Recent content by RavenBlackwolf

Where as in Δx? Don't I need the vf to find that though? That's the only reason I wanted to find vf at all. All I have is the .16m when the spring is compressed to the fullest and v=0 I believe ax is zero there too but that can't be the answer. I feel like I'm missing something but I can't...

The acceleration in the x direction would be 0 correct? I thought of that before but I got lost because wouldn't I need time if I'm using the kinematics equations? Are they what I should use to find vf or should I be using an energy concept?

Homework Statement A 4.0 kg block starts at rest and slides a distance d down a frictionless 35.0 incline, where it runs into a spring. The block slides an additional 16.0 cm before it is brought to rest momentarily by compressing the spring, whose spring constant is 429...

12.9 is actually the number I was looking for so I think I found the correct process for what my professor wanted. Thanks for your help.

T=.72m*15kg*9.8m/s2*sin(17)=30.94 30.94=I(2.4rad/s2) so I=12.9

Tnet=I∝ you mean this one? But then what is the net torque? I know torque is usually force multiplied by how far away from the axis you are so then would the force be the force of gravity at this angle multiplied by .72m?

No, there's more. I'm going to take a minute to try out a different method of solving.

My guess is the one with tangential acceleration. And I only use that symbol because that's what my professor uses. So you're saying if I do what I did but using the correct form of acceleration I should get the correct answer?

I didn't really know what I was doing, so I tried out multiplying the mass by acceleration for a force F=(15)(2.4)=36 then I did T=(.72)(36sin(17))=7.6 7.6=I(2.4) I=3.2 which is, not correct. And I'm also not clear as to the thought process.

Homework Statement A small pine tree has a mass of 15kg. Its center of mass is located at .72m from the ground. Its trunk is sawed through at ground level, causing the tree to fall, with the severed trunk acting as the pivot point. At the instant the falling tree makes a 17° angle with the...

Oh I got it. The variable k is wavenumber, my professor never made that clear. So using: ω=2π/T, T being the inverse of 39Hz or T=.0256 ω=245.04 v=ω/k=43m/s so the answer is c. This would've been a whole lot easier if that had been clear in my notes. Oops.

What do you mean by "the good one." I don't understand.

wave speed? I have v=ω/k=λ/T=λf I believe but what I don't know is whether or not that's what propagation speed is.

Oh I didn't know there was one, sorry.

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown > A transverse wave on a string has an amplitude of 16cm, a wavenumber of 5.7m-1, and a frequency of 39Hz. What is the propagation speed of that wave? (a) 6.84 m/s (b) 39.2 m/s (c) 43 m/s (d) 6.24...