Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C
Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference
As we know the capacity is value that depend on the geometry of the capacitor, in our case two concentric spheres.
So Potential Difference between the spheres is given by:
ΔV=-
Where E = k*Q/ r^2
so we have 
then
Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)
Finally using C=Q/ΔV=ab/(k(b-a))
To caclulate the electric firld we first obtain the charge
Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C
so E=KQ/r^2 for both values of r
r=12.8 cm ( in meters)
r2=14.7 cm
E(r1)=4.83* 10^3 N/C
E(r2)=3.66 *10^3 N/C
<h3><u>Answer;</u></h3>
Mid-ocean ridges
<h3><u>Explanation</u>;</h3>
- Sea-floor spreading is the process by which molten material adds new oceanic crust to the ocean floor.
- Mid-ocean ridge is an undersea mountain chain where new ocean floor is produced at a divergent plate boundary.
- Mid ocean ridge occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.
Answer:4s^-1
Explanation:Tangential Acceleration=v^2/r=(20)²/100=400/100
Answer
i'm not 100% sure but 1764
Explanation:
Work done = gravitational potential energy
Gravitational potential energy = mass(kg) × height(m) × gravitational field strength(N/kg)
We can assume that the student is on earth so the gravitational field strength is 9.8N/kg
So work done = 60 × 3×9.8
=1764
(if you need help calculating power but if you do just divide your answer by 12 and you will get 147)
At certain altitude, the temperature of air decrease, The air becomes saturated and water vapour molecules starts condensing.
As the altitude of air increase, the atmospheric pressure decrease due to which the temperature of the air decrease. The water molecules in the atmosphere start condensing, which saturate the air (that is air can no hold water molecules), due to which the water vapour molecules starts condensing and falls on the earth in the form of rain.