Answer:
Electric field at a point ( x , y , z) is 
.
Explanation:
Given :
Electric potential in the region is , 
We need to find the electric field .
We know , electric field , 
{ Here r is distance }
In coordinate system ,
{ 
is partial derivative }
Putting all values we get ,
Hence , this is the required solution.
Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83 Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640 m^3/s for in and 1000 m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
Answer:
C. both a speed and a direction
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
