Recent content by StrawHat

I put in F~10 25N.

The answer gives me this error: "Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

Homework Statement Homework Equations \vec{F} = k\stackrel{q1q2}{r^{2}} The Attempt at a Solution 6.022e23*(0.505) = 3.041e23C <-- electrons 6.022e23*(0.495) = 2.981e23C <-- protons 3.041e23 - 2.981e23 = 6e21C <-- the difference between the two charges \vec{F}_{e} =...

m(dv/dt)=k(dx/dt) m(dv/dt)=kv?

dy/dt=cos(t), d2y/dt2=-sin(t)? Thank you for your help thus far, but it's 4AM over here in the EST timezone, so I must go to bed. I will check back on this thread in five hours or so.

F = -kx W = Fd =∫Fnetdx Wtotal = ΔK ΔU = 1/2kx2

x=ma/k? I am totally lost... I have ax=4. Am I on the right track?

d/dt(kx)=d/dt(ma) k(dx/dt)=m(da/dt) kv(t)=m(da/dt)?

kx=ma, perhaps? If so, should I use kx = m(dv/dt)? But then how will I obtain a value for velocity?

Homework Statement A 250 gram mass is connected to a spring and executes simple harmonic motion. The period of motion is 0.5 seconds and the total mechanical energy is 0.50J. What is the amplitude of motion? Homework Equations ΔU = 1/2kx2 The Attempt at a Solution I get 1/2kx2 =...

Okay, I've tried part B of the question and this is what I've done so far. Kfinal = 1/2Iω2 Kinitial = 1/2mv2 I divided both of those to get (Iω2)/(mvinitial) Calculating that further, and I get this: I'm guessing that it's not Kfinal/Kinitial, but (Kinitial - Kfinal)/Kfinal? EDIT: Okay, I...

Oh okay, I thought r was supposed to be the distance between the rotation point and the end of the rod. I got the right answer of ω = (mvi)/(1/6Md+1/2md) Thanks for your help!

Why wouldn't it be d/2? If the diameter of the circular motion is d, then wouldn't the radius be d/2?

1/12mr2

[SOVLED] Angular speed of rod after projectile collides into it Homework Statement Homework Equations L = r x p Iparallel = ICM + md2 The Attempt at a Solution L = r x p = mvir = mvid/2 mvi(d/2) = (Iparallel + md2)ω mvi(d/2) = (1/12M(d/2)^2 + m(d/2)^2)ω Plugging and solving...