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Natali5045456 [20]
2 years ago
12

Lena is playing with a remote-controlled car in her backyard. She knows that the car uses a wheel and axle to move. What is the

primary source of power for the wheel and axle in the remote-controlled car?
A.
friction between the ground and the tires

B.
radio waves from the remote control

C.
chemical energy from the batteries

D.
potential energy with the car sitting at the top of a small hill
Physics
1 answer:
melomori [17]2 years ago
7 0

Answer:

C. chemical energy from the batteries

Explanation:

Remote-controlled car or commonly called RC cars are toy vehicles that are controlled or propelled to move from a distance by a wire-connected remote that sends radiowaves to the receiver. According to this question involving the remote-controlled car Lena is playing with, a transmitter from the remote propels the wheel and axle to make the car move.

The car contains a battery cell that uses it's chemical energy as a source of power for the wheel and axle system. Hence, the chemical energy from the batteries of a remote-controlled car powers the wheel and axle to make the car move.

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If mass A is 1.0 kg, mass B is 5.0 kg, and the frictional pulley has a mass of 0.5 kg and a radius of 0.15 m, what is the veloci
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b. A = 3.14 m/s, B = -3.14 m/s

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Assuming this refers to an Atwood machine, draw free body diagrams for each mass.

For mass A, there are two forces: tension T₁ pulling up and weight m₁g pulling down.

For mass B, there are two forces: tension T₂ pulling up and weight m₂g pulling down.

For the pulley, there are two torques: tension T₁r pulling counterclockwise and tension T₂r pulling clockwise.

Sum of forces on A in the +y direction:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of forces on B in the -y direction:

∑F = ma

m₂g − T₂ = m₂a

T₂ = m₂g − m₂a

Sum of torques on the pulley in the clockwise direction:

∑τ = Iα

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m₂g − m₂a − (m₁g + m₁a) = ½ ma

m₂g − m₂a − m₁g − m₁a = ½ ma

m₂g − m₁g = m₂a + m₁a + ½ ma

g (m₂ − m₁) = a (m₂ + m₁ + ½ m)

a = g (m₂ − m₁) / (m₂ + m₁ + ½ m)

a = (9.8 m/s²) (5.0 kg − 1.0 kg) / (5.0 kg + 1.0 kg + ½ (0.5 kg))

a = 6.272 m/s²

Given:

v₀ = 0 m/s

a = 6.272 m/s²

t = 0.5 s

Find: v

v = at + v₀

v = (6.272 m/s²) (0.5 s) + 0 m/s

v = 3.14 m/s

Therefore, mass A will rise at 3.14 m/s, and mass B will fall at 3.14 m/s.

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