-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s
-- The acceleration of gravity is 9.8 m/s².
-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.
-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================
-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
Answer:
i do belive its C
Explanation:
i remeber this question from somewhere also it makes the most sense
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
UUA, UUG, CUU, CUC, CUA and CUG
So the code CUU could mutate to CUC, CUA or CUG