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Levart [38]
3 years ago
8

A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each

with a radius of 5.00 mm. If water flows through the single pipe at 1.45 m/s, calculate the speed (in m/s) of the water in the narrower pipes.
Physics
1 answer:
irinina [24]3 years ago
8 0

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s

So the flow speed of each of the narrower pipe is:

v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}

v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s

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Answer:

84.4 %

Explanation:

Mechanical efficiency = output work/input work × 100 %

output work = 432 J of work for the bike to turn the gears

input work = 512 J of work to ride.

Mechanical efficiency =  432 J/512 J × 100 %

= 0.844 × 100%

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3 years ago
Suppose the U.S. national debt is about $14 trillion. If payments were made at the rate of $3,500 per second, how many years wou
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Answer:

It will take 126.84 years to pay off the debt

Explanation:

Total debt = $14,000,000,000,000.00

Paid $3,500 per second

Number of seconds to pay off the debt will be:

14 ×10^12 /3500

Number of seconds = 4× 10^9 seconds

Converting seconds to year:

I second = 3.171 ×10^-8 calendar year

Therefore, number of years it will take to pay off $14 Trillion =( 4 ×10^9 ) × ( 3.171 × 10^-8)

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The element radon is at the opposite end of the range, with the lowest specific heat of all naturally occurring elements. Radon'
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A 1300 kg steel beam is supported by two ropes. (Figure
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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

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R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

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R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

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R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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2 years ago
In a compression wave, particles in the medium move
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