Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :
![v=u-gt](https://tex.z-dn.net/?f=v%3Du-gt)
![u=gt](https://tex.z-dn.net/?f=u%3Dgt)
![t=\dfrac{u}{g}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bu%7D%7Bg%7D)
![t=\dfrac{13.7}{9.8}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B13.7%7D%7B9.8%7D)
t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :
![h=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, a = -g
![h=ut-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dut-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2](https://tex.z-dn.net/?f=h%3D13.7%5Ctimes%201.39-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%20%281.39%29%5E2)
h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J
From the first Law of thermodynamics, obtain useful work performed as
W = Qin - Qout
= 0.5 x 10⁵ J
By definition, the efficiency is
η = W/Qin
= 100*(0.5 x 10⁵/4 x 10⁵)
= 12.5%
Answer: The efficiency is 12.5%
Ocean currents are formed by a type of heat transfer that is convection
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