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2 years ago

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Rainwater draining from a neighborhood street initially travels at 4 ft/s through a pipe with a cross-sectional area of 15.7 ft2

. This pipe connects down the street to another pipe with a cross-sectional area of 65.4 ft2. What would be the speed of the water as it moves through this larger pipe?

1 answer:

Fudgin [204]2 years ago

6 0

Answer:

The velocity is $v_2 = 0.96 \ ft/s$

Explanation:

From the question we are told that

The initial speed is $v_1 = 4 \ ft/s$

The cross -sectional area of the first pipe is $A_1 = 15.7 \ ft$

The cross -sectional area of the second pipe is $A_2 = 65.4 \ ft$

Generally from continuity equation we have that

$A_1 * v_1 = A_2 * v_2$

So

$v_2 = \frac{A_1 * v_1 }{A_2 }$

=> $v_2 = \frac{15.7 * 4 }{65.4 }$

=> $v_2 = 0.96 \ ft/s$

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Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

$R\propto\dfrac{1}{D}$

The power is inversely proportional to the resistance.

$P\propto\dfrac{1}{R}$

$P\propto D^2$

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

$\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}$

Put the value into the formula

$\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}$

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Hence, The factor of the diameter is 0.95.

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Hanging from a horizontal beam are nine simple pendulums of the following lengths:

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Answer:

Options d and e

Explanation:

The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.

We can get the length of the pendulums likely to oscillate with the formula;

$L =\frac{g}{w^{2} }$

where g=9.8m/s

ω= 2rad/s to 4rad/sec

when ω= 2rad/sec

$L= \frac{9.8}{2^{2} }$

L = 2.45m

when ω= 4rad/sec

$L=\frac{9.8}{4^{2} }$

L = 9.8/16

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L is between 0.6125m and 2.45m.

This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.

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An adult human is 60 percent water; a third of this water is in extracellular fluid, and 20 percent of extracellular fluid is in

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Answer:

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The amount of water in the 200 pound person is

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Of the 120 pounds, a third of this is extracellular fluid, the amount of extracellular fluid is

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The rest is interstitial fluid, which is

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2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

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Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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Answer:

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Explanation:

5(2)+5(7)=10+35=45

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