Answer:
x = 2381.2 m , t = 14.43 s , I = 8.05 10⁶ N s
Explanation:
a) to find the distance to break the sound barrier (v = 330 m / s) we use the kinematic relationship
v² = v₀² + 2 a x
To find the acceleration suppose the rocket does not break the wire that holds the load, so we can use the second Newton law to find the maximum acceleration
T = m a
a = T / m
Body mass and weight is related
W = mg
m = W / g
a = T g / W
a = 44.1 9.8 / 18.9
a = 22,867 m / s²
Let's look for the distance if part of the rest
x = v² / 2 a
x = 330² /(2 22,867)
x = 2381.2 m
.b) the time to break this barrier
v = v₀ + a t
t = v / a
t = 330 / 22,867
t = 14.43 s
c) the thrust is
I = Δp
I = m –m v₀
I = 2.44 10⁴ 330 -0
I = 8.05 10⁶ N s
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.