Explain how the thermal energy of a swimming pool compares to the thermal energy of a lake at the same temperature.
1 answer:
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Answer:
The tank is losing
Explanation:
According to the Bernoulli’s equation:
We are being informed that both the tank and the hole is being exposed to air :
∴ P₁ = P₂
Also as the tank is voluminous ; we take the initial volume ≅ 0 ;
then can be determined as:
h₁ = 5 + 15 = 20 m;
h₂ = 15 m
as it leaves the hole at the base.
radius r = d/2 = 4/2 = 2.0 mm
(a) From the law of continuity; its equation can be expressed as:
J =
J = πr²
J =
J =
b)
How fast is the water from the hole moving just as it reaches the ground?
In order to determine that; we use the relation of the velocity from the equation of motion which says:
v² = u² + 2gh ₂
v² = 9.9² + 2×9.81×15
v² = 392.31
The velocity of how fast the water from the hole is moving just as it reaches the ground is :
Answer:
The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s
Explanation:
The rotational inertia of the merry-go-round = 600 kg·m²
The radius of the merry-go-round = 3.0 m
The mass of the boy = 20 kg
The speed with which the boy approaches the merry-go-round = 5.0 m/s
Where;
= The tangential force
I = The rotational inertia
m = The mass
α = The angular acceleration
r = The radius of the merry-go-round
For the merry go round, we have;
= The rotational inertia of the merry-go-round
= The angular acceleration of the merry-go-round
= The linear velocity of the merry-go-round
t = The time of motion
For the boy, we have;
Where;
= The rotational inertia of the boy
= The angular acceleration of the boy
= The linear velocity of the boy
t = The time of motion
When the boy jumps on the merry-go-round, we have;
Which gives;
From which we have;
The velocity of the merry-go-round, , after the boy hops on the merry-go-round = 1.5 m/s.