Question

During a lightning flash, there exists a potential difference of Vcloud ‒ Vground = 1.2×109 V between a cloud and the ground. As a result, a charge of ‒25 C is transferred from the ground to the cloud. (a) How much work W ground–cloud is done on the charge by the electric force? (b) If the work done by the electric force were used to accelerate a 1100-kg automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat, how many kilograms of water at 0 °C could be heated to 100 °C?​

Answer 1

Explanation:

(a)

∆u=q∆v & ∆u=w

therefore,

w=(-25)(1.2×10^9). [is it 109 or 10^9]

w=-3×10^10

(-ve sign tells direction)

(b)

w=Fd=ma d

3×10^10=1100×ad

a=3×10^8/11d

put in Newton's 3 equation of motion

${v}^{2} = {u}^{2} + 2ad$

${v}^{2} = 2 \times \frac{3 \times {10}^{8} }{11d} \times d \\ v = \sqrt{ \frac{6}{11} \times {10}^{8 } } \\ v = \sqrt{ \frac{6}{11} } \times {10}^{4}$

(c)

Answer 2

Answer:

During a lightning flash, there exists a potential difference of Vcloud ‒ Vground = 1.2×109 V

between a cloud and the ground. As a result, a charge of ‒25 C is transferred from the

ground to the cloud.

(a) How much work W ground–cloud is done on the charge by the electric force?

(b) If the work done by the electric force were used to accelerate a 1100-kg automobile

from rest, what would be its final speed?

(c) If the work done by the electric force were converted into heat, how many kilograms

of water at 0 °C could be heated to 100 °C?

Explanation:

zero