Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. 
2. 
Since both cars will travel the same distance x, we can equal both formulas and solve for t:

We simplify the fraction present and rearrange for our formula so that it equals 0:

In the very last step we factored a common factor t. There is two possible solutions to the equation at
and:

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at
(when the SUV passed the police car) and
(when the police car catches up to the SUV)
Longitudinal waves transfer energy parallel to the direction of the wave motion
Answer
Given,
refractive index of film, n = 1.6
refractive index of air, n' = 1
angle of incidence, i = 35°
angle of refraction, r = ?
Using Snell's law
n' sin i = n sin r
1 x sin 35° = 1.6 x sin r
r = 21°
Angle of refraction is equal to 21°.
Now,
distance at which refractive angle comes out
d = 2.5 mm
α be the angle with horizontal surface and incident ray.
α = 90°-21° = 69°
t be the thickness of the film.
So,


t = 2.26 mm
Hence, the thickness of the film is equal to 2.26 mm.
Naturally we assume that 10000 km/hr is initial velocity (same as being shot from a cannon), and no air resistance. With so high a velocity, the effect of diminishing gravity with increasing radius must be taken into account, so you use an energy solution. M is earth mass, r is earth radius.
KE/m = (9000000/3600)^2/2 = 3858025 J/kg
ΔPE/m = (PE(at height) - PE(at surface))/m = -GM/(r+h) + GM/r
KE/m = ΔPE/m
KE/m - GM/r = -GM/(r+h)
h = -GM / (KE/m - GM/r) - r = 335665.44 m
(Using G = 6.673E-11 Nm^2/kg^2, M = 5.9742E24 kg, r = 6378100 m)