Physics Question Thread 2

[Pembleton]

Hi guys,

I don't have my physics textbook nearby and I was wondering if someone could remind me how to figure out the center of mass of two objects on a plank, for instance.

For example a 10 kg box is 1 m from the edge of a 10 m plank and a 20 kg box is 2 m from the other edge of the plank. Where is the center of mass on the plank?

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[Just Joshin]

I know the equation for frequency is f = 1/T, but here's a problem in Nova that I'm confused about.

It asks if a 60 g mass is connected to a spring with a frequency of 30 Hz, what will a mass of 240 g do to the frequency?

Why can't you set up a proportion here?

60/30 = 240/x

I like proportions because they're easy to do in my head so I'm always looking to do them when I can. Why can't I here?

 

[hung]

I know the equation for frequency is f = 1/T, but here's a problem in Nova that I'm confused about.

It asks if a 60 g mass is connected to a spring with a frequency of 30 Hz, what will a mass of 240 g do to the frequency?

Why can't you set up a proportion here?

60/30 = 240/x

I like proportions because they're easy to do in my head so I'm always looking to do them when I can. Why can't I here?

that proportion wouldn't work. if u increase the mass, the frequency doesn't increase in the same proportion (it shouldn't). examine their relationship.

try the simple harmonic equation: 2pi x f= square root (k/m)
new freq is 15Hz, so decrease freq by factor of 2.

 

[pky463]

For this question, the answer was I and III:

A certain metal plate is completely illuminated by a monochromatic light source. Which of the following would increase the number of electrons ejected from the surface of the metal?

• Increasing the intensity of the light source
• Increasing the frequency of the light source
• Increasing the surface area of the metal plate

I don't understand how increasing the SA of the metal will increase the number of electrons. The number of electrons ejected is related only to the number of photons incident (Intensity), right? If the power of the light source is kept constant, and the area is increased, wouldn't intensity be lowered (P=IA)?

 

[Doc Henry]

I don't ever remember learning this stuff in my physics class...anyone know how much this is actually tested on the MCAT?

 

[armybound]

For this question, the answer was I and III:

A certain metal plate is completely illuminated by a monochromatic light source. Which of the following would increase the number of electrons ejected from the surface of the metal?

• Increasing the intensity of the light source
• Increasing the frequency of the light source
• Increasing the surface area of the metal plate

I don't understand how increasing the SA of the metal will increase the number of electrons. The number of electrons ejected is related only to the number of photons incident (Intensity), right? If the power of the light source is kept constant, and the area is increased, wouldn't intensity be lowered (P=IA)?

well I'm just thinking that a bigger surface area = more light hitting the metal plate = more electrons ejected

is the P = IA equation for the area of the light source, not the area of the target?

 

[pky463]

well I'm just thinking that a bigger surface area = more light hitting the metal plate = more electrons ejected

is the P = IA equation for the area of the light source, not the area of the target?

Hmm, maybe. Thanks!

 

[homelesssimpson]

Tricky Physics question:

Molecules conserve energy and mometum when they absorb photons. An H2 molecule at 250K absorbs a UV photon with a wavelength of 120nm. What is the change in H2's velocity?

 

[BlackSails]

Tricky Physics question:

Molecules conserve energy and mometum when they absorb photons. An H2 molecule at 250K absorbs a UV photon with a wavelength of 120nm. What is the change in H2's velocity?

Its not possible to figure out. A molecule absorbing a UV photon will probably lose an electron, or go into some other excited mode and gain some kinetic energy.

 

[olygt]

I'm going over the simple machine physics lecture and was just curious about the relationship to the machines in a gym. Since the machines are reducing the force you have to exert to lift the weight, it would be easier to lift a 200lb on a machine than 200 lbs of free weight, right? Also, since the work is the same would you be building your muscle up the same way as if you were lifting free weights? And where does the energy transfer of work take place, in your body or the energy it takes to lift the weight? Whew, that was a lot of questions.

 

[Phlame217]

The weights in the gym are built to compensate for the difference in weight actually being lifted. Therefore lifting 200 pounds on a machine should equal 200 pounds of freeweight.

