A tire rotates 50 times in 2.5 seconds what is its period?
1 answer:
You might be interested in
Answer:
The maximum height covered is 3.25 m.
The horizontal distance covered is 9.81 m.
The total time in the air is 1.63 seconds.
Explanation:
The launch speed, .
Angle of launch with the horizontal,
So, the vertical component of the initial velocity,
.
The horizontal component of the initial velocity,
Let, t be the time of flight, to the horizontal distance covered
.
Not, applying the equation of motion in the vertical direction.
Where s is the displacement in time t, u is the initial velocity and a is the acceleration.
In this case, (from equation (i), s=0 (as the final height is same as the launch height) and
(negative sign is due to the downward direction).
seconds.
So, the total time in the air is 1.63 seconds.
From equation (i),
Total horizontal distance covered is
.
Now, for the maximum height, H, applying the equation of motion as
Here, v is the final velocity and v=0 (at the maximum height), and h=H.
So,
.
Hence, the maximum height covered is 3.25 m.
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.