To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.
Ff = 0.5 * 16 * 9.8 = 78.4 N
a = 4.9 m/s^2
If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?
In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.
Ff = 0.28 * 16 * 9.8 = 43.904 N
Net force = 78.4 – 43.904 = 34.496 N
To determine the acceleration, divide by the mass of the crate.
a = 34.496 ÷ 16 = 2.156 m/s^2
Answer:
354.72 m/s
Explanation:
= mass of lead bullet
= specific heat of lead = 128 J/(kg °C)
= Latent heat of fusion of lead = 24500 J/kg
= initial temperature = 27.4 °C
= final temperature = melting point of lead = 327.5 °C
= Speed of lead bullet
Using conservation of energy
Kinetic energy of bullet = Heat required for change of temperature + Heat of melting

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