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luda_lava [24]
3 years ago
14

A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) W

hat impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0156 s, what is the magnitude of the average force on the floor from the ball?
Physics
1 answer:
Anarel [89]3 years ago
3 0

Answer:

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Explanation:

Given that

V_1=21 m/s,V_2=9.8m/s

We know that impulse(J) impulse is a vector quantity

J=P_2-P_1

We know that P=mV

SoJ=m(V_2-V_1)

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

F=\dfrac{dP}{dt}

or we can say that

F=m\left(\dfrac{v_2-V_1}{dt}\right)

So F=\dfrac{40.04}{0.0156}

F=2566.66 N

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1.1 m/s²

Explanation:

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F -mgμ = ma.................... Equation 1

Where F = applied force, m = mass of the apple cart, g = acceleration due to gravity, μ =  coefficient of friction., a = acceleration of the apple cart.

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