<span>Subrahmanyan Chandrasekhar
</span>
In 1920, after returning from Army service, he produced a successful model and in 1923 turned it over to the Northeast Electric Company of Rochester for development.
I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.
I am not sure what this is
Answer:
h = 3.3 m (Look at the explanation below, please)
Explanation:
This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy
Kinetic energy =
m
Plug in the numbers =
(4.0)(
)
Solve = 2(64) = 128 J
Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.
Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)
Kinetic energy = Potential Energy
128 J = 39.2h
h = 3.26 m
h= 3.3 m (because of significant figures)