what do you see when you have a plane mirror? You see the exact inverse image, so what are you going to have there by the Law of Reflection? -- D.0940

All right. Number 51 -- I have sound waves of some wavelength incident on two slits in a box with non-reflecting walls and we are going to try and find -- well it tells us the first order maximum, where it is from the central maximum, and we want to find the distance between the slits.0952

It is a double slit diffraction problem, therefore, D sin(θ) = M(λ), and what we are really looking for here is D.0971

So let us solve that for D = M(λ)/sin(θ), but as you look at your diagram there, sin(θ) is opposite/hypotenuse.0981

The opposite side is 3 m, the hypotenuse is 5 m, so sin(θ) is 3/5, so as I plug in my values here I would say that D is equal to -- well, M = 1, our wavelength = .12 and sin(θ) 3/5...0996 3/5 down here or I can put that 5 up there -- 5 × .12 = .6/3 or .2 m -- Choice D.1018

Number 52 is another PV diagram where we have an ideal gas that corresponds to .1 on the graph and what we are looking for is which of the following relationships are true?1037

Is T1 less than T3? -- Well that cannot be the case because it is isothermal, constant temperature, so it is not choice A.1049

B -- T1 is less than T2 -- absolutely, as you go up and to the right, you see higher temperatures. 1057

T2 is greater than T1, so that must be our answer, answer B because T2 is up and to the right, the furthest most up and to the right point on our graph.1063

Number 53 is another ideal gas problem. The absolute temperature of a sample of an ideal gas is doubled, but the volume is held constant.1080

We know that PV = NRT and we want to know what effect that it is going to have on the pressure and density.1089

So I am going to get pressure on a side by itself and pressure = NRT/V, and in this problem itself -- well N is constant, R is constant and V is constant, and what we are going to do is we are going to double the temperature.1096

If I double the temperature, multiply the right side by 2, I have to double the left side.1112

We must have twice the pressure, which means P doubles.1117

How about density? Well density remember is mass divided by volume. 1123

Mass is not changing and the volume in this problem is constant, that is a given, therefore, density must be constant in this problem, so the choice where we have double the pressure and density remains the same is choice C.1129

Number 54 -- This is just a nasty question. 1150

I am really not sure why they would include something like this on an AP exam, but now that it is here let us just take a quick look at how you would solve something like this without having any background in it.1154

We are trying to find the number of atoms that you could stick on the top of a pinhead, which is really an order of magnitude estimation problem.1166

Well the radius of an atom is about 1 angstrom or 10-10 m, just as a very, very, very rough estimate.1174

Therefore the area of the atom -- we will call πr2, which is going to be π × (10-10)2 or π × 10-20 square meters -- rough, rough, rough.1185

Let us do the same thing for the pinhead. 1201

If we want the area of the pinhead, then that is going to be πr2 and if the diameter is 1 mm, half a millimeter is its radius...1204 0.5 × 10-3 m2, is going to give us 25 π × 10-8 square meters.1217

If we want the number of atoms, I will take the area of the pinhead, which is 25 π × 10-8 m2 divided by one single atom's area (π times 10-20)...1231

...and π's will cancel out and we have 10-8 up here, so that will be 10-12 down here, so I get something like 25/10-12, which will be 25 × 1012... 1253

...or about 2.5 × 1013, which the closest answer or the only thing even remotely close to that is choice B -- 1014.1267

That is how you could estimate something like this, but if this is the problem that gave you trouble, you are doing just fine.1279

Let us take a look here at Number 55 -- In an experiment we have light of a particular wavelength incident on a metal surface and we have electrons being emitted.1289

This sounds like the photoelectric effect.1298

If we want more electrons per unit time, but we want less kinetic energy with each electron, what should we do?1301

Well if we want more electrons, that means we need more light waves, more intensity.1309

But we also want them to have less kinetic energy, so if we want less kinetic energy that means that the light waves that hit them have to have less energy; those photons have to have less energy, they have to have a lower frequency and if they have to have a lower frequency that means they must have a longer wavelength.1319

The correct answer here would be B, increase the intensity and the wavelength of the light.1339

Now notice if you increase the wavelengths too much, those photons will not have enough energy to free any photoelectrons, so at a certain point, you keep reducing that wavelength, and you are not going to get any emitted photoelectrons. 1346

Number 56 -- All right, we have an object moving up and down with an acceleration given as a function of time by A = A sin(ω)T, where A and ω are constants, so what is the period?1361

Well, period -- just right back from simple harmonic motion -- is just 2 π/ω -- one of our formulas, so there is our choice right there.1381

All we have to really do is know the formula for 56.1391

Number 57 -- we have a ball at rest and is then kicked.1397

It is given some speed and a direction and we want to find the impulse imparted to the ball by the foot.1401

