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.
</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>
</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>
You might be interested in
Answer:
a) v = 21.34 m/s
b) v = 21.34 m/s
c) v = 21.34 m/s
Explanation:
Mass of the snowball, m = 0.560 kg
Height of the cliff, h = 14.2 m
Initial velocity of the ball, u = 13.3 m/s
θ = 26°
The speed of the slow ball as it reaches the ground, v = ?
The initial Kinetic energy of the snow ball,
Potential energy of the snow ball at the given height, PE = mgh
Final Kinetic energy of the ball as it reaches the ground,
a) Using the principle of energy conservation,
b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch
c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass
v = 21. 34 m/s