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Answer:
(a) The time the ball stays in the air is approximately 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters
(c) The maximum altitude of the ball is approximately 26.62 m
Explanation:
The given parameters of the question are;
The initial velocity at which a baseball cannon fires balls, u = 29 m/s
The angle at which the cannon is tilted, θ = 52°
(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;
Where;
t = The time it takes the baseball to reach maximum height
2·t = The total time of flight = The time the ball stays in the air
θ = The angle at which the ball is tilted = 52°
g = The acceleration due to gravity ≈ 9.81 m/s²
u = The initial velocity = 29 m/s
Therefore, we have;
The time the ball stays in the air, 2·t ≈ 4.66 seconds
(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;
∴ The distance from the cannon at which the ball will hit the ground = R
The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters
(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;
Therefore, we have;
The maximum altitude of the ball ≈ 26.62 m
Answer:
a)n =7 , b) n = 5
Explanation:
The energy levels of the hydrogen atom is described by the Bohr model
= - 13.606 1/n²
This equation energy is given in elector volts and n is an integer
A transition occurs when the electro sees from a superior to a lower state
E₀ - = -13.606 (1/
² - 1/n₀²)
Let's apply this expression
n₀ = 2
Let's look for the energy of the different levels and subtract it
n₀ (eV)
1 -13,606
2 -3.4015
3 - 1.5118
4 -0.850375
5 -0.54424
6 -0.3779
7 -0.2777
The wavelength of the transition is 397 nm = 397 10⁻⁹ m
The speed of light is related to wavelength and frequency
c = λ f
The Planck equation gives the energy of a transition
E = h f
E = h c /λ
Let's calculate
E = 6.63 10⁻³⁴ 3 10⁸/397 10⁻⁹
E = 5.01 10⁻¹⁹ J
Let's reduce to eV
E = 5.01 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 3.1313 eV
Let's examine the possible transitions from the initial level ni = 2
ΔE = -
= -3.1313
En = 3.13 -3.4015
= 0.2702 eV
When examining the table we see that the level that this energy has is the level of n = 7
Part B the transition is in the infrared
The frequency is 74 10¹² Hz
We use the Planck equation
E = h f
E = 6.63 10⁻³⁴ 74 10¹²
E = 4.9062 10⁻²⁰ J
E = 4.9062 10⁻²⁰ / 1.6 10⁻¹⁹
E = 0.3066 eV
We look for the level with the energy difference
ΔE = E₄- = 0.3066
= 0.3066 - 0.85037
= -0.54376 eV
When examining the table this energy has the level n = 5, therefore from this level the transition occurs