As to the muscle question, the targetted muscles (such as doing curls freeweight wise vs machine ) would result in very similar if not the same muscle improvements, but freeweights are not static where as machines are static except for one direction therefore with freeweights you'd recieve the benefits of building surrounding muscles at the same time that aid in balance and control.

And i suppose the energy takes place in the body because you dispensing chemical energy to contract muscle fibers which lower and raise weights.

 

[SketchLazy]

Tricky Physics question:

Molecules conserve energy and mometum when they absorb photons. An H2 molecule at 250K absorbs a UV photon with a wavelength of 120nm. What is the change in H2's velocity?

Its not possible to figure out. A molecule absorbing a UV photon will probably lose an electron, or go into some other excited mode and gain some kinetic energy.

Except in this case, the question states that energy and momentum are conserved. This is just basically a conservation of momentum problem.

Even though photons have no mass, they still have momentum which is given by the equation:

P=h/λ

So when the hydrogen molecule absorbs the photon the equation for momentum is:

mv(i) + h/λ = mv(f)

h=planck's constant
m = mass of hydrogen molecule
λ = wavelength of the photon
v(i) = initial velocity of hydrogen molecule
v(f) = final velocity of hydrogen molecule

You're given the mass of the hydrogen molecule (2 amu) and the wavelength so all you need to solve for v(f) is v(i).

v(i) is given using the kinetic theory of gases:

.5mv^2 = 1.5RT

m = mass of molecule
v = velocity
R = gas constant
T = temperature in K

Once you solve for v, you get the velocity of the hydrogen molecule at 250K, and then you can solve for its velocity after absorbing a photon using the conservation of momentum equation.

 

[BlackSails]

Except in this case, the question states that energy and momentum are conserved. This is just basically a conservation of momentum problem.

Well its not my fault that the premise is wrong!

 

[NAVYLABTECH08]

Point charges of +2.00 nC and - 5.00 nC are arranged on two of the vertices of an equilateral triangle. Find the magnitude and sign of the third charge which must be placed at the 3rd vertex of the triangle so that the potential at a point on the side midway between the +2.00 nC and - 5.00 nC is equal to zero.

What do you guys think...

 

[physics junkie]

Point charges of +2.00 nC and - 5.00 nC are arranged on two of the vertices of an equilateral triangle. Find the magnitude and sign of the third charge which must be placed at the 3rd vertex of the triangle so that the potential at a point on the side midway between the +2.00 nC and - 5.00 nC is equal to zero.

What do you guys think...

Does anyone know if this forum has LaTeX support? I hate my notation coming out all nasty...

Well, since you asked for potential and potential drops off with 1/r where r is the distance from the source charge we can solve this problem without too much work. If I'm too formal then I would recommend studying the solution I provide and trying to solve it yourself at a later time to see if you really get it.

Let L = length of one side of the triangle. You can set this to one for convenience if you'd like.
Let q = a unit of charge
Let n = the charge of the unknown particle that you will place at the third vertice.
Let d = distance from the third point to the midpoint of the other two points. Notice for any configuration this will be the same since its an equilateral triangle.

You have the equation

Since you can add potentials(since they are linear and drop off with r and not r^2 or log(r) or r^3) you see that the potential at the midpoint is -3q/(0.5*L). This is basically adding the two potentials since the point you are trying to find the potential at is equidistant between two sides.

The third charge point is not 0.5*L away. We can find the distance using the pythagorean theorem. (L)^2 - (0.5L)^2 = d^2. If you use 1 for L you get 4 - 1 = squareroot(3). So d = sqrt(3). Or we can just recall this figure from trigonometry:

So we need a charge that will balance the potential to zero when the current potential is -3q/(0.5*L). This can be represented more clearly as -6q/L. The charge you need has to make the opposite sign of the same magnitude potential so it would equal (sqrt(3)*6*q)/(sqrt(3)*L). That makes n=6*sqrt(3). For some reason this answer doesn't feel right so maybe its wrong but all the methods you need to solve it are right here in this post.

Hopefully this was helpful. I wouldn't actually solve the problem like this on the MCAT. I would most likely just do it in my head after realizing the symmetry of the problem.

