Answer:
a lever and a wheel and axle (i guess)
Explanation:
don't have any
The gravitational pull of Earth is stronger in satellite A
Answer:
1.
2.
Explanation:
Polarizes axis can create two possible angles with the vertical.
first we have to find the intensity of first polarizer
which is given as



For a smaller angle for the first polarizer:
According to Malus Law



taking square root on both sides



For a larger angle for the first polarizer:
According to Malus Law


taking square root on both sides



Answer:
Atomic name is your answer.