The acceleration due to gravity near the surface of the planet is 27.38 m/s².
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Acceleration due to gravity near the surface of the planet</h3>
g = GM/R²
where;
- G is universal gravitation constant
- M is mass of the planet
- R is radius of the planet
- g is acceleration due to gravity = ?
g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²
g = 27.38 m/s²
Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².
Learn more about acceleration due to gravity here: brainly.com/question/88039
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D. Ted associated being asked a question with embarrassment.
I think you need to add more.. but I may know where you are leading
Was he 200 m away and made the trip in 200 seconds?
If yes...
2 m/s was his speed and 0 velocity
It’s a vector quantity, which means it possesses both magnitude and direction. So the SI unit would be B)kg•m/s
Kinetic energy is related to velocity by:
KE = (1/2)mv^2
solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m
at 20 km/hr
KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J
