Answer:
A: In all cases, the acceleration was the same.
Explanation:
I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.
All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared
Answer:
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Explanation:
Affirmations
a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics
b) True. If both give the same results and use the same quantum number (n)
c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it
d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero
Answer:
g(h) = g ( 1 - 2(h/R) )
<em>*At first order on h/R*</em>
Explanation:
Hi!
We can derive this expression for distances h small compared to the earth's radius R.
In order to do this, we must expand the newton's law of universal gravitation around r=R
Remember that this law is:

In the present case m1 will be the mass of the earth.
Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

Therefore, we can see that

With a the acceleration due to the earth's mass.
Now, the taylor series is going to be (at first order in h/R):

a(R) is actually the constant acceleration at sea level
and

Therefore:

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:
