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kipiarov [429]
3 years ago
8

A garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but t

he nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.78 m.
What is the pressure inside the hose?
Physics
1 answer:
GarryVolchara [31]3 years ago
5 0

To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as

P = P_{atm} +\rho gh

Where,

P_{atm} = Atmospheric Pressure

\rho =Density, water at this case

g = Gravity

h = Height

The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.

Our values are given as

P_{atm} = 1.013*10^5Pa

\rho = 1000Kg/m^3

g = 9.8m/s^2

h = 0.78m

Replacing at the equation we have,

P = P_{atm} +\rho gh

P = 1.013*10^5 +(1000)(9.8)(0.78)

P = 108944Pa

P = 0.1089Mpa

Therefore the pressure inside the hose is 0.1089Mpa

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Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

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∴                               V₂ =  V₂₁ + V₁

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                                 V₂ = 3 + 2

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8 0
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equal and opposite reaction.

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A TV set is pushed a distance of 2 m with a force of 20 N how much work is done on the set ​
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A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

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The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

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