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lalalah
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1. A fountain sends a stream of water straight up into the air to a maximum height of 4.23 m. The effective area of the pipe feeding the fountain is 5.38 x 10^-4 m^2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10^-3 m^3)
F = (mv^2)/r
F = P * A
P = density * g * h
m = density*(Ah)
P= 1.013x10^5 * pgh
P= 1000*9.8*4.23 + 1.013x10^5
P= 142754 Pa
PA = F
PA = (mv^2)/r
r = (Area/pi)^(1/2)
r= .013086 m
mass = 1000 * area * 4.23
mass = 2.27574 kg
142754 Pa * (5.38 *10^-4 m^2) = (2.27574 kg * v^2)/.013086 m
v = .66455 (m/s multiply by area
.66455 * (5.38 * 10^-4) = .0001926 m^3/s
divide v by 3.79 * 10^-3 m^3, then multiply by 60 seconds
5.66 gal/min
this is so obviously wrong, but I've tried this prob beyond five times already and am at wits end... can anyone steer me in the right direction? thankkkks!
Homework Equations
F = (mv^2)/r
F = P * A
P = density * g * h
m = density*(Ah)
The Attempt at a Solution
P= 1.013x10^5 * pgh
P= 1000*9.8*4.23 + 1.013x10^5
P= 142754 Pa
PA = F
PA = (mv^2)/r
r = (Area/pi)^(1/2)
r= .013086 m
mass = 1000 * area * 4.23
mass = 2.27574 kg
142754 Pa * (5.38 *10^-4 m^2) = (2.27574 kg * v^2)/.013086 m
v = .66455 (m/s multiply by area
.66455 * (5.38 * 10^-4) = .0001926 m^3/s
divide v by 3.79 * 10^-3 m^3, then multiply by 60 seconds
5.66 gal/min
this is so obviously wrong, but I've tried this prob beyond five times already and am at wits end... can anyone steer me in the right direction? thankkkks!