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:
53.895 m.
Explanation:
Using the equation of motion,
v² = u² + 2as .............. Equation 1
Where v = final velocity of the swan, u = initial velocity of the swan, a = acceleration of the swan, s = distance covered by the swan.
make s the subject of the equation,
s = (v² - u²)/2a----------- Equation 2
Given: v = 6.4 m/s, u = 0 m/s ( from rest) a = 0.380 m/s².
Substitute into equation 2
s = (6.4²-0²)/(2×0.380)
s = 40.96/0.76
s = 53.895 m.
Hence the swan will travel 53.895 m before becoming airborne.
Answer:
A
Explanation:
Did on a quiz and got it right