Note that the simulation allows you to also display the force of the smaller moon
1 answer:
the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet
Explanation:
In this problem we are analzying the gravitational force acting between a planet and its moon.
The magnitude of the gravitational attraction between two objects is given by
where :
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, we are considering a planet and its moon. According to Newton's third law of motion,
"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"
If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.
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As we know that electrostatic force between two charges is given as
here we know that electrostatic repulsion force is balanced by the gravitational force between them
so here force of attraction due to gravitation is given as
here we can assume that both will have equal charge of magnitude "q"
now we have
now we have
Answer:
measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen,
Explanation:
The expression for the diffraction phenomenon is
a sin θ = m λ
for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles
tan θ = y / L
in these experiments the angles are small
tan θ = sin θ / cos θ = sin θ
sunt θ = y / L
we substitute
a = m λ
y = m L λ / a
therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.
The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.