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:
Quick maths
First you find the fafarick and the lalickc and the caprisum and the joinnt
The second major reason for the difference in gravity at differentlatitudes is that the Earth's equatorial bulge (itself also caused by centrifugalforce from rotation) causes objects at the Equator to be farther from the planet's centre than objects at the poles.
Answer:
Explanation:
given,
width of door dimension = 1 m
mass of door = 15 Kg
mass of bullet = 10 g = 0.001 Kg
speed of bullet = 400 m/s
a) from conservation of angular momentum
Answer:
0.45 seconds
Explanation:
Letting the value of g = 10 m/s/s
final velocity (v) = 0 m/s (since the egg will come to rest at the maximum height)
initial velocity(u) = 4.5 m/s
acceleration = -10 m/s/s (since the gravity is acting against the egg)
time = t seconds
From the first equation of motion:
<em>v = u + at</em>
<em>0 = 4.5 + (-10)t</em>
<em>t = -4.5 / -10</em>
t = 0.45 seconds
