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jek_recluse [69]
3 years ago
6

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

uld the car have to accelerate so that a quarter placed on the back wall would remain in place?
Physics
1 answer:
Helga [31]3 years ago
6 0

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

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Gnesinka [82]

Answer:

113.53 g

Explanation:

Please see attached photo for explanation.

In the attached photo, M is the mass of the meter stick.

The value of M can be obtained as shown below:

Clockwise moment = M × 10.5

Anticlockwise moment = 65.5 × 18.2

Anticlockwise moment = Clockwise moment

65.5 × 18.2 = M × 10.5

1192.1 = M × 10.5

Divide both side by 10.5

M = 1192.1 / 10.5

M = 113.53 g

Thus, the mass of the meter stick is 113.53 g

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I hope it's helpful!

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