 

[SketchLazy]

So we need a charge that will balance the potential to zero when the current potential is -3q/(0.5*L). This can be represented more clearly as -6q/L. The charge you need has to make the opposite sign of the same magnitude potential so it would equal (sqrt(3)*6*q)/(sqrt(3)*L). That makes n=6*sqrt(3). For some reason this answer doesn't feel right so maybe its wrong but all the methods you need to solve it are right here in this post.

My answer was a little different. I used the 30-60-90 triangle just like you but I set up my equations differently.

Here's how I went about it. Hope you don't mind that I'm using your diagram.

Since electric potential is a scalar quantity, direction isn't important and the total potential for a certain point is just based on the sum of the potentials exerted by each charge in the field. This is given by the equation:

V=kQ/r + kQ/r +...etc...


Now the question wants to know what charge at point A will make the potential at D zero. Using the 30-60-90, I made the distance between D and B equal to r, so the distance between D and C also has to be equal to r. If the distance between D and B is r, then the distance between D and A would be equal to sqrt(3) * r. So here's how I set it up:

0 = k(2 nC)/r + k(-5 nC)/r + kQ/((sqrt3)*r)

Solving for Q, the k's and r's all cancel out, so you're left with:

3 nC = Q/(sqrt(3))

So the charge at point A would have to be 3 * sqrt(3) nC in order for the potential at D to be equal to 0.

 

[physics junkie]

My answer was a little different. I used the 30-60-90 triangle just like you but I set up my equations differently.

Here's how I went about it. Hope you don't mind that I'm using your diagram.

Since electric potential is a scalar quantity, direction isn't important and the total potential for a certain point is just based on the sum of the potentials exerted by each charge in the field. This is given by the equation:

V=kQ/r + kQ/r +...etc...


Now the question wants to know what charge at point A will make the potential at D zero. Using the 30-60-90, I made the distance between D and B equal to r, so the distance between D and C also has to be equal to r. If the distance between D and B is r, then the distance between D and A would be equal to sqrt(3) * r. So here's how I set it up:

0 = k(2 nC)/r + k(-5 nC)/r + kQ/((sqrt3)*r)

Solving for Q, the k's and r's all cancel out, so you're left with:

3 nC = Q/(sqrt(3))

So the charge at point A would have to be 3 * sqrt(3) nC in order for the potential at D to be equal to 0.

Nice. You figured out what tripped me up. I knew the numbers didn't "flow"

 

[scarterinscrubs]

How in depth does Quantum Mechanics get on the MCAT?

 

[Begaster]

No depth.

 

[werd]

indeed. no quantum mechanics on the mcat. calculus also doesn't exist. string theory, on the other hand... i would know it pretty solid.

 

[physics junkie]

Be familiar with the photoelectric effect and what the equivalent forms are of energy and momentum. Be aware that energy is a function of frequency and not wavelength. Know what the different quantum numbers are. Understand beta decay, gamma decay, positron decay, etc. Understand that electron orbits(energy levels) are quantized and how to calculate the energy released when an electron jumps orbitals.

Oh and then review your copies of Quantum Field Theory Volumes I-III. If you need some help brushing up on the abstract algebra then maybe you should consider another profession.

 

[physics junkie]

Anyone who goes to the gym regularly knows that there is no set formula for being able to tell whether its heavier or lighter because not all machines are built equal. If you do tricep pulldowns with the rope you'll notice that its much tougher on a machine where there are two pulleys or a thicker band connecting the rope in your hands to the weight. It's much easier when the rope is thinner or there is only one pulley. Even though the second pulley shouldn't make a difference(one reason being the physics you've done is in an ideal world where there is no rotational inertia to the pulley) it definitely does.

Also, free weights can be a lot more difficult to lift because the same lift might require more balance or stabilizer muscles. Try bench pressing with 80lb dumbbells in each hand. Then try bench pressing 160lb on a machine. Then 160lb with a bar. The machine will be the easiest, the bar will be the second easiest, and the dumbbells will be the hardest. Why? Because even though its a similar exercise its not the same exercise.

Ideal physics does not translate to real life situations. There is actually a completely different field called engineering that deals with applying physics to the real world. Free weights are the best. You can see me over at forums.bodybuilding.com with the same name btw.

