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Agata [3.3K]
3 years ago
10

A man claims that he can hold onto a 16.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.05 s.
Find the magnitude of the average force needed to hold onto the child.
N
Physics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

Explanation:

Given;

Mass of child m = 16 kg

Speed of each car v = 59.0 mi/h = 26.37536 m/s

Time t = 0.05s

Applying the impulse momentum equation;

Impulse = change in momentum

Ft = ∆(mv)

F = ∆(mv)/t

F = m(∆v)/t

Where;

F = force

t = time

m = mass

v = velocity

Since the final speed of the car is zero(at rest) then;

∆v = 0 - v = -26.37536 m/s

Substituting the given values;

F = 16×-26.37536/0.05

F = -8440.1152 N

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

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A horizontal 745 N merry-go-round of radius
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The kinetic energy of the merry-go round after 3.62  = ?

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m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

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Substituting the values

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Substituting the values,

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Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

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Now KE (100% rotational) after 3.62s is:

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KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

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