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
Forces in the universe that act over long distance, meaning the distance is greater than the diameter of the nucleus of the atom are:
1. Electrostatic force or Coulomb force: Fc=(k*Q₁*Q₂)/r²,
2. Gravitational force: Fg=(G*m₁*m₂)/r²,
3. Magnetic force: Fm=qvB,
4. London dispersion force, also known as one of the van der Waals forces.
Explanation:
zoom I'd - 9038735228 pwd -97cEeT I m getting bored wanna come for intresting talk
The resistance in this circuit is 39.8 ohms.
Explanation:
Any circuit having resistor, battery and ammeter connected in series will obey the ohm's law in basic case. So according to the Ohm's law, the current flowing in the circuit through the ammeter will be equal to the voltage shown in the voltmeter or battery and resistor is the proportionality constant. So with this law
V = IR
So, Resistance R = V/I
As the voltage is given as 23.90 V and the current is given as 0.6 A, then resistance is
R = 23.90/0.6 = 39.8 ohms.
So, the resistance in this circuit is 39.8 ohms.
Can you give us the options…?