An airplane is flying through a thundercloud at a height of 2 000 m. (this is a very dangerous thing to do because of updrafts,
turbulence, and the possibility of electric discharge.) if there is a charge concentration of +40 c at height 3 000 m within the cloud and ï40 c at height 1 000 m, what is the magnitude of the electric field e at the aircraft? (ke = 8.99 ï´ 109 n·m2/c2)
2 answers:
It is given that an<span> airplane is flying through a thundercloud at a height of 2000 m.
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Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:
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</span><span>
Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:
</span>
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Answer:
a) 0 metres
b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s
c) 20 m/s
Explanation:
a) <em>Formula of displacement= velocity x time</em>
time=40 s
velocity =0 m/s
∴ displacement= 0 x 40 = 0 m
Magnitude of displacement is 0 m
b) The increase in velocity shows that there has been acceleration.
c) The average velocity of the car is = {initial velocity + final velocity}
=
=20
Therefore, the magnitude of the average velocity of the car is 20 m/s