Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Answer is
9.773m/s^2
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Given,
h=8848m
The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.
g′=g(1 − 2h/h)
=9.8(1 - 6400000/17696)
=9.8(1 − 0.00276)
9.8×0.99724
=9.773m/s^2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2
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hope this helps :)
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status: mineral
Answer:
Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits. Bohr built upon Rutherford's model of the atom. ... So it was not possible for electrons to occupy just any energy level.
Explanation:
