<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>
The bouncy ball experiences the greater momentum change.
To understand why, you need to remember that momentum is actually
a vector quantity ... it has a size AND it has a direction too.
The putty and the ball have the same mass, and you throw them
with the same speed. So, on the way from your hand to the wall,
they both have the same momentum.
Call it " M in the direction toward the wall ".
After they both hit the wall:
-- The putty has zero momentum.
Its momentum changed by an amount of M .
-- The ball has momentum of " M in the direction away from the wall ".
Its momentum changed by an amount of 2M .
Answer:
The himalayan mountain are at a divergent boundary
Answer:
If external is force applied
Explanation:
