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 
It is itself. This question does not make sense.
First, let's put 22 km/h in m/s:

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

Notice we don't use the kinetic coefficient even though the bike is moving. This is because when the tires meet the road they are momentarily stationary with the road surface. Otherwise the bike is skidding.
Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:
Answer:
in the parallel connection the light bulbs shine less than in the series connection
Explanation:
In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is
P = i² R
where R is the resistance of each bulb and i the current of the circuit.
If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,
i = i₁ + i₂
if the two light bulbs are the same
i = 2 i₁
i₁ = i / 2
so the power of each bulb is is
P = i₁² R
P = R i² / 4
P = ¼ P_initial
Therefore we see that in the parallel connection the light bulbs shine less than in the series connection
You can use mostly anything as long as it is circular. Depending on how big it is, you could use sturdy paper plates and use a stick/rod and tape to hold it together, or you could use bottle caps if the car you are trying to make is really small.