You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 m/s relative to the earth at an angle of 35.0∘ above the horizontal. Your mass is 72.0 kg and the rock’s mass is 3.50 kg . What is your speed after you throw the rock?

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.