In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy. That idea is easy to apply to this question.
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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Answer:
<h2>Magnitude of the second charge is

</h2>
Explanation:
According to columbs law;
F = 
F is the attractive or repulsive force between the charges = 12N
q1 and q2 are the charges
let q1 = - 8.0 x 10^-6 C
q2=?
r is the distance between the charges = 0.050m
k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²
On substituting the given values
12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²
Cross multiplying

Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball
Answer:
a = 0,1[m/s^2]
Explanation:
First we need to indentify the initial data.
And using this kinematic equation we have:
![v = 4[m/s]\\v_{0}= 2 [m/s] \\t = 20[s]\\\\v= v_{0}+a*t\\a=\frac{v-v_{0}}{t} \\a= \frac{4-2}{20} \\a=0.1[m/s^{2}]](https://tex.z-dn.net/?f=v%20%3D%204%5Bm%2Fs%5D%5C%5Cv_%7B0%7D%3D%202%20%5Bm%2Fs%5D%20%5C%5Ct%20%3D%2020%5Bs%5D%5C%5C%5C%5Cv%3D%20v_%7B0%7D%2Ba%2At%5C%5Ca%3D%5Cfrac%7Bv-v_%7B0%7D%7D%7Bt%7D%20%5C%5Ca%3D%20%5Cfrac%7B4-2%7D%7B20%7D%20%5C%5Ca%3D0.1%5Bm%2Fs%5E%7B2%7D%5D)