You've given the answer, right there in your question.
The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.
We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about
(1 / √0.66) = 1.23 times
the Earth's radius, so the height is about 910 miles above the surface.
I dont know but i know i dont lnow if this is true but the gravity is slowly going away every 4 year i dont know i think.
Answer:
Explanation:
Some exoplanets may depending on the climate and vicinity from the sun.
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