 

[Medman27]

I'm currently using the Ek physics book and I'm not finding it very useful. It seems like I understand what I'm reading but then am tested on things that are out of scope. Does anyone recommend another approach to Ek, or any other supplemental material that might be helpful. I appreciate it

 

[NAVYLABTECH08]

This problem is giving me so hard time. Does anyone know how to set this problem up?

Point charges of +2.00 nC and - 5.00 nC are arranged on two of the vertices of an equilateral triangle. Find the magnitude and sign of the third charge which must be placed at the 3rd vertex of the triangle so that the potential at a point on the side midway between the +2.00 nC and - 5.00 nC is equal to zero.

 

[NAVYLABTECH08]

Sorry, I posted the question twice....

 

[SketchLazy]

This exact question was setup in this thread:

http://forums.studentdoctor.net/showthread.php?t=415792&page=6

 

[BerkReviewTeach]

I'm currently using the Ek physics book and I'm not finding it very useful. It seems like I understand what I'm reading but then am tested on things that are out of scope. Does anyone recommend another approach to Ek, or any other supplemental material that might be helpful. I appreciate it

On this site, it seems that people who use NOVA for physics like it and people who use BR for physics like it. Both of those should prove useful.

 

[Lucylu123]

I have a question about a simple problem from EK 1001. Its problem # 26.

If a particle moves along path C at a constant speed of 1m/s, what is the avg acceleration of the particle as it moves from position 1 to position 2?

Path C is a curved pathway.

I thought that whenever something moved at constant speed, acceleration is automatically 0. However the answer is .4/pi m/s^2.

The explanation says that the original velocity is 1m/s and the final velocity is 1m/s, so the change in velocity is 2m/s. I guess I'm just confused because I thought it would be zero.

Can anyone explain? In what cases would acceleration not be zero when there is no change in velocity?

 

[Lucylu123]

Nevermind...figured it out.

 

[majahops]

I understand, qualitatively, when an image is inverted or upright and real or virtual and where (i.e. left or right) with respect to a lens or mirror the image (and obviously the object) is located.

However, I fail to see a unifying logic behind the sign conventions used for lenses and mirrors. I suppose my questions are:

1) Does the MCAT require that you memorize sign conventions for each variable in mirror and lens systems?

2) If so, which convention is the "MCAT"-testable convention? I've seen Examkrackers and Kaplans conventions and it seems like they just want you to memorize a table. I want to understand the sign conventions, so that I can re-derive them based on logic and basic understanding if need be.

Thanks guys.

 

[Waterlilies]

I understand, qualitatively, when an image is inverted or upright and real or virtual and where (i.e. left or right) with respect to a lens or mirror the image (and obviously the object) is located.

However, I fail to see a unifying logic behind the sign conventions used for lenses and mirrors. I suppose my questions are:

1) Does the MCAT require that you memorize sign conventions for each variable in mirror and lens systems?

2) If so, which convention is the "MCAT"-testable convention? I've seen Examkrackers and Kaplans conventions and it seems like they just want you to memorize a table. I want to understand the sign conventions, so that I can re-derive them based on logic and basic understanding if need be.

Thanks guys.

Hi Tim,



I've taken many MCAT practice tests/sections (EK, Kaplan, PR, AAMC), and generally I can say it doesn't dwell much on optics and the questions aren't difficult either. It's really some basic memorization and lens and mirrors follow very similar sign conventions, with a few exceptions. Some frequent "testable" conventions are:

1. diverging lens and convex mirrors have negative focal lengths​

2. real images (+) = where light goes through after interacting with lens or mirrors, and since lens refract real is the other side, while mirrors reflect so real is in front of mirror​

[virtual images (-) will then follow the opposite idea)​

3. objects b/w converging mirror & focal point produces virtual, upright images (enlarged); at focal point produce no images; and beyond focal point produces real, inverted images (small).​

I just memorized some basic conventions for the test, and didn't feel that you need to really learn in depth (like drawing those ray diagrams - though it's great if you can!). I think if you want to truly understand them, then you should learn the ray diagrams. EK have great mnemonics for sign conventions. Hope this helps!

 

[BerkReviewTeach]

1) Does the MCAT require that you memorize sign conventions for each variable in mirror and lens systems?

