Answer:

Explanation:
From the question we are told that:
Distance 
Angle 
Force 
Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by



Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components


The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components


So we have enough information to solve for the components of the acceleration vector,
and
:


The acceleration vector then has direction
where

Answer:
91.84 m/s²
Explanation:
velocity, v = 600 m/s
acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2
Let the radius of the loop is r.
he experiences a centripetal force.
centripetal acceleration,
a = v² / r
39.2 x r = 600 x 600
r = 3600 / 39.2
r = 91.84 m/s²
Thus, the radius of the loop is 91.84 m/s².
Answer:
It gives our light which we need for probably everything.
Explanation:
Answer:
M g H = 1/2 M v^2 potential energy = kinetic energy
v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2
v = 10.8 m/s
(C)