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LekaFEV [45]
3 years ago
6

Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes,

subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is Q=ΔV/Δt what is the flow rate in pipe A? Viscosity: Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes, subject to the same pressure difference across the lengths of the pipes. If the flow rate in pipe B is what is the flow rate in pipe A?
a) Q√2
b) 16Q
c) 2Q
d) 4Q
e) 8Q
Physics
1 answer:
zheka24 [161]3 years ago
6 0

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

\Delta P = \dfrac{128 \mu L Q}{\pi D^4}

Since the diameter in both pipes is the same, we can say:

D = D_A = D_B

where;

length of the first pipe A L_A = L and the length of the second pipe B L_B = 2L

Since the difference in pressure is equivalent in both pipes:

Then:

\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}

\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}

\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}

\mathbf{Q_1 = 2Q}

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

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Separation = 4.22 cm

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Solution:

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m_{p}a = q_{p}E

a = \frac{q_{p}.E }{m_{p} }      Equation 1

So,

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S = ut + 1/2at^{2}

Here, u = 0.

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Put equation 1 into the above equation:

S = 1/2 x (\frac{q_{p}.E }{m_{p} }  )t^{2}      Equation 2

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Similarly, the distance covered by electron will be:

(D-S) = 1/2 x (\frac{q_{e}.E }{m_{e} }  )t^{2}    Equation 3

We know that the charge of electron is equal to the charge of proton so,

q_{p} = q_{e} = q

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S = \frac{m_{e}.D }{(m_{e} + m_{p})  }

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we already have the equation, we need to put the values in it.

So,

S = \frac{m_{Cl}.D }{(m_{Cl} + m_{Na})  }

As we know the mass of chlorine is 35.5 and of sodium is 23

S = \frac{35.5 . 4.22}{(35.5 + 23)}

S = 2.56 cm

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