Can you please stop pasting this question, just go to his profile and ask him.
Answer:

Explanation:
The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.
Let us find the acceleration:


Electric force is given as the product of charge and electric field strength:
F = qE
where q = electric charge
E = Electric field strength
Force is generally given as:
F = ma
where m = mass
a = acceleration
Equating both:
ma = qE
E = ma / q
For an electron:
m = 9.11 × 10^{-31} kg
q = 1.602 × 10^{-19} C
Therefore, the electric field strength of the electron is:

Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, 
Area of cross section, 
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
Weight of the object = drag force
R = W
or

Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,

So, the drag coefficient is 1.01.
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa
100/5= 20, 20 seconds is the time needed.