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olga55 [171]
3 years ago
7

The truck is accelerating at a rate of +1.50 m/s2. The mass of the crate is 120-kg and it does not slip. The magnitude of the di

splacement is 65 m. What is the total work done on the crate by the frictional force?
Physics
1 answer:
trapecia [35]3 years ago
8 0

Answer:

The work done by the friction force is 11700J

Explanation:

The work done by the friction force is given by:

Wfs=fs*cos(\theta)*d\\Wfs=fs*cos(0)*65m

According to Newton's second law:

f=m*a\\fs=120kg*1.50m/s^2=180N

So:

Wfs=180N*65m\\Wfs=11700J

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Vikentia [17]

Yo sup??

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3 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

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Here, \theta=90

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v=\dfrac{ma}{qB}

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or

v=3.41\times 10^5\ m/s

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3 years ago
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Answer:

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Explanation:

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Even one drink and make you an unsafe driver
sp2606 [1]
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