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
D. 289
Take the formula:
K=5/9(Fahrenheit-32)+273
Plug in Fahrenheit
K=5/9 (60-32)+273
From here it is simple math and you can plug it into your calculator getting 288.5555556 and round to 289
Answer:
T = 60 s
Explanation:
There are 6 poles on the track which are equally spaced
so the angular separation between the poles is given as
so the angular speed of the train is given as
now we have time period of the train given as
Yellow and red hope that helped
The force on the ship is more than a car
