## 1998 Ap Biology Essay

*Hello, everyone, and welcome back to www.educator.com.*0000

*I'm Dan Fullerton, and in this final lesson we are going to continue our work on the 1998 practice exam,*0003

*this time focusing on the free response questions.*0010

*Let us dive right in and make sure you have got the test printed out.*0014

*I highly recommend that you take a shot at these questions before coming back to the video.*0017

*Starting with number 1, let us look at part A.*0022

*We have that sphere B1 and a point/ A and that B1 sphere is held in place.*0027

*We area asked to find the charge on sphere B1.*0034

*As I look at this, the first thing I think I'm going to do is draw a free body diagram for B1 to see all the forces that are on it.*0038

*Let us take a look here.*0046

*If there is our sphere, we will put our Y axis and X axis.*0048

*And labeling their forces, we have an electrical force repulsion to the right.*0057

*We have a force of tension from our string, we will call that T.*0062

*We have the weight of the sphere MG.*0069

*Using Newton’s second law, we can write that the net force in the Y direction is going to be T sin 70°.*0075

*If 70° here, if that is 20 to match what they have on our diagram would give us 4 θ, - MG = 0,*0088

*which implies then that T must equal MG/ sin 70° which is our mass 0.025 × the acceleration*0094

*due to gravity on the surface of the Earth 9.8 m / s² / the sin of 70° or about 0.26 N.*0111

*We try the same thing in the X direction.*0123

*The net force in the X direction, we have the electrical force to the right - T cos 70, the X component of that tension.*0127

*Those all have to be = to 0 because it is in equilibrium, it is not accelerating.*0139

*Which implies that the electrical force = T cos 70° is going to be, we just found T was 0.26 N cos 70° or 0.089 N.*0144

*If we know the electrical force, we should be able to back out that charge using in Coulomb’s law.*0162

*Our electrical force is K QA QB / R² which implies that the charge on sphere B is just going to be our electrical force × R² ÷ K QA.*0167

*Our electrical force we just found was 0.089 N.*0187

*The distance between the centers of our charged particles is 1.5 m, K 9 × 10⁹ N meters²/ C² which is the same as 1/ 4 π ε₀.*0196

*The charge on A is 120 µc.*0212

*Put that all into my calculator and I find the value for QB of 1.86 × 10⁻⁷ C.*0218

*Part A, check.*0231

*Part B asks, suppose this b1 is replaced by a second suspended sphere B2 that has the same mass but this one is conducting.*0235

*If we establish equilibrium again, what happens to that equilibrium angle?*0246

*That angle has going to have to be less, that angle θ is going to be less than 20°*0251

*because the charges are going to move in the conductor leaving more positive charge on the far side of your B2 sphere*0257

*and less on the near side near A.*0264

*Effectively, that distance between charges increases so the electrical force between them is going to decrease by Coulomb’s law.*0266

*Electrical force decreases, it is going to come down a little bit.*0274

*I would say that that is going to be less than 20°.*0277

*Here for part C, the sphere B2 is now replaced by a very long horizontal none conducting tube.*0282

*The tube is hollow with thin walls, gives us the radius and the uniform positive charge per unit length λ.*0290

*Use Gauss’s law to show that the electric field is given by that expression.*0298

*I’m really good at Gauss’s law, so we will try that.*0303

*Here is our tube and the Gaussian surface I'm going to choose is a cylinder around that.*0306

*Choose a Gaussian surface over here at some distance R from its center.*0317

*The radius of the little one inside this R, the radius of our Gaussian surface is r.*0324

*We can write Gauss’s law, integral / the closed surface of E ⋅ DA is equal to the total enclosed charge ÷ ε 0.*0333

*The left hand side electric field should be constant because we chose this with symmetry, that in mind.*0346

*And the area of our close surface is going to be 2 π R × its length L and*0353

*that is going to be equal to the enclosed charge which is the linear charge density × our length ÷ ε₀.*0361

*Therefore, our electric field is going to be λ L / 2 π ε₀ RL.*0370

*Our L's make a ratio of 1 so that is just going to the λ / 2 π ε₀ R.*0379

*And if we start putting our values in here, λ is 0.1 × 10⁻⁶ C / m, 2 π ε₀ is 8.54 × 10 ⁻12 C²/ N-m².*0386

