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lisov135 [29]
3 years ago
13

A .25 kg ball initially at rest is hit with a 460 N impact. What is the impulse for the ball if it ends up moving at 40 m/s?

Physics
1 answer:
9966 [12]3 years ago
3 0

Answer:

it required 0.022 s of contact time

<h2>pls branieslt! c:</h2>
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F_e=qE=ma             (1)

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You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

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x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

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