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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?

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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LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

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Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

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At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la

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Answer:

Explanation:

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Velocity of sound in water at 20 degree = 1470 m/s

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Rest of the time is

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This time is taken to cover distance in water. If d be the depth of lake

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