*And we also have our R down here.*0406

*This implies then that the electric field is going to be, I come up with about 1797/ R N-m/ C,*0410

*which is approximately 1800/ assuming R is in meters, N/ C with R in meters.*0422

*We prove that 1800/ R N/ C.*0437

*Very good, moving on to part B.*0443

*A small sphere A with charged 120 µc is now brought in the vicinity of the tube and held at distance of 1.5 m.*0448

*Find the force the tube exerts on the sphere.*0455

*That should be pretty easy now that we have the electric field strength.*0458

*The force is just charge × electric field which is going to be our 120 × 10⁻⁶ C × our 1800 N/ C / R 1.5 m or 0.144 N.*0463

*And part E, let us take a look at E here.*0486

*Calculate the work done against the electrostatic repulsion to move sphere A toward*0494

*the tube from a distance R = 1.5 m to a distance R= 0.3 m from the tube.*0500

*That should be a pretty straightforward calculation of work.*0506

*Work = the integral of F ⋅ DR, which is going to be the opposite of the integral as we go from R = 1.5 m to 0.3 m*0509

*throughout the work done against the electrostatic repulsion of, we have our 120 × 10⁻⁶ our charge × our 1800 N/ C ÷ R*0521

*to give us our force QE DR, which implies then that the work is going to be equal to -0.216, when I pull the constants out.*0539

*Integral from R = 1.5 to 0.3 of DR/ R.*0552

*Integral of DR/ R is nat log of R.*0559

*W = -0.216 log of R evaluated from 1.5 to 0.3, which is -0.216 log of 0.3 - log of 1.5,*0563

*which is when I plugged into my calculator about 0.348 J.*0583

*That finishes up free response problem number 1.*0595

*Let us move on to number 2, a circuit problem with the capacitor and inductor.*0598

*As we look here at number 2, we are going to start off with the circuit, the switches initially open,*0605

*the capacitor is uncharged, and there is a voltmeter but they are not showing the measure of potential difference across R1.*0612

*On the diagram above, draw the voltmeter with a proper connections for measuring the potential difference.*0618

*That was nice and simple, what you will only do is you take your drawing that you have there, there is your 20V.*0625

*We have R1 just put your voltmeter in parallel V with R1.*0629

*Taking a look at part B, at time T = 0, the switch is moved to position A so we now have the capacitor in the circuit.*0640

*Find the voltmeter reading for the time right after you do.*0649

*The trick here is the moment you close that, that capacitor initially is going to act like a wire.*0653

*Effectively, you have a circuit that looks like this.*0658

*There is R1, there is R2.*0662

*R1 is 10 ohms, R 2 is 20 ohms, and if we want to know the voltmeter reading,*0670

*we could do this as a voltage divider and probably not too tough to say that it is going to drop 1/3 of the voltage across,*0677

*1/3 of the resistance to 10 ohms which will be 1/3 of 20 or 6.67V.*0685

*But more formally, we can do this by saying the total current in the circuit is E/ R which is going to be 20 V/ your total resistance 30 ohms or 0.667 amps.*0690

*The voltage drop across R1 is just going to be current × resistor 1 which is 0.667 amps × our 10 ohms which is about 6.67 V.*0704

*Moving on to part C, after a long time the measurement of potential difference across R1 is again taken,*0721

*determine for this later time the voltmeter reading.*0731

*After a long time, the capacitor is going to act like an open.*0736

*That is going to break your circuit, you are not going to have any current flowing through R1.*0741

*Therefore, there is no voltage drop across R1 so the voltmeter reading is going to be equal to 0.*0747

*And C2, the charge on the capacitor.*0757

*The entire voltage will be across the capacitor at that point.*0759

*If C = Q/ V, that implies that Q = CV, which is going to be your capacitance 15 µf or 15 × 10⁻⁶ F ×*0763

*your potential difference across your capacitor 20 V or 3 × 10⁻⁴ C.*0775

*There is part C, let us take a look at D.*0787

*At a still later time t = T, the switch is moved to position B.*0795

*Determine the voltmeter reading right after that happens.*0800

*Now we have got a circuit that looks kind of like this, for 2D we got out 20V.*0803

