The refractive index of a material is a dimensionless number that describes how fast light travels through the material. It is defined as n={\frac {c}{v}}, where c is the speed of light in vacuum and v is the phase velocity of light in the medium.
the ratio of the velocity of light in a vacuum to its velocity in a specified medium.
The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>
<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
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let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
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t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
Following the Law of Conservation of Energy, energy cannot be destroyed nor created.
So you would do 75 N x 10m x Cos (0 degrees)= 750 J
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx