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Homework Statement Determine the specific latent heat of fusion of ice given the amount of time it takes the ice to melt (150 seconds). Energy is supplied to the ice at 530 W. The sample is .25 kg. Homework Equations Q= ml The Attempt at a Solution (Time * 530W) = .25kg * L 150...

Homework Statement It takes 15 seconds to warm ice 15 degrees ( -15 degrees celsius to 0 degrees celsius). The ice has a mass of .25 kg. The energy is supplied at 530 W to the ice. Homework Equations Q = mc delta T The Attempt at a Solution I plugged in all the information for Q...

Would you mind helping me with the follow up question? It is: The resistor R is replaced with an electromagnet and a switch. When the current is switched on, small pieces of iron, initially on the ground below the electromagnet, are attracted to, and stick to, the electromagnet. (A) State...

I was thinking though, normally to find (I) don't you need to have the total resistance in the circuit? Since r and R are in series it should be V = I (R + r) right? 1.4 = .2(6 + r) 1.4/.2 = 6 + r 7 = 6 + r 1 = r is that correct?

alright so using that, 1.2/6= .2 so I is .2. In the equation its then E = .2 + .2r In order to solve for r though I need to know what E is. What does E stand for?

so I found V=IR for the 6 ohm resistor 1.4= I (6) so I equal .2333 so in E = DeltaV + Ir E = .2 + (.2333)(r) Is this correct? What does E stand for again?

Homework Statement The cell supplies 8.1×10^3 J of energy when 5.8×10^3 C of charge moves completely round the circuit. The current in the circuit is constant. A. Deduce that the e.m.f of the cell E is about 1.4 V B. The resistor R has resistance 6.0 Ω. The potential difference between its...

I would need to convert g=9.81 m/s to g=0.00981 km/s to get coulombs correct? If that does work, T= (0.00981 * 1)/cos(11.14) T=.009998389 T*sin(11.14) = F(elec) = qE .009998389 * sin(11.14) =qE [.009998389 * sin(11.14)]/E = q .0019317587 / E = q .0019317587 / 9000 = q 2.146398 e-7 Correct?

Ok so cos-1 (52/53) = 11.14 degrees So can I plug that into the equations I was given up above? T*cos(11.14) = mg T*sin(11.14) = F(elec) = qE

I tried this, but the answer was incorrect. Can you see what I did wrong? | \ | \ | \ 52 | \ 53 | \ | T \ * So the point charge is basically the dobber of a simple pendulum. In this case you have two forces acting on it...

Homework Statement I need some help solving this question, it is the last one I have to do and I'm not sure how to solve it. A point charge (m = 1.0 g) at the end of an insulating string of length 53 cm is observed to be in equilibrium in a uniform horizontal electric field of 9000 N/C...

{(8.99*10^9)(-9*10^-6)}/.015^2 = -359,600,000 {(8.99*10^9)(6*10^-6)}/.015^2 = 239, 733, 333 Adding these two up -359,600,000 + 239, 733, 333 = -119, 866, 666 -119, 866, 666 N/C seems ridiculous, what did I do here?

You say "for the 1q" What is the 1q? Is it the one quadrant? You gave me two equations for the vectors for attractive and repulsive forces, do I need to add these two vectors? I don't know what I need to do. I can plug the numbers in for these equations: q2/d2 i - q2/d2 j q2/2d2*cosθ i +...

so sigma is the sum right? E = k(6x10^-6)/r2 is added with: E = k(-9*10^-6)/r2 once I've added these two Es up is that my final answer?

I'm still a bit confused by what your telling me, I talked to a friend and here is what he said: To calculate the forces you really only need to do it once and then apply the direction as if you were doing it in each quadrant. . K= 9x109. r = .9m for doing the adj and r = √.92+.92 for the ones...

So there will therefore be two answers? I'm confused by what your saying "For E you are using the q of the charges you are measuring." I realize this, but do I need to set up two equations then? Or do I have to add the charges?

So for each charge F = q* E And E is the sum of the forces? And where are the X and Y planes drawn for this situation?

What Q value would you use though?

Homework Statement What are the magnitude and direction of the electric field at a point midway between a -9.0 µC and a +6.0 µC charge 3.0 cm apart? Assume no other charges are nearby. Homework Equations Fe= {K(q1)(q2)}/d^2 k = (8.99 x 10^9) The Attempt at a Solution First I...

Homework Statement Two Q = +5.86 mC charges are placed at opposite corners of a square d = 0.9 m on each side, and -Q = -5.86 mC charges are placed at the remaining corners. Determine the magnitude and direction of the force on each charge. Homework Equations Coulomb's Law The...

Homework Statement If provided only the spring constant K and the mass of an object undergoing simple harmonic motion how do you find the amplitude? Homework Equations Angular Frequency = sqrt (k/m) Period = 1/t The Attempt at a Solution Using the two equations above you can derive...

Homework Statement After falling 5m a rock has a velocity of v. What is the total distance the rock must fall to get a speed of 2v? Homework Equations GPE=mgh Ke=1/2mv^2 The Attempt at a Solution Should I set both equations equal to each other? I'm not even sure how to begin this one.

Alright thanks a lot. I really appreciate the time you spent helping me.

Right, I understand that bit but if Fnet=R*sine(14) then I need to use the R value from above right? So using the two sides I know 8500N and then 2100N (the rounded up horizontal component of R) so I use the Pythagorean theorum right? a^2 + b^2 =C^2 to find the hypotenuse?

to find mass 8500N / 9.81 = 866 kgs so I multiply that by V^2/r The problem is that when this question is asked, they have not yet provided the radius so I don't have that, nor do I have the velocity

Oh alright then, that makes sense to me. Do you understand what the other question is asking about "state the magnitude and direction of the resultant force acting on the car" What is this resultant force?

