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lyudmila [28]
3 years ago
7

A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts fo

r a total of 17.0 ms. What is the magnitude of the average force exerted on the sandbag by the ground during the collision?
Physics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

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So the frequency is equal to the reciprocal of the period:

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The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

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T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

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T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

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