Water flowing through a garden hose of diameter 2.77 cm fills a 25.0-L bucket in 1.40 min.
(a) What is the speed of the water leaving the end of the hose?
Answer in 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?
Answer in m/s
area = A = pi D^2/4 = pi (.0277^2)/4 = 6.03*10^-4 m^2
Q = volume/second 25 * 10^-3 m^3 / 84 s
= .298 *10^-3 m^3/s
A v = Q
v = Q/A = .298*10^-3/6.03*10^-4
= .0494 *10^1 = .494 m/s
To find the speed of the water leaving the end of the hose, we can use the equation for the volume flow rate:
Q = A * v
where Q is the volume flow rate, A is the cross-sectional area of the hose, and v is the speed of the water.
(a) First, let's find the cross-sectional area of the hose. The hose has a diameter of 2.77 cm, so its radius is half of that:
r = 2.77 cm / 2 = 1.385 cm
Now, we can use the formula for the area of a circle:
A = π * r^2
Plugging in the values:
A = π * (1.385 cm)^2
Now, we need to convert the area to square meters, as the speed should be in m/s. There are 10,000 square centimeters in a square meter, so:
A = π * (1.385 cm / 100)^2 m^2
Next, we need to find the volume flow rate. The problem states that the hose fills a 25.0-L bucket in 1.40 min. One liter is equal to one cubic decimeter (dm^3), so 25.0 L = 25.0 dm^3.
Since 1 dm^3 is equal to 0.001 m^3, we can convert the volume to cubic meters:
V = 25.0 dm^3 * (0.001 m^3 / 1 dm^3) = 0.025 m^3
The time is given in minutes, so we need to convert it to seconds:
t = 1.40 min * (60 s / 1 min) = 84 s
The volume flow rate is given by:
Q = V / t
Plugging in the values:
Q = 0.025 m^3 / 84 s
Finally, we can solve for the speed (v) by rearranging the equation:
v = Q / A
Plugging in the values:
v = (0.025 m^3 / 84 s) / (π * (1.385 cm / 100)^2 m^2)
Simplifying the expression:
v ≈ 0.047 m/s
Therefore, the speed of the water leaving the end of the hose is approximately 0.047 m/s.
(b) Now, let's calculate the speed of the water leaving the nozzle. The problem states that the nozzle diameter is one-third the diameter of the hose, so its radius is one-third the radius of the hose:
r_nozzle = (1/3) * 1.385 cm = 0.462 cm
Again, we need to convert the radius to meters:
r_nozzle = 0.462 cm / 100 = 0.00462 m
Now we can find the cross-sectional area of the nozzle using the same formula:
A_nozzle = π * (0.00462 m)^2
To find the new volume flow rate (Q_nozzle), we assume that the volume flow rate is conserved, so Q = Q_nozzle. Therefore, we can write:
Q_nozzle = A_nozzle * v_nozzle
where v_nozzle is the speed of the water leaving the nozzle. Then we can solve for v_nozzle:
v_nozzle = Q_nozzle / A_nozzle
Plugging in the values:
v_nozzle = (0.025 m^3 / 84 s) / (π * (0.00462 m)^2)
Simplifying the expression:
v_nozzle ≈ 0.637 m/s
Therefore, the speed of the water leaving the nozzle is approximately 0.637 m/s.