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Hunter-Best [27]
3 years ago
7

Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

and (c) ft3/s ?
Physics
1 answer:
Paladinen [302]3 years ago
6 0

Answer:

0.05997\ m^3/s

3596.1395\ L/min

2.11647\ ft^3/s

Explanation:

1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s

Volume rate of flow is 0.05997\ m^3/s

1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min

Volume rate of flow is 3596.1395\ L/min

1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s

Volume rate of flow is 2.11647\ ft^3/s

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
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Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

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Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

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Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

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