"The 8500N looks like the weight (m*g). If you set the 2 terms equal to each other then all you don't know is v and happily that's all they are asking. m*g*sin14 is gravity (m*v2Cos14)/r is the centrifugal force (acting outward against the bank)" But Rl.Bhat's explanation works as well...

Ok, I actually think I understand it now. I was just confusing myself with the angles. This is so similar to the problems we did in dynamics I feel kind of stupid now. We were talking about before how the horizontal component of the force has to be sin(14) * 8500. I understand that bit but I...

"The downward force of gravity along the banked incline must be balanced by the upward along the bank component of the centrifugal force shouldn't it?" I agree with this, but all this means is that the vertical component of R is 8500N. I just don't see how this ties into finding the maximum...

Thank you so much. So the m*g*sin14 is my centripetal force?

I have no clue how to find centrifugal force, if you would provide me with the sine/cosine explanation I'd be forever grateful. I did draw a force diagram. Is the pull of gravity sine(14) then you multiply this by something to get the Force? I would be extremely grateful if you'd post a...

Homework Statement Homework Equations The Attempt at a Solution I only know how to do this with the coefficient of friction, and since there is no friction here I don't know how to do it. Normally the equation is = Fstatic friction max = v^2/r mu * g * r = v^2 and then finally...

Well I don't know how to calculate the r as a whole I think I know how to get the component of r and that is the vertical component: Cos(theta) * r = 8500. Is the angle 14 degrees? If it's not it must be 76 because all the angles must add up to 90 Hope that helps

I was trying to find acceleration when I found 9.28 1.4*(2*pi*.6)=5.277 m/s (velocity) v^2/r=a so 27.84/.600=46.41m/s/s (acceleration) and I found it when I put it in the Fnet=ma equation I mutliplied mass by the acceleration I had found (46.41*.20) which equals 9.28 I found 1.962 as mg...

Well the website I had to submit it to said it was wrong, so maybe it's wrong?

Wait, I have to use tan^-1 right? if that's the case I get 78.06 degrees which seems a lot more reasonable. However, I checked and the answer is wrong, does anyone know what I did incorrectly?

9.28= FT*sin (theta) 1.96200=ft*cos (theta) so Ft cancels and i divide 9.28/1.962=4.72986748 and then tan (4.72986748) to get theta? The problem is that only gives me .0827 and i have a feeling that is not what my angle should be.

^^^^^ Bump I really need this for my quiz review packet, please help!

I'm not sure how to do that, I understand both of the equations but how do you divide one equation by the other? Are you telling me to substitute? And are you sure about your two equations? I have something similar in my first post but the sin and cos are switched you said: m*v^2/r = FT*cos...

Homework Statement Estimate the force a person must exert on a string attached to a 0.200 kg ball to make the ball revolve in a horizontal circle of radius 0.600 m. The ball makes 1.40 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the...

Homework Statement A ball on the end of a string is cleverly revolved at a uniform rate in a vertical circle of radius 65.0 cm, as shown in Fig. 5-33. Its speed is 4.00 m/s and its mass is 0.300 kg. (a) Calculate the tension in the string when the ball is at the top of its path. (b)...

Thanks a lot for the help, I've got it now.

Wow that was an infuriating struggle for such a simple answer. I figured out that the only thing that was wrong with my answer was the negative sign. I tried -484 instead of 484, I'm not sure why it has to be positive though. Could it be negative?

This is fairly frustrating, I have no idea what to do. Well this is what I think the change in momentum is (the problem is that I have no idea if this is right): (M1*V1) + (M2*V2) (110 x 4.4) + (M2*V2) = change in P so I have 484 + M2V2 = delta P how do I find the tacklers mass or...

Sorry, I was worried that the thread was too far down for it to be seen again. I don't seem to quite get it, I know that X is equal to force but isn't the force equal to impulse? Impulse = FΔt = Δ(mv). Do I need to multiply the Force that I find by the change in time? .75X = -484

Right but I thought that this: .75X= 110 * 0 + 110 * 4.14=Impulse = FΔt = Δ(mv) The .75 is delta t the X is the force the change in momentum is (110 *0) + (110*4.14) I don't see what I'm doing wrong, that is the whole problem

Impulse and Conservation of Momentum Homework Statement A 110 kg fullback is running at 4.4 m/s to the east and is stopped in 0.75 s by a head-on tackle by a tackler running due west. A. Calculate the impulse exerted on the fullback. B. Calculate the impulse exerted on the tackler. C...

Homework Statement A 110 kg fullback is running at 4.4 m/s to the east and is stopped in 0.75 s by a head-on tackle by a tackler running due west. A. Calculate the impulse exerted on the fullback. B. Calculate the impulse exerted on the tackler. C. Calculate the average force exerted on...

Alright well that's wrong and I'm not sure why Cos(26.5) * H = A Cos(26.5) * (60 x 9.81) = A which ends up being 118.71N When I plug that in I get Fpush - (.17 x 118.97) = 4.14 Fpush = 24.36 And I this doesn't work, so I'm not sure what's going wrong

So then do I just do Cosine (26.5) = A/H and in this case adjacent is the force I am looking for and H is gravity Is the gravity Fg? so I can substitute Fg=ma 60 * 9.81 = Fg Then I have Cosine (26.5) * (60 * 9.81) = A Is this correct?