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gtnhenbr [62]
3 years ago
9

CAN ANYONE PLEASE HELP ME WITH THIS ONE ITS URGENTLY NEEDED

Physics
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

Explanation:

conservation of momentum

initial momentum is zero

Lets say that m₁ moves along the x axis

and m₂ moves along the y axis

in the y direction

0 = m₁(0) + m₂(30) + m₃(vy)

0 = m₁(0) + m₂(30) + 3m₂(vy)

-3m₂(vy) = m₂(30)

vy = -10 m/s

in the x direction

0 = m₁(30) + m₂(0) + m₃(vx)

0 = 2m₂(30) + m₂(0) + 3m₂(vx)

-3m₂(vx) = 2m₂(30)

vx = -20 m/s

v = √(-10² + -20²) = 10√5 m/s ≈ 22.36 m/s

θ = arctan(-10/-20) = 206.56505... ≈ 206.6°  CCW from the + x axis

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The water is flowing through the horizontal constricted pipe. The pressure at one end is 4500Pa, speed is 3m/s and area of cross
Oksana_A [137]

The speed and pressure at another end will be 9 m/sec and -31500 pascals.

<h3>What is gauge pressure?</h3>

The difference between absolute pressure and atmospheric pressure is known as gauge pressure. Relative pressure is another name for gauge pressure.

Given data in problem is;

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Area of cross-section at end 1,A₁ =A

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From Bernoulli's equation;

\rm \frac{P_1}{\rho g} + \frac{v_1^2}{2g} +Z_1 = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} +Z_2  \\\\ \frac{P_1-P_2}{\rho g } = \frac{v_2^2-v_1^2}{2g} \\\\ \frac{P_1-P_2}{\rho} =  \frac{v_2^2-v_1^2}{2} \\\\ \frac{4500-P_2}{1000} =  \frac{9^2-3^2}{2} \\\\ 4500 -P_2 = 36000 \\\\ P_2 =  - \ 31,500 \ pa

Hence the speed and pressure at another end will be 9 m/sec and -31500 pascals.

To learn more about the gauge pressure, refer to the link;

brainly.com/question/14012416

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7 0
2 years ago
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