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In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
(1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
(2)
If we use the formula to rewrite
, eq.(2) becomes
Which we can solve to find a, the width of the slit:
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
D is the distance of the screen from the slit
a is the width of the slit
In our problem,
while the distance between the first and the fifth minima is
If we use the formula to rewrite
Which we can solve to find a, the width of the slit:
Answer: the picture is the answer
Explanation:
The air drag is a force that depends on the speed of an object relative to the wind. Under certain conditions, it can be modeled as:
Where b is a constant.
As a falling object reaches a speed so that its weight is cancelled out by the air drag, the object will reach a maximum velocity.
In a speed vs time gaph, the speed would approach the maximum speed like an asymptote.
On the other hand, since the object falls from rest, the initial speed on the graph must be zero.
Taking these considerations into account, the correct graph for the movement of an object that falls from rest if air drag is not ignored, is option B.
Gravity pulls on the object in the downward direction. The normal force to the inclined plane points in a direction perpendicular to the surface of the plane. Then for the object to be at rest, the frictional force must act parallel to the inclined plane, pointing in the upward direction.
