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Pavel [41]
3 years ago
10

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

ee smaller pipes are 27, 16, and 11 L/min, what is the flow rate in the 2.0 cm pipe? (b) What is the ratio of the speed of water in the 2.0 cm pipe to that in the pipe carrying 27 L/min?
Physics
1 answer:
Verdich [7]3 years ago
8 0

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

b) A x V=54L/min => \frac{\pi }{4} d^2 x v

   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

\frac{v}{v_1}= 0.845

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A 295-kg object and a 595-kg object are separated by 4.10 m.
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a)F=3 x 10⁻⁷ N

b)x=2.405 m

Explanation:

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m₁=295 kg

m₂=595 kg

d= 4.1 m

a)

m₃=63 kg

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by putting the values

F_{13}=\dfrac{Gm_1m_3}{r^2}

F_{13}=\dfrac{6.67\times 10^{-11}\times 295\times 63 }{2.05^2}

F₁₃=2.94 x 10⁻⁷ N

The force  between the mass m₂ and m₃

by putting the values

F_{23}=\dfrac{Gm_2m_3}{r^2}

F_{23}=\dfrac{6.67\times 10^{-11}\times 595\times 63 }{2.05^2}

F₂₃=5.94 x 10⁻⁷ N

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F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N

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b)

Lest take at distance x from mass m₂ net force is zero.

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F_{13}=\dfrac{Gm_1m_3}{(4.1-x)^2}

Form above two equation

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\dfrac{295}{(4.1-x)^2}=\dfrac{595}{x^2}

x²=2.01(4.1-x)²

x=1.42 (4.1-x)

x=5.82 - 1.42x

x=2.405 m

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