Answer:
Explanation:
For acceleration due to gravity g , the expression is
g = GM / R² , where G is gravitational constant M is mass of the earth and R is radius of the earth .
At height h , let the value of it becomes g / 4 , so
g / 4 = GM / ( R + h )²
dividing
4 = [( R+ h)² / R² ]
2 = (R + h) / R
2 = 1 + h / R
h / R = 1
h = R
So at height equal to radius of the earth , acceleration due to gravity becomes 1 /4 of value on the surface of the earth .
The entire flight time is three hours the flight time up to erewhon is 2 hours making flight time for remander of the trip is 1 hour so as long as i understood what your asking that is your answer
In physics we refer to how heavy the object is as the mass.
Fg = mass x acceleration due to gravity
Force is directly proportional to mass, thus if force of gravity increases mass also increases.
Approximate the instantaneous velocity by computing the average velocity of the car over the interval [1.95, 2.05] (i.e. the interval centered at <em>t</em> = 2 s with length 0.1 s).
By definition, average velocity is given by
<em>v</em> = ∆<em>x</em> / ∆<em>t</em>
So we have
<em>v</em>(2 s) ≈ (<em>x</em>(2.05 s) - <em>x</em>(1.95 s)) / (0.1 s) ≈ 13.94 m/s
Answer:
Explanation:
Velocity at the bottom of height h
= √2gh
deceleration on rough horizontal surface
= μg , μ is coefficient of friction
= .27 x 9.8
= 2.646 m / s²
v² = u² - 2as
0 = 2gh - 2 x 2.646 x 19
h = 2 x 2.646 x 19 / 2 x 9.8
= 5.13 m