Water flowing through a garden hose of diameter 2.72 cm fills a 25.0-L bucket in 1.10 min.
(a) What is the speed of the water leaving the end of the hose?
m/s
(b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?
m/s
you get 25L/1.1min
volume of lenght d of hose: diameter*length
= PI*(.0272dm)^2(L)
speed=L/time=25/(PI*.0272)^2 *1/1.10
speed is in dm/minute, so to change that to m/s, divide by 10, and multiply by 60.
To calculate the speed of the water leaving the end of the hose and the nozzle, we can use the principle of conservation of mass. According to this principle, the mass of water entering the hose per unit time should be equal to the mass of water leaving the hose per unit time.
We can use the equation of continuity to relate the speed of the water to the cross-sectional area of the hose or nozzle.
The equation of continuity is given by:
A_1 * v_1 = A_2 * v_2
where A_1 and A_2 are the cross-sectional areas of the hose and nozzle, and v_1 and v_2 are the speeds of the water in the hose and nozzle, respectively.
a) To calculate the speed of the water leaving the end of the hose, we need to find the cross-sectional area of the hose. The formula for the area of a circle is given by:
A = π * r^2
where A is the area and r is the radius of the circle. Given that the diameter of the hose is 2.72 cm, we can calculate the radius by dividing the diameter by 2:
r = (2.72 cm) / 2 = 1.36 cm = 0.0136 m
Substituting this radius into the area formula, we get:
A_1 = π * (0.0136 m)^2 ≈ 0.00058 m^2
Now, we can substitute the given values into the equation of continuity to solve for the speed v_1:
A_1 * v_1 = A_2 * v_2
(0.00058 m^2) * v_1 = (unknown) * (unknown)
Since we don't have the information about the nozzle's diameter in part (a), we cannot determine the speed of the water leaving the end of the hose. We can only find it in terms of the unknown diameter and velocity in part (b).
b) The nozzle diameter is given as one-third the diameter of the hose. Therefore, the radius of the nozzle is one-third of the radius of the hose:
r_n = (1/3) * (0.0136 m) = 0.0045 m
Using the same formula for the cross-sectional area, we can calculate the area of the nozzle:
A_2 = π * (0.0045 m)^2 ≈ 0.000064 m^2
Substituting this area, the known area of the hose (A_1), and the known speed of water in the hose (v_1), we can solve for the speed of the water leaving the nozzle (v_2):
(0.00058 m^2) * v_1 = (0.000064 m^2) * v_2
Dividing both sides by (0.000064 m^2), we get:
v_2 = (0.00058 m^2 * v_1) / (0.000064 m^2)
Now, you can plug in the values for v_1 (the speed of water in the hose) and calculate v_2.