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Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
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Answer:
if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
Explanation:
The air in the tube can be considered an ideal gas,
P V = nR T
In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H
For pressure the open end of the tube is
P₂ = P_atm + ρ g H
Let's write the gas equation for the colon
P₁ V₁ = P₂ V₂
P_atm V₁ = (P_atm + ρ g H) V₂
V₂ = V₁ P_atm / (P_atm + ρ g h)
If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.
The main assumption is that the temperature during the experiment does not change
