A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. When the system reaches its
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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?
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Answer:
1) The time it takes the diver to move off the end of the diving board to the pool surface, , is approximately 1.392 seconds
2) The horizontal distance from the edge of the pool to where the diver enters the water, , is approximately 5.76 meters
Explanation:
1) The given parameters are;
The height of the diving board above the pool's surface, h = 9.5 m
The length by which the diving board over hangs the pool L = 2 m
The speed with which the diver runs horizontally along the diving board, v₀ = 2.7 m/s
Taking = The time it takes the diver to move off the end of the diving board to the pool surface
Therefore, we have from the equation of free fall;
h = 1/2 × g × ²
Where;
g = The acceleration due to gravity = 9.81 m/s²
Substituting the values, gives;
9.5 = 1/2 × 9.81 × ²
= √(9.5/(1/2 × 9.81)) ≈ 1.392 s
The time it takes the diver to move off the end of the diving board to the pool surface = ≈ 1.392 s
2) The horizontal distance, , in meters from the edge of the pool to where the diver enters the water is given as follows;
= L + v₀ ×
= 2 + 2.7× 1.392 ≈ 5.76 m
∴ The horizontal distance from the edge of the pool to where the diver enters the water ≈ 5.76 meters.