A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled
1 answer:
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Answer:
Approximately .
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case, is the same as the change in the ball's height:
. By assumption, this ball was dropped with no initial velocity. As a result,
. Since the ball is accelerating due to gravity,
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.
The refractive index for glycerine is 
, while for air it is 
.
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
Answer:
Explanation:
a ) Slit separation d = .1 x 10⁻³ m
Screen distance D = 4 m
wave length of light λ = 650 x 10⁻⁹ m
Width of central fringe = λ D / d
=
= 26 mm
b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm
c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to
λ / d
=
= 6.5 x 10⁻³ radian.