Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Answer:
I think it is velocity.
because velocity shows the direction and magnitude .
Answer:
<em>0.153 m/s</em>
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Explanation:
The flowrate Q = 0.625 x 10-3 m^3-/s
The diameter of the nozzle d = 5.19 x 10^-3 m
the velocity V = ?
The cross-sectional area of the flow A = 
==> (3.142 x 5.19 x 10^-3)/4 = 4.077 x 10^-3 m^2
From the continuity equation,
Q = AV
V = Q/A = (0.625 x 10-3)/(4.077 x 10^-3) = <em>0.153 m/s</em>