To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,



PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

At the same time we have that centripetal acceleration is given as

Replacing



Answer:
The answer is
A. Pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.
Explanation:
The question is incomplete, here is a complete question with full options
You are caulking a window. The caulk is rather thick and, to lay the bead correctly, the exit nozzle is small. A caulking gun uses a plunger which is operated by pulling back on a handle. You must squeeze the handle very hard to get the caulk to come out of the narrow opening because:_________.
A. pressure is distributed uniformly throughout the fluid and the area of the plunger is much larger than the area of the opening.
B. viscous drag between the walls of the tip and the caulk causes the caulk to swirl around chaotically.
C. Newton’s third law requires most of the energy in the caulk to be used to push back on the plunger rather than moving it through the tip.
D. the high density of the caulk impedes its flow through the small opening.
Since the caulk is thick and the exit nozzle is small, the pressure needed to deliver the caulk will be very high as pressure is uniformly distributed at the plunger side at every part of the caulk, hence very high pressure is needed to deliver the caulk which is why the handle needed the very hard squeeze
Answer: This is called backscatter which refers to the ability of big waves to reflect the energy in order to give back the signal .
Explanation:
What is meant by backscatter?
Backscatter is the process where by the waves or signal is reflected back to the original direction and get scattered in all directions.
Backscatter allows us to receive signal and be able to view all the channels that are connected through the satellite.
~686newtons on earth and
~1617 newtons on jupiter
the formula is weight = gravitational acceleration * mass of the object
Answer:
by lying down on a nice and soft quilted matress