Explanation:
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The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
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Answer:
2.000 s
Explanation:
Given:
a = 50.00 m/s²
v = 100.0 m/s
v₀ = 0 m/s
Find: t
v = at + v₀
(100.0 m/s) = (50.00 m/s²) t + (0 m/s)
t = 2.000 s
Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s