2) If so, which convention is the "MCAT"-testable convention? I've seen conventions and it seems like they just want you to memorize a table. I want to understand the sign conventions, so that I can re-derive them based on logic and basic understanding if need be.

1) You should definitely know the sign conventions as it pertains to object and image position, because answer choices could describe positions in terms of right/left with distance or (+) and (-) with distance.

2) The best combination of sign conventions applied to a visual aid (lens/mirror summary) is the optics summary diagram in the BR physics supplement, pages 138 and 139. If you think about lens and mirror problems like that, you can answer any multiple-choice question (even double lens ones) in less than fifteen seconds.

 

[Vihsadas]

As a physics student, I can tell you that you are dead on when you say that 'there is no logic to the sign conventions'. That's exactly right, they are arbitrarily agreed upon by physicists and thus we use them. But in reality, there are many such examples like this. Another one that comes to mind is that 'current' is actually idealized as 'positive charges moving' even though we now know that its actually the movement of negative charges. So, these kinds of things you do have to memorize.

 

[wes431]

A 16 kg loudspeaker is suspended 2.0 m below the ceiling by two 3.0 m long cables that angle outward at equal angles. What is the tension in the cables?

 

[wes431]

A 1150 kg car pushes a 1700 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 3800 N.
(a) What is the magnitude of the force of the car on the truck?
N


(b) What is the magnitude of the force of the truck on the car?
N

 

[Vihsadas]

A 16 kg loudspeaker is suspended 2.0 m below the ceiling by two 3.0 m long cables that angle outward at equal angles. What is the tension in the cables?

Doing this super quick in my head and it's late so someone correct me if I'm wrong:

Total load is 16kg*g = -16g.
The sum of the tension along the y-axis in both cables has to be equal and opposite the total load. Since both cables are at the same angle their vertical loads are the same. Each vertical tension is = 8g.

I think if you draw a triangle (doing it in my head), the hypotenuse is 3m, and the opposite side to theta is 2m. Thus, Sin(theta) = 2/3.

Also, since the opposite side to theta is the y-axs, it corresponds to 8g.

So if the Tension in the string (hypotenuse) is "T", we can write the expression:

T*sin(theta) = 8g.

So we have:

(1) sin(theta) = 2/3
(2) T*sin(theta) = 8g

(2/3)*T = 8g

T = 12g.

The tension in each rope is 12*g

 

[MercuryX]

Ahh im very confused can anyone please help me solve these 2 problems:

1. A passenger at the rear of a train traveling at 15 m/s relative to Earth throws a baseball with a speed of 15 m/s in the direction opposite the motion of the train. what is the velocity of the baseball relative to Earth as it leaves the thrower's hand?

2. A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0* (degrees) east of north. what is the velocity of the dog relative to the road?

 

[HCW212]

Ahh im very confused can anyone please help me solve these 2 problems:

1. A passenger at the rear of a train traveling at 15 m/s relative to Earth throws a baseball with a speed of 15 m/s in the direction opposite the motion of the train. what is the velocity of the baseball relative to Earth as it leaves the thrower's hand?

2. A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0* (degrees) east of north. what is the velocity of the dog relative to the road?

Don't quote me on the answers but here's what I think:

1. 0 m/s.

2. 25 m/s + 1.75cos35

Do you have the answers?

 

[TomSawyer]

Hey guys, i'm solving the MCAT 1001 book, and i've been stuck on this problem for a while. I'd appreicate any help possible.

215) The following experiment takes place on the moon. A 2kg block rests on a flat board. One end of the board is slowly lifted until the block begins to slide. If the block begins to slide when the board is at an angle of 30 degrees with the horizontal, what is the coefficient of static friction between the block and the board?
A. 0.1
B. 0.6
C. 1
D. 2

I know Ff=Fn*uk but since this is on the moon i thought i'd have to find the acceleration due to gravity on the moon. I used

Fg=Gm1m2/r^2 and since Fg=mg i thought i was able to find the acc. of grav on the moon plugging in the masses of the earth and moon and dividing it by the mean distance between them. Unfortunately my answer doesn't coincide with any of the choices. I'd appreciate any help. thanks in advance.

