State two uses of total internal reflection
1 answer:
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Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F =
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2
(d-r) ² = r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - = 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r =
r = [2d ± 2d ] / 2a
r = (1 ± √ (1.65 10⁻³)) =
(1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = 1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
Answer:
a) Initial speed of the ball = 14.45 m/s
b) At height 6 m speed of ball = 9.55 m/s
c) Maximum height reached = 10.65 m
Explanation:
a) We have equation of motion , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.
s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8
Substituting
Initial speed of the ball = 14.45 m/s
b) We have equation of motion , where v is the final velocity
s = 6 m, u = 14.45 m/s, a = -9.8
Substituting
So at height 6 m speed of ball = 9.55 m/s
c) We have equation of motion , where v is the final velocity
u = 14.45 m/s, v =0 , a = -9.8
Substituting
Maximum height reached = 10.65 m