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Sedbober [7]
3 years ago
7

A hole of 1 mm² is punched in a full 1-liter milk carton, 5 cm from the bottom. What is the initial speed of the outflow?

Physics
1 answer:
Oksanka [162]3 years ago
6 0
Use Bernoulli equation with the following assumptions.

1) velocity of the milk on the top is near 0, because the hole is so small that the level of milk will descend slowly, compared with the velocity of the outflow.

2) Pressure on the top = pressure on the bottom. This means that we consider that the top is open to the atmosphere.

Then, the terms of pressure cancel, and Bernoulli's equation leads to:

Z1 + p1/(d*g) + (V1)^2/(2g)  = Z2 + p2 /(d*g) + (V2)^2 /(2g)

=> (V2)^2 = 2g (Z1 - Z2)

g = 9.8 m/s^2

Z1 - Z2 = 5cm

(V2)^2 = 2*(9.8 m/s^2)*(0.05m) = 0.98 m^2/s^2

V2 = 0.99 m/s  


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Question 9 of 27
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The power of man performing 500 J of work in 8 seconds is 62.5 J/s.

Power can be defined as the pace at which work is completed in a given amount of time.

Horsepower is sometimes used to describe the power of motor vehicles and other machinery.

The pace at which work is done on an item is defined as its power. Power is a temporal quantity.

Which is connected to how quickly a project is completed.

The power formula is shown below.

Power = Energy / Time

Power = E / T

Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.

Here we have given Energy as 500 J and Time as 8 second.

Power = Energy / Time

Power = 500 / 8  Joule / sec

Power = 250 / 4  Joule / sec

Power = 125 / 2 Joule / sec

Power = 62.5 Joule / sec  or  62.5 watt

Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.

So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.

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brainly.com/question/1634438

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