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vitfil [10]
3 years ago
12

The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

bearings and the air brings the wheel to rest in 1.0 $h$. What is the magnitude of the constant angular acceleration of the wheel in $rev/min^2$? Do not enter the units.
Physics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

⇒     α = - 1.883 rev/min²

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Answer:

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Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

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mA = mass of cart A = 0.500 kg

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mB = mass of cart B = 1.50 kg.

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Since the momentum of system remains constant:

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(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

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