Well, impulse is change in momentum, and change in momentum is going to be mass × change in velocity.1406

Our mass (0.4 kg) and our change in velocity from 0 to 5 m/s is just going to be 5 m/s or 2 N-s, so the answer is C.1414

Number 58 -- We have a wheel with a couple masses on it and we want to know what the masses are so it remains in static equilibrium, so that it is not rotating about that center point.1430

That sounds like a net torque equals zero problem, a rotational equilibrium problem.1440

If our net torque = 0 -- well our net torque we have from the left-hand side, that (m) -- we have a torque of mgR, from the (M) on the upper left...1445

...that torque is going to be MgR × cos(60 degrees) and causing the clockwise rotation -- so a negative torque -- on the right-hand side we have 2MgR.1461

And again all of that must be equal to 0, so really what we are doing here is we are solving for (m).1477

This implies then that -- let us say -- we can do some cancellations here.1482

We have (g) and (R) in all of these so gR, gR, gR -- so this implies then that m + M cos(60 degrees) - 2M = 0 or m = 2M - M cos(60 degrees).1489

But we know the cos(60 degrees) is 1/2, so m = 2M - M/2 or 3/2(M) -- answer C.1514

Number 59 -- We have a rock thrown horizontally off a building from some height (H).1535

The speed of the rock as it leaves the hand horizontally is V0.1542

How much time does it take to travel to the ground?1546

To travel to the ground -- we do not care how fast it is going horizontally, that is a vertical kinematics sort of problem.1550

Vertically we know that the initial velocity in the y-direction is 0, we know the change in displacement, the change in distance, or the distance traveled is (H), and its acceleration is going to be G, assuming we are calling down the positive y-direction.1556

In order to find the time, I would use the equation δy = V0t +1/2at2, and since V0 is 0, that term goes away...1573

...and I can rearrange this to say that T = 2H/G square root where H was my δy and (a) was my G, so the answer there must be E.1586

Now Number 60 -- a connected problem, says what is its kinetic energy just before it hits the ground?1605

All right, well the kinetic energy -- we are going to have kinetic energy from its horizontal velocity and its vertical velocity.1612

We could do this by conservation of energy if we wanted to make it really easy, because its horizontal velocity is not changing, so the contribution horizontally is going to be 1/2 mv02.1619

Now vertically, it initially starts with potential energy mgh, right?1632

Well when it gets down to the ground, what is its kinetic energy going to be?1637

All that potential has become kinetic, so vertically that has become (mgh), so kinetic energy is the horizontal contribution plus the vertical contribution.1641

And of course you could go through and solve for the velocity vertically, plug it into 1/2 mv2 and work through it that way, but the easy way -- we have kinetic energy horizontally and you have this contribution from the vertical component.1653

By conservation of energy, you are all set right there, answer D.1667

Number 61 -- We are looking for which of the statements is not the correct assumption about ideal gases.1676

Well molecules are in random motion -- Yes, that is one of our assumptions.1682

The volume of the molecules is negligible compared to the volume occupied by the gas, or you have a lot of free space in there -- Yes.1685

The molecules obey Newton's Laws of Motion -- I certainly hope so. Yes.1694

The collision between molecules are inelastic -- Absolutely not. 1698

We have to assume that all the collisions are elastic, so the correct answer here, the one that is not an assumption that we need to be in ideal gas is D.1702

Number 62 is another ideal gas problem. 1715

Here we have an ideal gas in a tank at constant volume; it is going to absorb heat energy so that its temperature doubles and we know V1 is its initial average gas molecule speed and V2 is the speed after it absorbs that heat.1717

Let us find the ratio of V2/V1. 1733

First thing is let us figure out what happens to the kinetic energy when we double the temperature.1736

We know that the average kinetic energy is 3/2 times Boltzmann's constant times the temperature. 1740

If we double the temperature, we have to double the kinetic energy, so K average is doubling.1748

Now we also know kinetic energy is 1/2 mv2 and if we are looking for velocity, we could rearrange that to say that velocity is 2K/m square root.1757

So what do you actually do if you want to put a 2 in here if you double the kinetic energy?1770

Well you have multiplied the right side by square root of 2, so you have to multiply the left side by square root of 2, so really what do we do? 1775

We have square root of 2V, so the correct answer must be C.1783

All right, moving on to Number 63 -- We have two people, unequal mass, they are still on ice, they push each other and they go off in their separate directions.1791

What is true after they pushed each other?1805

The kinetic energies are equal? -- No, it does not have to be.1808

The speeds of the people are equal? -- Not necessarily.1811

The momenta of the two people are of equal magnitude? -- Yes, that is the Law of Conservation of Momentum, so C must be correct.1815

