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Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Answer:
1min since there is no gravity on the moon so it will take time to drop on the moon.
Explanation:
Answer:
The duration is ![T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m](https://tex.z-dn.net/?f=T%20%20%3D72%20%5C%20%20years%20%2Ftex%5D%3C%2Fp%3E%3Cp%3EExplanation%3A%3C%2Fp%3E%3Cp%3EFrom%20the%20question%20we%20are%20told%20that%20%3C%2Fp%3E%3Cp%3E%20%20%20%20The%20%20distance%20is%20%20%5Btex%5DD%20%20%3D%20%2035%20%5C%20light-years%20%3D%2035%20%2A%20%209.46%20%2A10%5E%7B15%7D%20%3D%203.311%20%2A10%5E%7B17%7D%20%5C%20%20m%20)
Generally the time it would take for the message to get the the other civilization is mathematically represented as
Here c is the speed of light with the value 
=> 
=> 
converting to years
Now the total time taken is mathematically represented as
=> 
=> [tex]T =72 \ years /tex]
