The formula we will be using is;F = G m1 m2 / r^2
F earth = 308N
F moon = G me m2 / (81.3 * 0.27^2 RE^2) = 1/5.927 G me m2 / RE^2 = F earth / 5.927 = 52 N is the force of the earth
So the answer in this question is 52 N.
Answer:

Explanation:
<u>Accelerated Motion
</u>
It refers to the motion of objects in which velocity is not constant over time. If the change of the velocity occurs at the same rate, then we say it's uniformly accelerated. Being
= initial speed,
= final speed, a= constant acceleration, x= distance traveled
Then, the scalar relation between them is

The aircraft needs to reach a liftoff speed of 53 m/s from rest (assumed) having only 420 meters to do so. We can compute the acceleration by solving for a



Answer:
B 4.18J\g degree C is the specific heat of water
Id say this is more of a biology question or even a psychology question , well its not a question at all but the endocrine system is a collection of glands that secrete hormones direct to the blood system (circulatory system) to be sent to the desired / target organ , an example of a gland could be the pituitary gland (located towards rear of brain)
Answer:
v=20m/S
p=-37.5kPa
Explanation:
Hello! This exercise should be resolved in the next two steps
1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed
Q=VA
for he exitt
Q=flow=5m^3/s
A=area=0.25m^2
V=Speed
solving for V

velocity at the exit=20m/s
for entry

2.
To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

where
P=presure
α=9.810KN/m^3 specific weight for water
V=speed
g=gravity
solving for P1

the pressure at exit is -37.5kPa