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klio [65]
3 years ago
13

A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall

of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall
Physics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

Explanation:

The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall

= mv = 6 x 13.3 = 79.8 kg m /s

If resisting force of wall be F , impulse on the wrench = F x time

= F x .07

Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s

So F x .07 = 79.8

F = 1140 N .

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Answer:

ω = 5.41 rad/s

Explanation:

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So, the initial angular momentum is

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The final angular momentum includes the rod and the bullet together. So,

L_2 = I\omega\\I = I_{rod} + I_{bullet} = \frac{1}{12}m_r L^2 + m_b(\frac{L}{4})^2 = \frac{1}{12}3(0.25)^2 + (4.5\times 10^{-3})(\frac{0.25}{4})^2 = 0.0156~{\rm kg.m^2}\\L_2  = L_1 = I\omega\\0.0844 = 0.0156\omega\\\omega = 5.41~{\rm rad/s}

5 0
3 years ago
An object is dropped and falls freely to the ground with an acceleration of g. If it is thrown upward at an angle instead, its a
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Answer:

g

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