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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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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

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\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

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☯ <u>Now, Finding the force exerted </u>

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\dashrightarrow\:\: \sf{F = ma}

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\dashrightarrow\:\: \sf{F = 40 \times 5}

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☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

8 0
2 years ago
A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the
11111nata11111 [884]

Answer:

T = 120.3 N

Explanation:

Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:

Tension = Centripetal Force + Weight of Stone

T = mv²/r + mg

where,

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r = radius of circle = length of string = 2.99 m

g = 9.8 m/s²

Therefore,

T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)

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