thephysicsman

PreAP–KEY–multiloop circuits quiz

Key is 2 pages: a blank copy followed by my work.

16:

You will construct a series circuit with the resistor you are given and another resistor, let’s call it X.  Draw a picture.  We know X has a value less than your given resistor (as per the problem) so we know the voltage across it is less than the that of the given resistor.  I’m going to be general since I am actually answer in this question for several people.  Let’s say you have a 6V battery and you need a 5V power supply.  What this means in practical purposes is that you need 5V across the given (larger) resistor.  The remaining voltage (1V in this case) will be across resistor X.  Use your values to find the current in your given resistor.  Since you have constructed a series circuit (a voltage divider) you can use this same current in resistor X.  You already know the voltage across resistor X is the difference between your power supply and what you needed (in this case 1V), so using your current and this voltage you can find the value of resistor X.

This is actually a common problem in the construction of electronics.  You need a particular voltage somewhere, but you only have a certain power supply.  This is the process you have to follow.

17c&d:

C: Since the hairdryer and the TV are plugged into the same outlet, this implies that they are in parallel (does your TV shut off if you turn off the hairdryer?  Hopefully not, although I do wonder why you are plugging your TV and hair dryer into the same outlet to begin with…a little odd).  So you simply need to find the equivalent resistance of the two items in parallel.

D: This is basically part B all over again, but using your equivalent resistance of the two appliances from part C.  You need count the contribution of resistance in the wires (for most people, 2-3 Ohm’s) once.

22:

You have 3 total volts, TWO times the internal resistance stated, plus the resistance of the bulb given in the problem.  A and B are easy.  Use ohm’s law to find the current (part A) and P=IV to find the power (part B).  Part C is the part I mentioned in class that is acting strange in webassign.  All you should have to do is get rid of the internal resistance mentioned in the problem and re-do parts A and B and find the difference between your original and new value, but webassign is spitting out 90% of the correct answers.

Toothpick Bridge project(Revised2013)–Forces of Destruction

Several folks have inquired about HW questions, particularly #’s 19-22.  These are certainly the more difficult problems on the HW assignment.  I will provide you with both some general information and some specific helps on each.

First, 19 & 20 are essentially the same.  19 has a North-South orientation, while 20 has an East-West orientation.  21 & 22 are similar except that you have more complicated 2-D vectors to consider.  Secondly, and perhaps the most important help I can give you: Ignore the signs when you use Coulomb’s Law.  Ignore the signs while you are doing your calculations and when you are done ask yourself, “What is the direction of this force?”  The remainder of the problem will depend on how you utilize that info.

Some background: If you keep the signs in the calculation when doing Coulomb’s Law Q1*Q2 give some interesting results.  If you have like charges (either + &+ or – & -) you will always end up with a positive force.  What does this mean?  Well, like charges repel, so a positive result must mean the charges push each other away.  Does the opposite hold?  Well if Q1 & Q2 have opposite signs (e.g., + & -) and you apply Coulomb’s Law your result will be negative.  This means an attractive force, or a force along a direction toward the second charge.

Let’s begin with the easier scenarios (19 & 20).

Three particles are placed in a line. The left particle has a charge of -67 µC, the middle, q_m = +42 µC, and the right, q_r = -87 µC. The middle particle is 72 cm from each of the others, as shown in Figure 20-13.

Figure 20-13.

(a) Find the net force on the middle particle.
Let’s get some labels here.  Call the Left particle A, middle B, and right C.

In order to find the force on B we only need to calculate the force of A on B and of C on B.  The force of A on C is irrelevant to this problem.

So F_net = F_AB + F_CB

F_AB = ?    B is pulled to the LEFT

F_CB = ?     B is pulled to the RIGHT

Since these forces are in opposite directions you need to treat them so, so when you calculate the net or total force you must SUBTRACT.

Part B)

Essentially the same idea, but a little different.  TO find the force on the right particle, you want

F_{net = FAC + FBC.}

You already know the value of F_BC from part A, but you must reconsider that the direction will be to the LEFT, relative to charge C.

F_AC will need to be calculated.  It’s direction is repulsive, so to the RIGHT on charge C.  Again, subtraction will be called for.  A final caution: When finding the magnitude of F_AC, be sure to consider the whole distance between particles A and C.

#’s 21 & 22:

We will look at number 21.  Like 19 and 20 you have a couple forces to calculate and consider.  The only difference here is that you must apply your rules for adding vectors at funky angles (find the x- and y-components for each vector, add them, find the resultant).  As always, one must consider direction whenever we “add” vectors.  Looking closely at one of your classmates versions of 21:

Three charged spheres are located at the positions shown in Figure 20-14. Find the total force on sphere B. (q_B = -7.4 µC, and q_C = +5.5 µC.)

First use Coulomb’s Law to find the two forces: The force of A on B, and the force of C on B.  Both are attractive, so charge B is pulled generally toward the left.
When I solved with your numbers, I found the following vectors:
187 N to the Left
146 N @ 36.9 degrees South of West

Breaking into components and adding the x- and y-directions respectively I get:
X: -304 (left)
Y: -87.7 (South)

Using trig an Pythagorean, I conclude with: 316 N @ 16.1 degrees South of West (take care in putting that into Webassign as you need to refernce the whole angle from 0-degrees).

1. Probably not.  If any image is visible it will be diffuse.  A surface that feels rough is too irregular to create a coherent reflection of light waves–the lumps and bumps are simply too large.

2.  sketch

3. 63 degrees

4. This can be done with a diagram and the premise of alternate interior angles since the opposing sides of the square block are themselves parallel.

5. You have two interfaces to consider.  Air –>Glass, then Glass–>Water.  The angle of refraction for the first interface becomes the angle of incidence for the second interface.
Theta-refracted-1 = 14.3 degrees = Theta-incident-2
Glass–>Water, theta refracted 2 = 16.4, which makes sense that the final bend is AWAY from normal since water has a slightly smaller index of refraction compared to glass.

6. Your eyes cannot comprehend the bend (refraction) of light cause by the change in medium from water to air, so to your eyes, sunlight reflecting off the fish comes straight out of the water towards you.  However, you know that since water is more optically dense than air, the light reflecting off the fish actually bends AWAY from normal as it leaves the water.  This fact means that the only possible path the light takes is the one closest to the dock.  This also means that you see the fish slightly higher and slightly farther away than it actually is (trace that ray back into the water).

7. Total internal reflection.  Theta-critical for water–>glass is 61 degrees

8.  Do Snell’s law, solve for n2.  n2 = 2.09 (this may not be accurate to real life, I think I made this problem up)

9. Assuming REAL images, image distance is positive, focal length is 12cm, assuming VIRTUAL image (negative i), f=60cm

10. Let’s go with a real image since the question does not specify.  M = hi/ho = -i/o.  If the image is real the image distance is positive so M = -30/20 = -1.5.  Remember the negative means inverted, which makes sense since real images are always upside down.
b.  3.3 m
c. Since the problem in #9 does not specify what type of lens it is, nor whether the image is real or virtual, no, there is no way to tell.  However, as indicated in part A, the negative magnification is correct on the assumption of a real image.  If we went with the virtual image case then the magnification would turn out postive, which makes sense since virtual images are always upright/erect.

***Disclaimer: I did these quickly on my iphone calc while riding in a car. I hope all the math is correct, but if you just cannot figure out why I’m getting my answers it is possible that my fat fingers hit cos instead of sin once or twice.