Recent content by Shadow Cloud

And pi/2 for cos is 0...yes you're right, the calculator did mess me up in this case. Thank you for the help.

Oh I haven't, I just forgot to include it when I posted the problem. I still do not understand why I am not getting the right answer when all I have to do is plug in 4.0 for T.

Oh I'm sorry, I forgot to include the pi in the equation.

The function x = (2.0 m) cos[(2pi rad/s)t + pi/2 rad] gives the simple harmonic motion of a body. Find the following values at t = 4.0 s. (a) the displacement: ____m Correct me if I am wrong, but to get x all I have to do is just plug 4.0 s in for t in that equation mentioned above right? I...

Oh nevermind, I finally figure out the necessity of the root 2. It's pertains to the length of a diagonal of a square.

I'm kind of following this, but why is the root 2 necessary and where would I put it?

What is rr in the equation of R = rr that you gave?

In Figure 13-34, a square of edge length 22.0 cm is formed by four spheres of masses m1 = 7.00 g, m2 = 3.50 g, m3 = 1.50 g, and m4 = 7.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.30 g? Okay this is what I did so far...

Figure 11-26 shows three particles of the same mass and the same constant speed moving as indicated by the velocity vectors. Points a, b, c, and d form a square, with point e at the center. Rank the points according to the magnitude of the net angular momentum of the three-particle system when...

I still don't have a clue as to what I should do. Okay providing I'm on the right track. I should start out with... T=Ia T = mr^2a a = T / mr^2 The problem is I don't know what else to do from here. Since the rod is massless does this mean that m is ignored or something? There's no force...

Oh I'm sorry by my last statement of "I know you need mass for t = Ia , but the fact that there is no value for m is what makes me confused as to what to do. I am surmizing that 9.8 m/s^2 factors into that problem somehow?" I was referring to this problem. Figure 10-43 shows particles 1 and...

Thank you for the reply. You are right though, I did get the correct answer for the tension. I have no idea what I did wrong before. I must had punched in the wrong values or something. How did I get the acceleration? Professly, I would like to understand this better. I originally got help with...

Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released. a.What is the magnitude of the initial acceleration of particle 1? b.What is...