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
To calculate the speed of water leaving the end of the hose, we can use the equation for volumetric flow rate:
Q = A * v
Where:
Q is the volumetric flow rate (in liters per minute)
A is the cross-sectional area of the hose (in square centimeters)
v is the speed of the water (in centimeters per minute)
First, let's convert the given measurements to the appropriate units. The diameter of the hose is 2.77 cm, so the radius is half of that, which is 1.385 cm (radius = diameter/2). To find the cross-sectional area of the hose, we use the formula for the area of a circle:
A_hose = π * r_hose^2
Substituting the values:
A_hose = π * (1.385 cm)^2
Next, we convert the bucket capacity from liters to cubic centimeters since the units of volume need to match:
V_bucket = 25.0 L = 25000 cm^3
With the given information, we can determine the volumetric flow rate using the equation:
Q = V_bucket / t
Where:
t is the time in minutes, which is given as 1.40 min.
Substituting the values:
Q = 25000 cm^3 / 1.4 min
Now, we can rearrange the equation to solve for the speed of the water:
v = Q / A_hose
Substituting the values:
v = (25000 cm^3 / 1.4 min) / (π * (1.385 cm)^2)
Simplifying the equation gives us the speed of the water leaving the end of the hose in centimeters per minute. To convert this to meters per second, we need to do one more unit conversion:
1 meter = 100 centimeters
1 second = 60 seconds
So, to convert from centimeters per minute to meters per second, we divide by 100 to get meters per minute, and then divide by 60 to get meters per second.
Now let's solve the equation to find the answer to part (a):
(a) The speed of the water leaving the end of the hose is approximately 0.283 m/s.
For part (b), we know that the diameter of the nozzle is one-third of the diameter of the hose. Therefore, the radius of the nozzle is one-third of the radius of the hose. We use the same formula for the cross-sectional area of a circle to find the area of the nozzle:
A_nozzle = π * r_nozzle^2
Since r_nozzle = 1/3 * r_hose, we can substitute this into the equation:
A_nozzle = π * (1/3 * r_hose)^2
To calculate the speed of the water leaving the nozzle, we use the same equation, but now with the cross-sectional area of the nozzle:
v = Q / A_nozzle
Substituting the values:
v = (25000 cm^3 / 1.4 min) / (π * (1/3 * 1.385 cm)^2)
Simplifying the equation gives us the speed of the water leaving the nozzle in centimeters per minute. We can then follow the same process as before to convert this to meters per second:
1 meter = 100 centimeters
1 second = 60 seconds
Now let's solve the equation to find the answer to part (b):
(b) The speed of the water leaving the nozzle is approximately 2.53 m/s.