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ziro4ka [17]
3 years ago
11

A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr

ack is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car's switch is turned on, it soon reaches a steady speed of 0.79 m/s relative to the track. What then is the track's angular velocity, in rpm?
Physics
1 answer:
lesya [120]3 years ago
8 0

Answer:

The track's angular velocity is W2 = 4.15 in rpm

Explanation:

Momentum angular can be find

I = m*r^2

P = I*W

So to use the conservation

P1 + P2 = 0

I1*W1 + I2*W2 = 0

Solve to w2 to find the angular velocity

0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2

W2 = 0.435 rad/s

W2 = 4.15 rpm

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Answer:

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Explanation:

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substituting the values of sinθ and tanθ we have

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(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

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