<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
v^2 = ((1.60 x 10^-19)^2 x 9.0 x 10^9) / (9.11 x 10^-31 x 5.3 x 10^-11 )
v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
Since one orbit is the distance,
one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
t = 33.3 x 10^-11 / 2.18 x 10^6 = 15.27 x 10^-17 s
Revolutions per sec = 1 / t = 1 / 15.27 x 10^-17 = 6.54 x 10^15 Hz</span>
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Answer:
The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.
Explanation:
I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.
Answer:
3600 J
Explanation:
According to given question
P(rated)=60w
V= 120
I =0.50 A
t=600 second
Now,
Energy can be calculated as :
Where,
V is voltage
I is current
t is time in second
Now,
Putting the all value in above equation E
So,
Therefore, 36000 J energy use up by light bulb
