Explanation:
(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.
(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.
First we must find the rocket speed when the engines stop:
This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:
So, the maximum height reached by the rocket is:
(c) In the first part we have:
And in the second part:
So, the time it takes to reach the maximum height is:
(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:
So, the total time is:
The projectile maintains its horizontal component of speed because there's nothing exerting any horizontal force on it. <em>(b) </em>
Gravity has no effect on horizontal motion.
The best conclusion to draw based on the description would be: <span>A.The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
This phenomenon happened because </span><span>The electric field from a positive charge will points away from the charge while the electric field from a negative charge will points toward the charge</span>
Answer:
n₁ = 3
Explanation:
The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,
ΔE = 
- E₀ = - k²e² / 2m (1 / 
²2 - 1 / n₀²)
The energy of this transition is given by the Planck equation
E = h f = h c / λ
h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)
1 / λ = Ry (1/ ² - 1 / n₀²)
Let's apply these equations to our case
λ = 821 nm = 821 10⁻⁹ m
E = h c / λ
E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹
E = 2.423 10⁻¹⁹ J
Now we can use the Bohr equation
Let's reduce to eV
E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV
- E₀ = -13.606 (1 / ² - 1 / n₀²) [eV]
Let's look for the energy of some levels
n 
(eV) - E
(eV)
1 -13,606 E₂-E₁ = 10.20
2 -3.4015 E₃-E₂ = 1.89
3 -1.512 E₄- E₃ = 0.662
4 -0.850375
We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value
