The fourth harmonic on a string fixed at both ends shows
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The magnitude of the gravitational force exerted on the satellite by the planet will be F = 11970.1 N
<h3>What is gravitational force?</h3>
The gravitational force is a force that attracts any two objects with mass.It is directly proportional to the masses and inverse of the distance between the two objects.
It is given that,
Mass of the satellite, m = 2400 kg
Speed of the satellite, v = 6670 m/s
Radius of the circular path,
Let F is the magnitude of the gravitational force exerted on the satellite by the planet. The centripetal force is equal to the gravitational force. It is equal to :
F = 11970.1 N
So, the magnitude of the gravitational force exerted on the satellite by the planet is 11970.1 N. Hence, this is the required solution.
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Explanation:
We have,
Slit separation, d = 0.329 mm
Distance between slit and screen, D = 2.20 m
The first bright fringe is formed at a distance off 2.9 mm from the center of the pattern. We need to find the wavelength.
For double slit experiment, the fringe width is given by :
is wavelength
So, wavelength is 433 nm.