A.) kiloliter. 1 kiloliter = 1,000 liters
c.) megaliter. 1 megaliter = 1,000,000 liters
hope this helps
-- 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
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-- 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>
negative acceleration- deceleration
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity
velocity after 5 s is 
Therefore acceleration during these 5 s


therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s


(b)total distance traveled before stoppage


s=136.89 m
Answer:
distance cover is = 102.53 m
Explanation:
Given data:
speed of object is 17.1 m/s


from equation of motion we know that

where d_1 is distance covered in time t1
so
=


where d_2 is distance covered in time t2


distance cover is = 213.31 - 110.78 = 102.53 m