4 Hz is the difference between 510 and the unknown.
<span>Therefore the unknown is either 514 or 506. </span>
<span>If 505 and 506 were struck together, the diff would be 1 Hz </span>
<span>If 505 and 514 were struck toghether, the diff would be 9 Hz which would be difficult to count accurately, </span>
<span>Therefore the unknown is 514</span>
Answer:
reduce the velocity of collision
Answer:
(a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 
Explanation:
Given that,
Charge q₁ = -4.00 μC
Inner radius = 3.13 m
Outer radius = 4.13 cm
Net charge q₂ = -6.43 μC
We need to calculate the charge on the outer surface
Using formula of charge



The charge on the inner surface is q.


We need to calculate the electric field outside the shell
Using formula of electric field

Put the value into the formula



Hence, (a). The charge on the outer surface is −2.43 μC.
(b). The charge on the inner surface is 4.00 μC.
(c). The electric field outside the shell is 