f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
1 answer:
Answer:
The velocity of the arrow after 3 seconds is 30.02 m/s.
Explanation:
It is given that,
An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:
We know that the rate of change of displacement is equal to the velocity of an object.
Velocity of the arrow after 3 seconds will be :
So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.
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The initial height of the first body is given by:
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
Answer:
Explanation:
The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.
Referring to table A.2 of modern compressible flow then
Also
and making
the subject of the formula then
Making reference to diagram then