Answer:
a). [tex]P=78.4W[/tex]
b). [tex]P=392kPa[/tex]
c.) It must be at the bottom
Explanation:
Given:
Volume flow [tex]V_f=2.0x10^{-4}m^3/s[/tex]
Well depp [tex]h=40.m[/tex]
a.
The power output of the pum
[tex]W=F*d[/tex]
[tex]F=m*g[/tex]
[tex]m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg[/tex]
[tex]W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2[/tex]
[tex]W=78.4J[/tex]
[tex]P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W[/tex]
b.
The pressure of difference the pum
Δ[tex]P=p*g*y'[/tex]
Δ[tex]P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa[/tex]
[tex]P=392kPa[/tex]
c.
It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way
