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
