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Drupady [299]
3 years ago
14

(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

e in cubic meters per second? (c) Would your answers be different if salt water replaced the fresh water in the fire hose?
Physics
1 answer:
son4ous [18]3 years ago
8 0

Answer:

12.5752053801 m/s

80\times 10^{-3}\ m^3/s

No.

Explanation:

Q = Volume flow rate = 80\ L/s=80\times 10^{-3}\ m^3/s

d = Diameter of pipe = 9 cm

A = Area = \dfrac{\pi}{4}d^2

Volume flow rate is given by

Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is 80\times 10^{-3}\ m^3/s

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

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25+35=60mins so 1.0 hours
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A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki
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The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
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Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

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Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

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3 years ago
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when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂  = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

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The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

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