Let us take a look at Number 64 -- We have two parallel conducting plates separated by a distance (d) and they are connected to a battery of some electromotive force.1826

What is correct if the plate separation is double while the battery remains constant?1837

We are looking for what happens to the electric charge or potential difference or capacitance.1843

Well let us start -- if they are parallel plates, the capacitance is ε0 A/d, and if we are doubling the distance, we have to cut the capacitance in half, so we know that C is halved.1848

Now if that is the case, we also know C = Q/V, therefore the charge equals CV.1863

If we cut C in half, what happens to Q if we are holding voltage constant?1870

We must have cut Q in half or we halved the charge, so the correct answer for 64 would be B, the electric charge on the plates is halved.1877

Number 65 -- We have some concentric loops that have different radii, B and 2B, so one has twice the radius of the other.1891

What is the resistance of the wire loop with the bigger radius if the resistance of the one with the initial radius is R?1901

Well if you double R and circumference = 2 π × R -- if you double R, you double the circumference, therefore what you are doing is you are doubling the length of your wire.1909

Now resistance, if you recall, is resistivity times length divided by area.1924

If we have doubled the length, we must have doubled the resistance, therefore our answer is D.1931

Number 66 -- We have a uniform magnetic field perpendicular to a plane that is passing through a loop and we have two different radii of wires and the induced EMF in the wire loop of radius B is given.1942

What is the induced EMF if we have twice the radius?1956

Well remember, EMF is the opposite of the change in magnetic flux divided by time, but as I look at this problem, the changing flux is the same in both loops.1969

As you go from the loop of radius B to the radius of 2B, there is no difference in the amount of flux or in the change of magnetic flux, therefore, it is going to be the same.1973

The correct answer is going to be C. 1980

There is your induced EMF in the wire loop of radius 2B, the same as you have in the wire of loop radius B.1988

Number 67 -- We have a stationary object exploding into three pieces of different masses, and we know when they explode that we have a couple pieces moving off at right angles to each other and it gives us their momentum.1998

We want to find the magnitude and direction of the piece that has the bigger mass.2011

Well if they started at rest and you had one piece that went that way, one piece that went that way, by conservation of momentum, your other piece must have gone that way so that their momentum all tally up.2015

So our direction is figured out.2029

Now to get its magnitude, let us use conservation of momentum.2031

In the x-direction, we know that the initial momentum 0 must equal the momentum of the pieces which is (mv) to the right plus 3m times our VX prime...2034

...the x-component of the velocity of our larger piece, so therefore VX prime must be equal to -V/3.2051

And if we did this in the y-direction, we are going to get the same math, only to find out that V prime Y equals -V/3 and we could do that by symmetry as well.2064

So what is the total V prime? 2072

Well to do that, we have to realize that our x-component is -V/3 and our y-component is -V/3 so we can use the Pythagorean Theorem to figure out the total.2075

That is going to be the square root of -V/32 + -V/32, which will give us the square root of V2/9 + V2/9...2088

...which is going to be the square root of 2V2/9 or square root of 2/3V.2102

The correct answer must be D.2113

Number 68 -- We have a rod on a table top pivoted at one end and it is free to rotate without friction.2121

We are applying a force at the end at some angle and what we want to do is see where would we have to apply that force if it were perpendicular?2129

That means that what we really want is we want the same torque.2137

All right, well, our torque is FR sin(θ), which implies -- well the initial torque when we apply (F) at the end of the rod, and that is our force (F), our arm is (L) times the sine of that angle θ.2140

And that is going to be equal to the torque if we move that force in, so if we apply the same force perpendicularly, but at some distance (x), (x) is going to tell us at what position we have to apply that.2161

If I solve for (x) and if I divide both sides by (F), I find that (x) is going to be equal to L sin(θ), so the correct answer is A.2172

Number 69 -- Which of the following imposes a limit on the number of electron in an energy state of an atom?2187

Well this one you just pretty much have to know. That is going to be B, the Pauli exclusion principle.2193

It says you can only have a certain number of electrons at each energy level.2200

And Number 70 -- It gives us a capacitor and its charged to a set potential difference. 2205

Find the energy stored in a capacitor.2212

This looks pretty straightforward. The energy in a capacitor is 1/2 CV2.2214

That will be 1/2 × 4 × 10-6 (our capacitance in farads) × 100 volts2 or 1/2 × 4 = 2 × 10-6 × 1002 = 104...2221 I am going to get 2 × 10-2 and the units of energy are joules, so the correct answer there is E.2242

All right, so we have gone through the multiple choice portion of the test.2253

Check your answers, see how you did and if there is anything you need to review, go back, take some time, check it out and then in our next session, we will start on the free response portion of the test.2258

Thanks so much for your time everyone and make it a great day.2267