Answer:
Explanation:
Electron information needed to solve the question:
![E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.11x10%7B-31%7Dkg%2A3.0x10%5E%7B12%7Dm%2Fs%5E2%7D%7B-1.6x10%7B-19%7DC%7D-%5B%2819.0x10%5E3mj%2B18.0x10%5E3m%29xi%28400x10%5E%7B-6%7DT%29%5D)
![E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C](https://tex.z-dn.net/?f=E%3D-i17.08N%2FC-%5B7.6%28-k%29%2B7.2%28j%29%5DN%2FC)
1) Length of the wire.
2) Thickness of the wire.
3) Temperature.
4) Type of metal.
Hope this helps!
-Payshence
The hypothesis because its very hard to make and it confounds me
<span>Answer:
Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52.
Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
