The hose shown below (point 1) has a radius of 1.25 cm, and the nozzle at the end (point 2) has a radius of 0.45 cm. Point 1 is on the ground and point 2 is 80 cm above the ground. If water is flowing out of the nozzle at 2 L/s,
Please show all math steps and how you set up the formulas to get the answers. I don't have the illustration but I don't think it's needed with the info available in the question. Thanks!
flow rate has been given to us = 2 L/s
A)
now convert Liter to m3
1 m3 = 1000 L
1 L = 1/ 1000 m3
so 2 L = 2/1000 m3 =2*10-3 m3
so flow rate = 2*10-3 m3 /s
B)
equation of continuity
A1*V1 = A2*V2 = 2*10^-3
r1 = 1.25 cm = 0.0125 m
V1 = 0.002/ A1 = 0.002/( pi*0.0125^2) =4.0743 m/s
and
r2= 0.45 cm = 0.0045 m
V2 = 0.002/A2 = 0.002 / ( pi*0.0045^2) =31.4380 m/s
density of water =ρ = 1000 kg/m3
NOW apply Bernoulli’s equation:
y1 = 0 m ( on ground)
y2 = 80 cm = 0.80 above ground
P1 + 0.5*1000*4.0743^2 + 1000*9.8*0 = P2 + 0.5*1000*31.438^2 + 1000*9.8*0.80
from here
P1-P2
= 0.5*1000*31.438^2 + 1000*9.8*0.80 - [ 0.5*1000*4.0743^2 + 1000*9.8*0 ]
= 493713.961755 Pa
= 4.937*105 pa answer
I hope this helps you
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Goodluck for exam Comment in case any doubt, will reply for
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