Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:

ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Finally, you obtain for E:

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
A light year is the DISTANCE light travels through vacuum in 1 year.
If light is traveling through vacuum, then it's traveling at the speed of light in vacuum. If a student at home at the beginning of the trip is holding the clock, then ...
Traveling 1 light year takes 1 year.
Traveling 2 light years takes 2 years.
Traveling 3 light years takes 3 years.
Traveling 10 light years takes 10 years.
If the light is traveling through some other substance, or if the clock is traveling along with the light, then these numbers all change.
YOU cannot travel at the speed of light. We have to just leave it at that
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a




h=5.15 m
The depth is 5.15 m.
Answer:
D
Explanation:
P=Work/Time
The rate at which work is done matches that.
Answer:
1 Newton
Explanation:
F=9*10^9*q0q1/r^2]]
F=9*10^9*(q0q1)/ r^2
r=3cm
F=4N
F=9*10^9*(q0q1)/3^2
4=9*10^9*(q0q1)/9
4=10^9 q0q1
q0q1=4/10^9
q0q1=4*10^-9
To calculate the force between the forces at a distance of 6 cm
F=9*10^9*(q0q1)/ r^2
=9*10^9*(4*10^-9)/6^2
=9*10^9*(4*10^-9)/36
=10^9*4*10^-9/4
=10^9*10^-9
=1 Newton