Answer: this answer is D. That's what I think it is
Explanation:
Answer:
6.86 * 10^8 m
Explanation:
Parameters given:
Mass of hot gas, m = 2 kg
Gravitational Force, F = 618.2 N
Mass of Alpha Centauri, M = 2.178 * 10^30 kg
The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:
F = (G*M*m) / r²
Where G = gravitational constant
Therefore,
618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²
=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2
r² = 4.699 * 10^17 m²
=> r = 6.86 * 10^8 m
We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.
The mean radius of Alpha Centauri is 6.86 * 10^8 m.
Recall that
where 
and 
are the initial and final velocities, respecitvely; 
is the acceleration; and 
is the change in position.
So we have
(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
