A large, cylindrical water tank with diameter 2.40 m is on a platform 2.00 m above the ground. The vertical tank is open to the
1 answer:
Answer:
a. h₁ =0.0008m = 0.8mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m
b. t =31.93 sec
Explanation:
a) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
given:,
cylindrical water tank with diameter, D - 2.40 m
the tank is 2.00 m above the ground.
depth of the water in the tank is 2.00 m.
diameter of hole, d - 0.520 cm
firstly, we need to calculate the volume of water in the tank:
Volume = πr²h
=(pi * D * D)/4 * height of water
=(3.142 * 2.4 * 2.4)/4 * 2 = 4.52*2 = 9.05m³
1.0 gal of water is equivalent to = 0.0038m³
the volume of 1gal is 0.0038 = A *h
the area of the tank is calculated above as 4.52m²
therefore, 0.0038 = 4.52 *h
h₁ =0.0008m = 0.8mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.0008 = 1.9992m
b) How long does it take you to collect 1.00 gal of water in the bucket?
1.0 gal of water is equivalent to = 0.0038m³
to calculate for Volume flow, Q for a draining tank
Q = Cd * A *
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole = (pi * D * D)/4 = (pi * 0.0052 * 0.0052)/4
A = 0.0000212m²
H= height of the hole from the tank water level= 2m - 0.0052 = 1.9948m
g = 9.8m/s2
Q = 0.9 * 0.0000212 *
Q = 0.0000191 * 6.253 = 0.000119 m³/s
Q = V/t
Qt = V
t = V/Q = 0.0038m³/0.000119 m³/s
t =31.93 sec
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Answer:
greater speed will be obtained for the elastic collision,
Explanation:
To answer this exercise we must find the speed that the sail acquires after each impact.
Let's start by hitting a ball of clay.
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initial instant. before the crash
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where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest
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p_f = (m + M) v
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p₀ = p_f
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for when n balls have collided
v = v₀
Now let's analyze the case of the bouncing ball (elastic)
initial instant
p₀ = m v₀
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p_f = m v_{1f} + M v_{2f}
p₀ = p_f
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m (v₀ - v_{1f}) = M v_{2f}
this case corresponds to an elastic collision whereby the kinetic energy is conserved
K₀ = K_f
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v₁ = v_{1f} v₂ = v_{2f}
m (v₀² - v₁²) = M v₂²
let's use the identity
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m (v₀ - v₁) = M v₂ (1)
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v₀
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v₂ =
Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.
In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.
Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.