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:
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Answer:
Explanation:
m = Mass attached to spring = 14 kg
g = Acceleration due to gravity =
x = Displacement of spring =
k = Spring constant
The force balance of the system is given by
The spring constant for that spring is .
Answer:
f = 1.69*10^5 Hz
Explanation:
In order to calculate the frequency of the sinusoidal voltage, you use the following formula:
(1)
V_L: voltage = 12.0V
i: current = 2.40mA = 2.40*10^-3 A
L: inductance = 4.70mH = 4.70*10^-3 H
f: frequency = ?
you solve the equation (1) for f and replace the values of the other parameters:
The frequency of the sinusoidal voltage is f
Explanation:
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