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adelina 88 [10]
3 years ago
8

The water flowing through a 1.8 cm (inside diameter) pipe flows out through three 1.2 cm pipes. (a) If the flow rates in the thr

ee smaller pipes are 27, 19, and 12 L/min, what is the flow rate in the 1.8 cm pipe? (b) What is the ratio of the speed of water in the 1.8 cm pipe to that in the pipe carrying 27 L/min?
Physics
1 answer:
Arisa [49]3 years ago
8 0

To solve this problem it is necessary to apply the Discharge of flow equations.

From the theory the flow rate is defined as

Q = AV

Where,

A =Area

V = Velocity

PART A) The question is telling us about the total fluid flow rate then

Q_T = Q_1+Q_2+Q_3

Q_T = 27+19+12

Q_T = 58L/min

PART B) The radius would be given between another pipe with a flow rate of 27L / min.

For proportionality ratio we have to

\frac{Q_T}{Q'} = \frac{A_TV_T}{A'V'}

\frac{V_T}{V'} = \frac{A_'Q_T}{A_TQ'}

\frac{V_T}{V'} = \frac{(\pi r_T^2)Q_T}{(\pi r'^2)Q'}

\frac{V_T}{V'} = \frac{1.2*^2 58}{1.8^2 27}

\frac{V_T}{V'} = 0.9547

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The time spent in the air by the ball at the given momentum is 6.43 s.

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Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

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next we will express the distance between the earth and the sun

r = \frac{a(1-E^2)}{1+Ecos\beta }   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

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input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

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Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

(V^2 = (\frac{4\pi^{2}  }{149.626*10^9})

therefore : V = 1.624* 10^-5 m/s

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