The potential difference between a storm cloud and the ground is 5.0 x 10^7 volts. During a lightning flash, 3.0 coulombs of charge are transferred to the ground.
A. How much energy is transferred to the ground in this lightning flash?
b. If this much evenrgy were used to accelerate a 3500-kg truck from rest, how fast would the truck end up going?
A. How much energy is transferred to the ground in this lightning flash?
b. If this much evenrgy were used to accelerate a 3500-kg truck from rest, how fast would the truck end up going?
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Hello
electric energy = E = U*I*∆t = voltage*current*time
and 1 Coulomb is is the charge obtained by a current of 1 Ampere in 1 second
3C = 3 A*s -->
E = 5*10^7 V*3A*s = 15*10^7 Ws = 15*10^7 J.
Converting this energy into kinetic energy of the truck gives
15*10^7 J = 1/2 mv^2
v^2 = 30*10^7 J/3500 kg
v = 292,8 m/s
Regards
electric energy = E = U*I*∆t = voltage*current*time
and 1 Coulomb is is the charge obtained by a current of 1 Ampere in 1 second
3C = 3 A*s -->
E = 5*10^7 V*3A*s = 15*10^7 Ws = 15*10^7 J.
Converting this energy into kinetic energy of the truck gives
15*10^7 J = 1/2 mv^2
v^2 = 30*10^7 J/3500 kg
v = 292,8 m/s
Regards