*At that next position for our switch, there is R1.*0812

*Here is our inductor and we have R2 in the mix, that is a 2 H conductor.*0816

*Initially, the inductor poses current flowing and acts like an open so potential across R1 is going to be 0.*0827

*Let us take a look at 2 E, a long time after t = T, the current in R1 reaches some constant final value I final.*0839

*Determine what that is going to be.*0848

*After a long time, that inductor just acts like a wire so I final is going to be equal to the potential ÷ the resistance in our circuit.*0851

*Which is again, 20 V/ 30 ohms or 0.667 amps.*0861

*Part 2, determine the energy stored in the inductor.*0872

*The energy in the inductor is given by ½ LI² and it is going to be ½ × our inductance 2 H × the square of our current 0.667 amps² or about 0.445 J.*0875

*We have got a part F, write but do not solve the differential equation for the current in R1 as a function of time.*0896

*We can use Faraday’s law to do this, the integral/ the closed loop of E ⋅ DL = - the time rate of change of the magnetic flux.*0908

*As we go around here, remember no electric field in there so we have got - E + I R1.*0921

*We have got the voltage drop across I R2 and all of that has to equal their change in flux - L DI DT.*0932

*There is no electric field here so no contribution to the left hand side from your inductor.*0944

*We got ID IDT so I suppose we have already written that differential equation.*0949

*But just to clean it up a touch, let us call that E – I R1 + R2 - L DI DT = 0.*0954

*There is the differential equation and we do not have to solve it this time.*0971

*That is a straightforward number 2 and I think we are going to make up for here in number 3.*0976

*This is one of the trickier free response problems, more involved free response problems I have seen on AP exam.*0982

*Here we have a conducting bar of mass M that is placed on some conducting rails*0992

*at distance L apart that is at an angle with respect to the horizontal.*0997

*We have got a magnetic field coming out of that ramp.*1001

*We are asked to determine the current on the circuit when the bar has reached the constant final speed, so 3A.*1005

*First thing I'm going to do is I'm going to make a free body diagram for that bar and I'm going to tilt my axis*1013

*so it is a little easier for me to see what is going.*1021

*On our bar we are going to have a magnetic force up the ramp and we are going to have gravitational force pulling it down.*1029

*We know that that force is going to be up the ramp, we can use the right hand rule Q × V × B*1040

*to find the force on the charges in the bar creating the current.*1046

*And then the charges in the bar move to find the force on the bar which is going to be up the incline.*1049

*As I do this, as I take a look here, determine the current in the circuit when the bar is reached a constant final speed.*1054

*Let us see, net force in the X direction, Newton’s second law point our free body diagram*1065

*is going to be the X component of its weight MG sin θ - FB the magnetic force, all has to equal MA.*1072

*Since V is constant, A = 0.*1085

*Therefore, we can say that the magnetic force has to equal MG sin θ, its constant final speed.*1089

*The magnetic force, my current flowing through the wire is going to be I LB.*1100

*Therefore, we can state that MG sin θ must equal I LB or the current must be equal to MG sin θ/ LB.*1107

*For part B, we are asked to determine the constant final speed of the bar.*1130

*For B, if current is potential / resistance which is – D φ B DT/ R.*1138

*Let us see here, we could do that and all of that must equal MG sin θ/ LB.*1151

*It would be helpful to know what this D φ B term is.*1162

*Φ B is going to be B × A or in this case B LX are geometry.*1167

*D φ B with respect to time is just going to be BL and X is the only thing that changes with time.*1177

*That is DX DT which is going to B BL × its velocity.*1185

*I can put that in up there for that.*1189

*This implies then that, we have B LV/ R = MG sin θ/ LB.*1194

*And solving just for B 4V, V would then equal, we will have MG R sin θ/ L² B².*1212

*Plugging away through this, part C.*1234

*Part C says, determine the rate at which energy is being dissipated when its reach its constant final speed.*1241

*Varying at which energy is dissipated.*1247

*A lots of fancy way of saying find the power.*1249

*We know I, we know R, so power = I² R is just going to be this² multiplied by R.*1251

