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
Water flowing through a garden hose of diameter 2.74 cm fills a 25-L bucket in 1.50 min. (a) What is the speed of the water leaving the end of the hose? (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?
To calculate the speed of the water leaving the end of the hose, we can use the equation:
Speed = Volume / Time
where the volume is the amount of water that fills the bucket and the time is the time it takes to fill the bucket.
(a) To find the speed of the water leaving the end of the hose:
1. Convert the diameter of the hose to meters:
Diameter = 2.77 cm = 0.0277 m (since 1 cm = 0.01 m)
2. Calculate the radius of the hose:
Radius = Diameter / 2 = 0.0277 m / 2 = 0.01385 m
3. Calculate the cross-sectional area of the hose:
Area = π * (Radius^2) = π * (0.01385^2) = 0.000602 m^2
4. Convert the bucket capacity to cubic meters:
Volume = 25.0 L = 0.0250 m^3 (since 1 L = 0.001 m^3)
5. Convert the filling time to seconds:
Time = 1.40 min = 84 seconds (since 1 min = 60 seconds)
6. Calculate the speed of the water leaving the end of the hose:
Speed = Volume / Time = 0.0250 m^3 / 84 s = 0.0002976 m^3/s
(b) To find the speed of the water leaving the nozzle:
1. Calculate the radius of the nozzle:
Nozzle Radius = (1/3) * Radius = (1/3) * 0.01385 m = 0.004617 m
2. Calculate the cross-sectional area of the nozzle:
Nozzle Area = π * (Nozzle Radius^2) = π * (0.004617^2) = 6.68472e-05 m^2
3. Calculate the speed of the water leaving the nozzle using the conservation of mass equation:
Speed = (Cross-sectional Area of the Hose / Cross-sectional Area of the Nozzle) * Speed of the Water leaving the Hose
Speed = (0.000602 m^2 / 6.68472e-05 m^2) * 0.0002976 m^3/s
Speed = 8.9998 m/s
Therefore, the speed of the water leaving the end of the hose is approximately 8.9998 m/s, and the speed of the water leaving the nozzle is approximately 8.9998 m/s.
To find the speed of the water leaving the end of the hose, we can use the equation of continuity. The equation of continuity states that the product of the cross-sectional area of a pipe (A) and the speed of the fluid (v) remains constant throughout the pipe.
(a) The given diameter of the hose is 2.77 cm. We can find the radius (r) by dividing the diameter by 2:
r = 2.77 cm / 2 = 1.385 cm = 0.01385 m (approximately)
The cross-sectional area (A) of the hose can be calculated using the formula for the area of a circle:
A = πr^2
A = π(0.01385 m)^2
A ≈ 6.01 × 10^-4 m^2 (approximately)
The volume of water that flows through the hose can be calculated using the formula:
V = A × v × t
Where V is the volume, A is the cross-sectional area, v is the speed of the water, and t is the time.
The given volume is 25.0 L, which is equal to 0.025 m^3. The time is 1.40 min, which is equal to 84 seconds.
0.025 m^3 = (6.01 × 10^-4 m^2) × v × 84 s
Rearranging the equation, we can solve for v:
v = (0.025 m^3) / [(6.01 × 10^-4 m^2) × 84 s]
v ≈ 4.93 m/s (approximately)
Therefore, the speed of the water leaving the end of the hose is approximately 4.93 m/s.
(b) Now let's determine the speed of the water leaving the nozzle. The diameter of the nozzle is one-third the diameter of the hose, which means its radius (rn) will be one-third the radius of the hose.
rn = 0.01385 m / 3
rn ≈ 0.00462 m (approximately)
The cross-sectional area of the nozzle (An) is given by:
An = πrn^2
An = π(0.00462 m)^2
An ≈ 6.70 × 10^-5 m^2 (approximately)
Using the equation of continuity again, with the new cross-sectional area, we can find the speed (vn) of the water leaving the nozzle:
0.025 m^3 = (6.70 × 10^-5 m^2) × vn × 84 s
vn = (0.025 m^3) / [(6.70 × 10^-5 m^2) × 84 s]
vn ≈ 4.97 m/s (approximately)
Therefore, the speed of the water leaving the nozzle is approximately 4.97 m/s.
25 L = 25 * 10^-3 m^3
2.77 cm = .0277 m
1.4 min = 84 seconds
flow rate * time = amount
pi r^2 v * 84 = .025 m^3
pi (.0139)^2 v = .025
v = 41.18 m/s
if 1/3 diameter then 1/9 of area so speed is nine times original
371 m/s