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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.
<h3>Pressure and temperature at equilibrium </h3>
The relationship between pressure and temperature can be used to determine the height risen by the water.
where;
- V₁ = AL
- V₂ = A(L - y)
- P₁ = Pa
- P₂ = Pa + ρgh
- T₁ = 20⁰C = 293 K
- T₂ = 10⁰ C = 283 k
Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.
The complete question is below:
A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?
Learn more about thermal equilibrium here: brainly.com/question/9459470
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