Complete Question
In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as 350 MV (35,000,000 V). The bottoms of the thunderclouds are typically 1500 m above the earth, and can have an area of 120 km^2. Modeling the earth/cloud system as a huge capacitor, calculate
a. the capacitance of the earth-cloud system
b. the charge stored in the "capacitor"
c. the energy stored in the capacitor
Answer:
a
b
c
Explanation:
From the question we are told that
The potential difference is 
The distance of the bottom of the thunderstorm from the earth is d = 1500 m
The area is 
Generally the capacitance of the earth cloud system is mathematically represented as
Here 
is the permitivity of free space with as value 
So
=>
Generally the charge stored in the capacitor (earth-cloud system) is mathematically represented as
=> 
=>
Generally the energy stored in the capacitor is mathematically represented as
=> 
=>
It has potential energy because of its position.
Answer:
a) v= 2.1 m/s
b) ω = 0.807 rad/s
Explanation
Conceptual analysis :
The dog and the merry-go- round describes a circular motion, then, the following formulas apply :
Formula (1)
v = ω *r Formula (2)
Where:
: Centripetal acceleration(m/s²)
v: linear speed or tangential (m/s)
r : radius of the circle (m)
ω : angular speed ( rad/s)
Data
r= 2.6 m
= 1.7 m/s²
Problem develpment
a) We replace data in the formula 1 to calculate the dog's linear speed(v):
v= 2.1 m/s
b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).
v = ω *r
2.1 = ω *2.6
ω = 2.1/2.6
ω = 0.807 rad/s
Answer:
The correct option is
(e)either (c) or (d) could be correct.
Explanation:
The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.
Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since
Work = Force × distance = 0 where distance ≠ 0.
The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.
Therefore it is either horizontally to the left, or horizontally to the right.
Can't really plot a graph here for question 1.
2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.
b) From the graph, at 10 seconds, the car is moving at 20 m/s.
