Answer:
Part A:

Part B:
Option B (Towards the South)
Explanation:
Part A:
Magnitude if electric field E:
E=Force/charge
Force=2.04×10−14 N
Charge=1.6×10−19 C

Part B:
Option B (Towards the South)
As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.
Answer:
(E) a greatly increased number of small particles in Earth’s orbit would result in a blanket of reflections that would make certain valuable telescope observations impossible
Explanation:
The trade is one strong reflection for many weak reflections (and more dangerous near-Earth space travel).
None of the answer choices except the last one has anything to do with the effect of exploding a satellite. When you are arguing that exploding a satellite is ill conceived, you need to address specifically the effects of exploding the satellite.
yes it does it because the material cause it to refract and the stronger it is the more it will refract
Explanation:
The given data is as follows.
m = 5000 kg, h = 800 km = 
, r = R + h = 
kg, G = 
As we know that,

v = 
And, it is known that formula to calculate angular velocity is as follows.

v = 
= 
= 
Thus, we can conclude that speed of the satellite is
.
Answer:
physical feature of a wave is related to the depth of the wave base is The circular orbital motion
B. The wave base is the depth, and the still water level is the horizontal level