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Gwar [14]
2 years ago
8

Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the

dam (in ft). Neglect friction. [h = 155 ft]
Engineering
1 answer:
Marysya12 [62]2 years ago
8 0

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

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Hopefully this helped.
4 0
3 years ago
A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

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2 years ago
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Answer:

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Explanation:

5 0
2 years ago
2) The switch in the circuit below has been closed a long time. At t=0, it is opened.
saul85 [17]

Answer:

  il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

  il(t) = e^(-t/.01)

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2 years ago
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