Answer:
Part a)

Part b)



Part c)



Explanation:
Part a)
As we know that charge density is the ratio of total charge and total volume
So here the volume of the charge ball is given as



now the charge density of the ball is given as

Part b)
Now the charge enclosed by the surface is given as

at radius of 5 cm


at radius of 10 cm


at radius of 20 cm

Part c)
As we know that electric field is given as

so we have electric field at r = 5 cm


electric field at r = 10 cm


electric field at r = 20 cm


I believe the answer is C
Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ
15 guesting i think it not right