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2 years ago

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A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is 'v', t

he diameter of the jet is 'd' and the density of water is 'p' , then what is the force exerted on the wall by the jet of water?

1 answer:

klasskru [66]2 years ago

3 0

Answer:

F = π/4 ρ d² v²

Explanation:

Force is mass times acceleration:

F = ma

Acceleration is change in velocity over change in time:

F = m Δv / Δt

Since there's no rebound, Δv = v.

F = m v / Δt

Mass is density times volume:

F = ρ V v / Δt

Volume per time is flow rate:

F = ρ Q v

Flow rate is velocity times cross sectional area:

F = ρ (v A) v

F = ρ A v²

Area of a circle is pi times radius squared, or pi/4 times diameter squared:

F = ρ (π/4 d²) v²

F = π/4 ρ d² v²

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Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

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To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

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$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

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$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

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2 years ago

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LekaFEV [45]

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Acceleration was not given in the problem, but we can solve for it by using the velocity and the time because by definition, acceleration is the change in velocity over time.

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The force is -2N. Now if you are looking for the magnitude of the force then you can just put 2N.

The negative sign indicates the direction of the force, if it is negative this means that the force applied was in the opposite direction.

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