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I didn't take AP Physics so I wouldn't know but I am taking an introductory physics class right now. I am not sure how hard AP Physics is but that sounds like a lot of materials. I would say optics is probably the easiest topic at least as far as math in concern. The formulas (at least in my...

I am sorry I can't really follow the work. Please include the units. I would approach this problem this way... First of all, I would sketch a P-V diagram. It starts at some pressure and volume and then undergoes an isobaric expansion tripling its side. With p on your vertical axis and v on...

If -2.62 is 1/f and the answer is indeed .34 for as in the answer key then o must be greater than 3 meters away. How is that possible? Do I somehow combine that 1.98 m with the other part a, which is .82 meters?

Homework Statement A small object of length dL is placed on the axis of a converging mirror of focal length f. The center of the object is a distance o from the mirror. Determine the length of the image and thus effective determine a longitudinal m' = dL'/dL for the lens. Homework...

Thanks for your help. I just realized that I need help for part b as well. b) 1/f = 1/o + 1/i The person's far point is .6 or .58 from the glasses. I want him to see at .25 m or .23 m from the glasses. 1/-f = 1/.58 + 1/-.23 1/-f = -2.62 1/f = 2.62 that is the power. I am...

Homework Statement a)A person has a near point of .6m and far point of 2m . What power of glasses would be needed for the person to see infinitely far away. Assume that glasses are 2cm from the person's eyes. If the person used those glasses, what would his new near point be? b) What...

Thank you very much for the help

Homework Statement .15 kg of steam at 100 degree Celcius is mixed with .2kg of water at 50 degree and .2 kg of ice at 0 degree. Determine the final temperature and phase(s) of the final state. Assume that no energy is lost to the surroundings. Homework Equations The Attempt at a...

Homework Statement Someone is putting a small shed in their backyard. The roof is not nailed down so gravity alone is holding it down. The wind suddenly flows across the top of the roogf at 20 m/s. The air inside the shed is 1.01 x 10^15 Pa, the normal atmospheric pressure. The roof has an...

Never mind I got it now. Thanks a lot for your help.

For the object at rest I got B - mg = 0 When it's pushed down B - mg -f = -ma Since B =mg f = ma I already know this though so how does it help prove that it's in simple harmonic motion with a repeating pattern?

When the object is at rest, I have mg pulling down and buoyant force pushing up. They are equal in magnitude. The net torque is also zero. When the object is pushed down I have f pushing down, mg pulling down, and buoyant force pushing up. This extra f is enough to push it down. My net...

Homework Statement A mass of density d floats in a liquid of density d_L. The mass is then pushed down a distance x and let go. Use Newton's Second Law to demonstrate that the mass will undergo simple harmonic motion. Recall that the SHM equation is d^2x/dt^2 + w^2*x = 0. Assume there is...

Thanks. I will try it again.

Homework Statement A 1kg hollow sphere of volume .00419m^3 is released from rest at the bottom of a 2m deep pool of water (density of water is 1000kg/m^3). The sphere accelerates upwards and flies out of the pool. How high does it get above the pool before coming down. Ignore the brief time...

Homework Statement A small child floats in a boat in a swimming pool. The boat contains an anchor The child drops the metal anchor into the water and watches it sink to the bottom. Meanwhile, the water level in the pool (measured by its height on the side of the pool) actually drops. Why...

It means that at most the pendulum travels 30 degrees in either direction from equilibrium point.

Homework Statement I have to do this for a lab. Draw and label the free body diaggram of a pendulum when it is at its maximum amplitude of 30 degrees. The magnitudes of the vectors must be correctly scaled and the directions correct. Homework Equations The Attempt at a Solution...

I have no idea on how to start these two problems. I haven't done any work, but please give me some hints on how to get started. The two problems are here http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1461065539 For problem 14, the anskwer are...

Thanks. For part b, I still don't know where v^2/gl come from. G is an acceleration due to gravity, l is a length, and v^2 is m^2/s^2. After division, v^2/gl is just a number with no unit. However, I don't know where the term come from.

I made some mistakes. L = 2r, r = (1/2)L w=v/r w^2=v^2/r^2 or v^2/(1/2*L)^2 = 4v^2/L Is M in this case 2M1 since we have two masses in the system before collision. Should the mass be 2M1 + M2 after collision?

Angular momentum = m*r*v I know that I for a rod is (1/12)ML^2 so L = Iw then L = (1/12)ML^2*v^2/L^2 L = (1/12)MV^2*L^2 r = 2L If I have IW = I_f*W_f W_f = I/I_f * W W_f = (1/12)ML^2/(I_f) * W I am stuck. I don't know what to do from here.

I am having problems with this part. I know that N_a + N_b = Mg, and I know that N_a is greater than N_b because the person is closer to A. N_a + N_b = 852.6 Since tension is internal as you said, I should not include it in the torque equation - right? So 0 = -T_na + T_nb -...

I know that net torque is zero 0 = -T_na + T_nb -T_mg + T_t I got 0 = -1.25N_a + 1.25N_b - .3123*mg + T_t I know that T_t is there, but I don't know the direction. Is it pushing right or left? After I get this, do I plug into those equations to get the rest. The answers are...

