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Anit [1.1K]
3 years ago
12

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

m ( 0.573 m) and the flow speed of the petroleum is 13.5 m/s. At the refinery, the petroleum flows at 5.83 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?
Physics
1 answer:
Mashcka [7]3 years ago
4 0

Answer with Explanation:

We are given that

Diameter of pipe,d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

Volume flow rate of the petroleum along the pipe=Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})

Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s

By equation of continuity

A_1v_1=A_2v_2

\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2

d^2_2=\frac{v_1}{v_2}d^2_1

d_2=\sqrt{\frac{v_1}{v_2}}d_1

d_2=0.573\sqrt{\frac{13.5}{5.83}}

d_2=0.87 m

d_2=0.87\time 100=87 cm

1 m=100 cm

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K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)

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Explanation:

16x^{2}y^{2}-25a^{2}b^{2}

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So, this factorized is:

(4xy+5ab)(4xy-5ab)

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