Answer:
a)ΔV = 6.48 KV
b)ΔU =18.79 mJ
Explanation:
Given that
E= 1.8 KV/m
a)
We know that
Electric potential difference ΔV given as
ΔV = E .d
Here
E= 1.8 KV/m
d= 3.6 m
ΔV = E .d
ΔV = 1.8 x 3.6 KV
ΔV = 6.48 KV
b)
Given that
q=+2.90 µC
Change in electric potential energy ΔU given as
ΔU = q .ΔV
ΔU =18.79 mJ
Answer:
Explanation:
Given
e=100 N/C
M=0.15 g
The ratio of the electric force on the bee to the bee's weight can be determined by the following formula
Solution :
Speed of the air craft, = 262 m/s
Fuel burns at the rate of, = 3.92 kg/s
Rate at which the engine takes in air, = 85.9 kg/s
Speed of the exhaust gas that are ejected relative to the aircraft, =921 m/s
Therefore, the upward thrust of the jet engine is given by
F = 85.9(921 - 262) + (3.92 x 921)
= 4862635.79 + 3610.32
=
Therefore thrust of the jet engine is .
