Recent content by maladroit

I figured out what I did wrong in this situation... cos-1(0) could be either pi/2 or -pi/2, and because the motion is coming to the end of a complete cycle I should have used -pi/2. To answer your question, I determined the phase shift=phi by solving the position equation. I knew at t=0...

Homework Statement At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the...

My mistake was in the bounds, which should have been from 1 to 34. Thank you for your help!

Homework Statement Find the area bounded between the two curves y=34ln(x) and y=xln(x) Homework Equations Integration by parts: \intudv= uv-\intvdu The Attempt at a Solution First I found the intersection points of the two equation to set the upper and lower bounds. The lower...

nevermind, I've got it! sorry for my faulty answer and thank you for correcting my mistake.

why would the limits of integration be from 1 to 3 if you are integrating with respect to the x axis?

I think the mistake you are making is in the formula that you are integrating. It seems like you were thinking washers while using the shell equation. The general equation to use for the shell method is 2\pi\int R dx(dx can change depending on which variable you are integrating with respect...

all you have to do is multiply it to the two solutions so say the factored quadratic is (x-4)(x+4), so the solutions are 4 and -4. just multiply the rad5 to those two answers.

Momentum is always conserved in a closed system-that is, if there are no external forces that could effect the situation. Hope that helps!

very happy to help:D

because the velocity is downward it is negative, as well as the acceleration. you can't interchange them. it is just important to note the direction of the velocity depending on the equation you decide to use!

also correct! acceleration is -9.8 m/s2. something to think about--what is the direction of the velocity?

exactly! 2 m/s would be the initial velocity.

Glad you understand it!

Is the initial velocity 66 or 68 m/s? If it is 68 m/s, your answer is very close to mine---I'm sure just rounding differences. Just double check the problem.