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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Everything is in the universe.
That's what the word means.
(a) 5.66 m/s
The flow rate of the water in the pipe is given by
where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have
the radius of the pipe is
r = 0.260 m
So the cross-sectional area is
So we can re-arrange the equation to find the speed of the water:
(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:
where we have
and where 
is the cross-sectional area of the pipe at the second point.
Solving for A2,
And finally we can find the radius of the pipe at that point:
Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Answer:
The option is B is not true for Hubble telescope.
