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
When a candle is burning the candle is releasing thermal and radiant energy
The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttles momentum
Answer:
A. attracted to the negative terminal of the voltage source.
Explanation:
When an electron is displaced in a semiconductor, the hole that's left behind is
A. attracted to the negative terminal of the voltage source.
The electron leaving leaves a net + charge, which is attracted to the negative terminal.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
