Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
Answer:
R = 0.21 Ω
Explanation:
the formula:
R = r x l/A
R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)
R = 6.6 x 10-⁷ / 3.14 x 10-⁶
R = 0.21 Ω
Answer:
Force, 
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

Finally, it stops, v = 0



F = -22713.92 N

So, the magnitude of the force that stops the bullet is 
light and radiation is able to experience by an observer watching the explosion from the vacuum.
<u>Explanation:</u>
- Basically in the nuclear explosion, there will be an enormous of energy released as noise, heat, visible light, radiation and atmospheric wave.
- Usually sound and wave propagate through medium but in vacuum, there will be no medium to transfer this type of energy. so there is no chance of sound and wave transfer.
- Light and radiation travel in vacuum because they didn't need the medium to transfer .
Water because the light is able to enter the water and allow heat to enter faster.