The problem is problem 6 at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460003525 A long thin rod of mass M and length L with two balls of mass M1 (same mass for both) attached is allowed to rotate about the horizontal axis shown. The bar is...

With net torque, I have mg*R + F(R-H). So F = mg*R/(R-H) From the way it seems, I got the F(R-H) part right, but mg*R is still not quite right. HOw do I get SQRT(2RH-H^2)?

Yes, http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460657842 I am sorry for leaving it out.

I am not sure about this...is the moment of inertia with respect to the contact point MR^2 + R^2 with the parallel axis theorem? I know that whatever force the wheel pushes against the contact point, the same force will push the wheel back. But I am not sure about how to find that force for...

Each side of a stepladder is 8m long and there is a hinge at the top. An 87kg man is 6m up one side and there is a 2.5m long rope connecting the two sides. The rope is exactly halfway up the ladder. Assume that the floor is frictionless and neglect the weight of the ladder. Find the tension...

What minimum force F applied horizontally at the axle of the wheel is necessary to raise the wheel of mass M and radius R over a step of height H. A picture for this is at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460005538 The answer...

Okay, thanks very much for your help tonight.

Oh I see now. Thanks very much.

I am not sure if this is right. Mg should be from the center of the pole. So the distance is 7.5/2*cos(37). If I plug this back in, I would get T2 = 48.38 N. Is this correct? I know that T2 = m2*g so I should be able to find m2. However, how do I find tension of the first rope since...

I would have mg pointing down from the center of the block and N pointing up. I would also have friction pointing to the left. The distance between the center and the rotational axis is H. However, why do we multiply this mg by L when L is the horizontal distance. Does this have anything to...

When I do that I get 8*9.8*6.31 + 5*9.8*7.5* cos(37) = 7.5cos(37) * T2 T2= 88.19 This answer is incorrect. The answer key says that the answers in this problems are 139N, 49N, 139N, 127.4 N for each of the forces. 88.19 is not any of these. What did I do wrong?

A force F applied at point A is just large enough to tip over a block of mass M. The block will rotate about its bottom right corner. Find the magnitude of F. The picture for this is problem 13 at this link...

A traffic light hangs from a structure shown at http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460004546 The uniform aluminum pole has a mass of 8 kg and is 7.5m long. The traffice light has a mass of 5kg. The cables are massless. Cable 1 is...

A ladder leans against a frictionlee wall. If the bottom of the ladder is pushed closer to the wall (so theta increases) which of the following change? Why or why not? Assume the ladder is in static equilibrium. a) Normal force exerted by the ground b) Normal force exerted by the wall...

Thank you.

A 20kg girl stands on a large turntable 4m from its center. The turntable has a moment of inertia (about its roation axis) of 200kgm^2. Assume that the girl is a point mass. a) Assume that the girl and turntable are rotating at 1 rad/s. If the girl walks inwards to a radius of 3m and...

A 2kg object has a position vector r = -4t i + 3t^2 j (m). Find its angular momentum relative to the origin as a function of time. The correct answer is -24t^2 k (kgm^2/s). I know that L = r*p and p is mv so I differentiate r to get v v = -4 i + 6t j do I just multiply 2 and the...

For part 2, I know that I = I_cm + mh^2, where h is the distance between the axes. So is h, L-r in this case? How can I get I about an axis through M2. Is it I = M2*L^2 + M1*(L-r)^2 ?

Wouldn't I use the formula I = Mx^2 then where x is the distance down from the rod?

Two masses (M1 and M2) are connected by a thin massless rod of length L. a) Determine the location of an axis perpendicular to the rod) which minimizes the moment of inertia of the system. The answer is "with M1 at the origin x = M2*L/(M1+M2) I know that I = Sum of ( M1*r1^2)...

Yes. So I have to include both in the problem. The given R is the outer radius, and the inner radius is not explicitly mentioned.

I know that the yoyo is a cylinder that spins counter clockwise going down and clockwise going up. It is attached to a fixed place a tthe top. I don't know what I am missing.

A yoyo is allowed to drop freely with a string held fixed in place at the top. Assume that the yoyo is a uniform disk of mass M and radius R. 1) Use Newton's Second Law to find the linear acceleration of the yoyo and the tension in the rope. My answer key says that the answers are 2/3 g...

I don't really understand what do. From what you told me, The net force in the horizontal direction is ma, with friction being mu*mg The net torque is I*(alpha), and I know that I for a sphere is (2/5)MR^2 alpha is a/R so I = (2/5)*MA*R and F = mu*Mg The answer has a 12 and...

Hello everyone. Thanks for all the help today. I know I have asked a lot of questions. This is my last one today. A spherical billard ball is sliding with a speed v_o. It has a radius R, mass M, and there is a friction coefficient mu. Determine the distance the ball travels before it...

Since T1 is trying to cause the pulley to rotate counterclockwise, while T2 is causing it to rotate clockwise, would it work if I have T1*R-T2*R = I (-alpha)