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
Answer:
The ball will be at 700 m above the ground.
Explanation:
We can use the following kinematic equation
.
where y(t) represent the height from the ground. For our problem, the initial height will be:
.
The initial velocity:
,
take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

So, the equation for our problem its:
.
Taking t=6 s:
.
.
.
.
.
So this its the height of the ball 6 seconds after being thrown.
<span>To find the gravitational potential energy of an object, we can use this equation:
GPE = mgh
m is the mass of the object in kg
g = 9.80 m/s^2
h is the height of the object in meters
GPE = mgh
GPE = (0.700 kg) (9.80 m/s^2) (1.5 m)
GPE = 10.3 J
The gravitational potential energy of this can is 10.3 J</span>