The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with
being the elementary charge, and
and
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
Answer: Escaped volume = 0.0612m^3
Explanation:
According to Boyle's law
P1V1 = P2V2
P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)
P2 = atmospheric pressure= 14.696psi
V1 = volume of tire =0.025m^3
V2 = escaped volume + V1 ( since air still remain in the tire)
V2 = P1V1/P2
V2 = 50.696×0.025/14.696
V2 = 0.0862m^3
Escaped volume = 0.0862 - 0.025 = 0.0612m^3
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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Explanation:
everything can be found in the picture
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854 *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854 *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854 *10 ^-12 equal to -1.908 *10^7.
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