most of the earth is covered with oceans
Answer:
3.33m
Explanation:
Given parameters:
Force applied =500N
Elastic constant = 150N/m
Unknown:
Amount of stretch or extension = ?
Solution:
To solve this problem use the expression below:
F = k e
F is the force applied
k is the elastic constant
e is the extension
So;
500 = 150 x e
e = 
= 3.33m
First blank is radiator
Second blank is radiator
The both have the unit (J) for Jules
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