 

[iA-MD2013]

Fg=Gm1m2/r^2 and since Fg=mg i thought i was able to find the acc. of grav on the moon plugging in the masses of the earth and moon and dividing it by the mean distance between them. Unfortunately my answer doesn't coincide with any of the choices. I'd appreciate any help. thanks in advance.

Your problem lies in calculating the force of gravity due to the moon. The masses you should be plugging into the equation are the mass of the moon and the mass of the object (not the mass of the earth). We want to find the force of gravity on the object not that on the moon due to the earth...You could also find the acceleration by g=Gm/r^2 where m is the mass of the moon.

 

[physics junkie]

Hey guys, i'm solving the MCAT 1001 book, and i've been stuck on this problem for a while. I'd appreicate any help possible.

215) The following experiment takes place on the moon. A 2kg block rests on a flat board. One end of the board is slowly lifted until the block begins to slide. If the block begins to slide when the board is at an angle of 30 degrees with the horizontal, what is the coefficient of static friction between the block and the board?
A. 0.1
B. 0.6
C. 1
D. 2

I know Ff=Fn*uk but since this is on the moon i thought i'd have to find the acceleration due to gravity on the moon. I used

Fg=Gm1m2/r^2 and since Fg=mg i thought i was able to find the acc. of grav on the moon plugging in the masses of the earth and moon and dividing it by the mean distance between them. Unfortunately my answer doesn't coincide with any of the choices. I'd appreciate any help. thanks in advance.

I swear there should be a thread on how to ask questions. You assume the mass of the moon is necessary for the problem but you don't include it? That's just laziness to expect someone to google that for you. Your notation needs work too--Fg=mg? Am I supposed to assume the first g is a subscript while the second g is the acceleration due to gravity?

Ok, enough bitching.

When you tilt the plank the force of gravity on the moon(which I'll denote with g) pushing the block down the incline is equal to mg*sin(30). The normal force acting perpendicular to the surface of the plank is equal to N=mg*cos(30). Since the frictional force is uN with u being the coefficient of static friction we can get somewhere. Draw a diagram of this before you continue and make sure you understand why the forces are adjusted by sin(30) and cos(30) respectively. If you still don't get it, ask.

Set the two equations equal to eachother.

mg*sin(30)=uN
or
mg*sin(30)=umg*cos(30)
Cancel out terms to get:
u=tan(30)
u=0.6

So what can we take away from the solution?

1.) the coefficient of friction is independent of the force of gravity.
2.) the coefficient of friction is independent of the mass of the object.
3.) it turns out the coefficient is intimately related(actually it's defined by) the tangent of the angle that an object will slip at.

You could have probably guessed this answer as the correct one from it's magnitude. Very few surfaces have a coefficient of friction of 1 or greater. In fact it's a widely believed myth that 1 is the upper limit for the coefficient of static friction. 0.1 is a very slick surface. If you've done a few friction problems you know your answer will probably be between 0.3 and 0.7.

 

[TomSawyer]

Thanks for the answers guys, and sorry if i wasn't too clear on my subscripts. I just have one last question which is question 219.

219) A 2kg object is placed on a plane inclined at an angle of 30 degrees. If the coefficient of static friction is 1, and the coefficient of kinetic friction is 0.1, what is the net force on the block?
A. 0N
B. 2.7 N
C. 17.3 N
D. 37.3 N

Once again i listed the givens as
m (mass)= 2kg
theta=30 degrees
Us(static)=1
Uk(kinetic)=.1

I found all the forces and tried finding the vector sum of all of them, but for some reason i never get a answer thats one of the choices. If anyone could help me one last time for this question i'd really appreciate it. thanks!

 

[tncekm]

Weight = mg = 20N
F_parallel = 20Nsin30 = 20N*0.5 = 10N

F_normal = 20Ncos30 = 17.4N
F_static ≤ 1(17.4) ≤ 17.4N

Since F_static is 17.4N it will resist a force up to 17.4N, so it can resist 10N; and mg = F_normal, the ΣF = 0 and the block doesn't move.

 

[TomSawyer]

365) A car moving at 35 m/s on dry pavement, skids to a stop in 7 seconds. What is the coefficient of friction between the car's tires and the pavement?

