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wel
2 years ago
13

What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

ressure of 35 lb/in2 Assume that the distance between each floors is 10ft.
Physics
1 answer:
irga5000 [103]2 years ago
3 0

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

  • <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
  • <em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

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27.6 m/s

Explanation:

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Which of the following is a form of pollution created when vehicle exhaust interacts with sunlight?
SVEN [57.7K]

Photochemical smog is formed when primary air pollutants interact with sunlight.

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8 0
1 year ago
What is the pathway of sound through fluids starting at the oval window through to dissipation of the sound waves at the round w
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6 0
2 years ago
A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
Whitepunk [10]
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
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5 0
3 years ago
Read 2 more answers
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
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