The vertical velocity is affected by the acceleration of gravity (ignoring the effects of air resistance usually)
Answer:
Explanation:
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The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
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Answer:
0 J
Explanation:
From the diagram below; we would notice that the Force (F) = Tension (T)
Also the angle θ adjacent to the perpendicular line = 90 °
The Workdone W = F. d
W = Fd cos θ
W = Fd cos 90°
W = Fd (0)
W = 0 J
Hence the force is perpendicular to the direction of displacement and the net work done in a circular motion in one complete revolution is = 0