A. 0.2
B. 0.5
C. 1
D. 2

I can't figure this out without a given mass. Is there a way to find the mass or is there a way to solve it without using the mass? Help please!

 

[olygt]

Fiber optic cables can transmit light, b/c light in the cable achieves total internal reflection as it travels. From which material could the BEST fiber optical cable be made for use in air, if the best cable is one that would let no light rays escape into the air, regardless of the degree to which the cable may be bent?

A) A aerogel, with n= 1.05
B) a polymer gel, with n = 1.35
C) Glass, with n= 1.50
D) Plastic, with n = 1.60

 

[RSAgator]

365) A car moving at 35 m/s on dry pavement, skids to a stop in 7 seconds. What is the coefficient of friction between the car's tires and the pavement?

A. 0.2
B. 0.5
C. 1
D. 2

I can't figure this out without a given mass. Is there a way to find the mass or is there a way to solve it without using the mass? Help please!

Well I would think the following:

35m/s, takes 7 seconds to come to a stop, so acceleration has to be 5m/s^2.

Since this acceleration is due to friction, the frictional force

F = m*a = u*m*g ==> a = u*g where a = 5, g = 10, thus u = 5/10 = .5

 

[RSAgator]

Fiber optic cables can transmit light, b/c light in the cable achieves total internal reflection as it travels. From which material could the BEST fiber optical cable be made for use in air, if the best cable is one that would let no light rays escape into the air, regardless of the degree to which the cable may be bent?

A) A aerogel, with n= 1.05
B) a polymer gel, with n = 1.35
C) Glass, with n= 1.50
D) Plastic, with n = 1.60

I would have to say D. Total internal reflection occurs at the critical angle and any angle above. Critical angle = arcsin(n_air/n_medium). Since n_air is going to be constant at 1, the angle will be smallest when n_medium is largest, which would make it a better material because total internal reflection occurs in a wider range of angles. Comparing A to D, you would find that A results in total internal reflection at angles above around 80 degrees whereas D occurs at angles around 40 degrees, so D would be better.

 

[tncekm]

I would have to say D. Total internal reflection occurs at the critical angle and any angle above. Critical angle = arcsin(n_air/n_medium). Since n_air is going to be constant at 1, the angle will be smallest when n_medium is largest, which would make it a better material because total internal reflection occurs in a wider range of angles. Comparing A to D, you would find that A results in total internal reflection at angles above around 80 degrees whereas D occurs at angles around 40 degrees, so D would be better.

Great technical explanation. However, to "simplify" it to a more intuitive answer, the bigger the disparity in the indecies of refraction, the larger the "bend". For a TIR problem you're thinking about bending away from the normal line. So, you want to go from high index of refraction (wave travels slower) to low index of refraction (wave travels faster) because that will cause a bend away from the normal line. And if you go from a very low index of refraction to a very high index of refraction, it will bend a great deal away from the normal.

 

[olygt]

I would have to say D. Total internal reflection occurs at the critical angle and any angle above. Critical angle = arcsin(n_air/n_medium). Since n_air is going to be constant at 1, the angle will be smallest when n_medium is largest, which would make it a better material because total internal reflection occurs in a wider range of angles. Comparing A to D, you would find that A results in total internal reflection at angles above around 80 degrees whereas D occurs at angles around 40 degrees, so D would be better.

The answer is D, but I'm still getting confused on this one. I can see why the sine of a refracted angle can never be greater than one b/c it will start internally reflecting. I drew a picture and the bigger the angle was to the normal the bigger the angle of reflection, wouldn't that make it reflect more? And the bigger the angle means the smaller the value of n.

 

[olygt]

When an object is moved from within the focal length ofa diverging lens to a distance beyond the focal length, the image is observed to:

a) change from upright to inverted, and move away from the lens.
B) change from inverted to upright and move closer to the lens
c) remain upright, and move away from the lens
d) remain inverted and move closer to the lens

I drew this out about five times and still am getting the wrong answer. My answer was remained upright and closer to the lens but this isn't even an answer choice.

 

[tncekm]

Make an arbitrary focal point for a diverging lens and then try doing the calculation at two points. Zero, and infinity. See if that helps