Answer:
The desire to continue with your exercise would be the correct answer! :D
Answer:
Following are the answer to this question:
Explanation:
In option (a):
- The principle of Snells informs us that as light travels from the less dense medium to a denser layer, like water to air or a thinner layer of the air to the thicker ones, it bent to usual — an abstract feature that would be on the surface of all objects. Mostly, on the contrary, glow shifts from a denser with a less dense medium. This angle between both the usual and the light conditions rays is referred to as the refractive angle.
- Throughout in scenario, the light from its stars in the upper orbit, the surface area of both the Earth tends to increase because as light flows from the outer atmosphere towards the Earth, it defined above, to a lesser angle.
In option (b):
- Rays of light, that go directly down wouldn't bend, whilst also sun source which joins the upper orbit was reflected light from either a thicker distance and flex to the usual, following roughly the direction of the curve of the earth.
- Throughout the zenith specific position earlier in this thread, astronomical bodies appear throughout the right position while those close to a horizon seem to have been brightest than any of those close to the sky, and please find the attachment of the diagram.
Answer:
a) FE = 0.764FG
b) a = 2.30 m/s^2
Explanation:
a) To compare the gravitational and electric force over the particle you calculate the following ratio:
(1)
FE: electric force
FG: gravitational force
q: charge of the particle = 1.6*10^-19 C
g: gravitational acceleration = 9.8 m/s^2
E: electric field = 103N/C
m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg
You replace the values of all parameters in the equation (1):

Then, the gravitational force is 0.764 times the electric force on the particle
b)
The acceleration of the particle is obtained by using the second Newton law:

you replace the values of all variables:

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.
Work = Force * distance
Force = 70 N
Work = 3500 J
3500 = 70d
d = 3500/70 = 50 m