Answer:
work done by electric field is 0.06 J
Explanation:
Given data:
Two point charge is
0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)
-1 charge positioned is (0 cm , -1 cm, 0.00 cm)
E = 3.0\times 10^6 N/C
From above information, the distance between given two charges d = 2 cm
then d = 0.02m
work needed is W = q E d
W = 0.06 J
Therefore work done by electric field is 0.06 J
Answer:
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
For Copper tube is 3/4 standard type K drawn tube
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene
Pressure difference given as
Where
L is length of tube
μ is dynamic viscosity
Q is volume flow rate
d is inner diameter of tube
ΔP is pressure drop
Now by putting the values
So flow rate is
Given:
Pressure, = 1300 kPa
Temperature, =
= 100 kPa
velocity, v = 40 m/s
A = 1
Solution:
For air propertiess at
= 1300 kPa
=
= 793kJ/K
=
and also at
= 100 kPa
= 401 KJ/K
=
a) Mass flow rate is given by:
Now,
= 46.51 kg/s
b) for the power produced by turbine,
= 18.231 MW