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:
the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.
w is angular frequency
formula:2π/T
now k is spring constant
F/R-->mw²
putting values:70*(2π/0.2)²
=4.9x10⁶
so we can say that SHM is not affected by the amplitude of the bounce.
Answer:
<h3>The answer is 0.67 m/s²</h3>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula

f is the force
m is the mass
From the question we have

We have the final answer as
<h3>0.67 m/s²</h3>
Hope this helps you
true if you are refering to the desing of the experimnt as it does identify the variable