*We will have M² G² R sin² θ / L² B².*1260

*Not too bad on C.*1279

*D, express the speed of the bar as a function of time T from the time it is released at T equal 0.*1283

*This one looks a bit more involved.*1292

*Let us see, we know MG sin θ - the magnetic force = MA Newton’s second law.*1297

*We said the magnetic force was I LB, we also know that A is the time rate of change of velocity DV DT.*1310

*MG sin θ – I LB must equal M DV DT.*1320

*But we also know that I = B LV / R, therefore, MG sin θ -, substituting in here for FB, B LV/ R.*1332

*We have still got our LB again, LV LB = M DV DT.*1355

*Let us move that M by all sides by M.*1370

*I get on the left hand side G sin θ - L² B² V/ MR = DV DT.*1375

*Which implies then, it is time to do our separation of variables.*1394

*This might get a little bit ugly, DV/ L² B² V/ MR - G sin θ = – DT.*1398

*We are going to try and integrate this.*1419

*We are going to integrate, the composite.*1422

*The integral from V equal 0 to some final value of V of, we have got DV / L² B² V / MR -G sin θ.*1425

*If I want that = the integral from t equal 0 to T of DT with a negative sign, I want this to fit the formula DU/ U.*1445

*I have got DV, I got V here.*1459

*I have got all this other stuff there as well.*1461

*I’m going to have to have that at the top to make that fit the form so that is L² B² / MR.*1464

*If I multiplying the left by L² B²/ MR, I have to multiply the right by L² B²/ MR × negative integral from 0 to T of DT.*1473

*That fits the form of DU/ U.*1486

*This implies then that this should be a nat log of our U which is down here,*1491

*L² B² V/ MR -G sin θ all evaluated from V = 0 to V, must equal.*1499

*This thankfully is little bit easier to integrate, - L² D² T/ MR.*1514

*We can expand this in the left as we substitute in our 0 and V.*1525

*The left hand side will be, we get L² B² V/ MR - G sin θ - the log of - G sin θ = - L² B² T/ MR.*1529

*We can compress this right hand side.*1557

*The difference of logs is the log of the quotient.*1558

*That is going to imply that the log of, we have L² B² V/ M R - G sin θ/ -G sin θ = - L² B² T/ MR.*1562

*Where do we go from here?*1594

*Let us see, it looks like we are going to have to do some simplifications.*1597

*I think we can get rid of that G sin θ or at least incorporate it a little bit better.*1601

*Which implies then that the log of - L² B² V/ MR G sin θ + 1 = - L² B² T/ MR, raising both sides to E*1606

*to get rid of that natural log, which implies in the left hand side becomes - L² B² V/ M R G sin θ +1*1632

*must equal E ^- L² B² T/ MR, which implies then, moving on to the next page.*1646

*L² B² V/ MR G sin θ = 1 - E ^- L² B² T/ MR.*1660

*Since we are solving for V, we can do one last step.*1678

*Get VL by itself and say that V =, we will have MR G sin θ/ L² B² × (1 - E ^- L² B² T/ MR).*1683

*Let us get that a nice big 3D box because good heaven knows we certainly earned that on that problem.*1714

*A lot involved on part D there.*1721

*Quite a tricky bit of work and a lot of math.*1725

*Let us finish off with part E, so 3 even.*1729

*Suppose the experiment is performed again, this time with the second identical resistor connecting the rails at the bottom.*1735

*Will this affect the final speed and how? Justify your answer.*1741

*If you have 2 resistors in parallel, your total equivalent resistance is going to go down.*1747

*We just said that the final velocity is MG R sin θ/ B² L².*1758

*If R decreases, our final velocity must go down.*1770

*I would say that the final speed increases because R is decreasing there.*1774

*Hopefully that gets you a great start to AP Physics C Electricity and Magnetism.*1785

*Thank you so much for joining us at www.educator.com.*1792

*Make it a great day everybody.*1794

А потом вы отдали кольцо какой-то девушке. - Я же говорила. От этого кольца мне было не по. На девушке было много украшений, и я подумала, что ей это кольцо понравится.

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