2) acceleration = final velocity - initial velocity / time —> V-U/T
Acceleration is the change in velocity over the change in time so it can be represented by the equation a = Δv/Δt.
3) first one- F=10.5 N
second one- 4 m/s^2
third one- 1200N
The expression for the magnitude of the electric field between two uniform conducting plates is

Here, V is potential difference between plates and d is separation between plates.
As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.
Therefore,

Thus, the electric field strength between the plates is 7000 V/ m
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
Answer: a= 52.02 m/s²
Fc= 78.03 N
Explanation: Solution attached: