A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13
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Answer:
Explanation:
u = 150 m/s
R = 800 m
The formula for the horizontal range is
where, θ is angle of projection.
Sin2θ = 0.348
2θ = 20.4°
θ = 10.2°
The range is same for the complementary angles.
So, the other angle is 90 - 10.2 = 79.8°
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :
So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.