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Natalija [7]
3 years ago
14

Una manguera de agua de 1.3 cm de diametro es utilizada para llenar una cubeta de 24 Litros. Si la cubeta se llena en 48 s. A) ¿

cual es la velocidad con la que el agua sale de la manguera?
B) si el diametro de la manguera se reduce a 0.63 cm y suponiendo el mismo flujo, ¿cual sera la velocidad del agua al salir de la manguera?
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

We must translate this:

a 1.3 cm diameter water hose is used to fill a 24-liter bucket. If the bucket is filled in 48 s.  

A) What is the speed with which the water leaves the hose?

B) if the diameter of the hose is reduced to 0.63 cm and assuming the same flow, what will be the speed of the water leaving the hose?

A) If the velocity of the water is Xcm/s

and the radius of the hose is equal to half its diameter, so it is 1.3cm/2

Then in one second we can considerate that a cylinder of:

V = pi*(1.3cm/2)^2*X cm^3 of water.

So we have that quantity in one second of flow.

where pi = 3.14

then in 48 seconds, the amount of water in the bucket is:

V = 48*pi*(1.3/2)^2*X = 24 L = 24,000 cm^3

Now we need to solve this for X.

48*3.14*(1.3/2)^2*X = 24000

63.679*x = 24000

x = 24000/63.679 = 376.89

So the velocity of the water is 376 cm per second.

B) if the diameter is 0.64cm, we have the equation:

48*3.14*(0.63/2)^2*x = 24000

14.955*X = 24000

X = 24000/14.955 = 1604.814 cm/s

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Input power = 15 KW

Output power given by the bulb  = Rate of doing work

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                                                   = 20000 ÷ 1800   <em>(since 1 min = 60 seconds)</em>

                                                   = 11.11 W

Noe, efficiency is the ratio of output power to input power

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