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Harrizon [31]
3 years ago
7

The n = 8 to n = 4 transition in the bohr hydrogen atom occurs in the __________ region of the electromagnetic spectrum.

Physics
2 answers:
tiny-mole [99]3 years ago
4 0
INFRARED REGION ..............
AleksandrR [38]3 years ago
3 0

<u>Answer</u>:

The n = 8 to n = 4 transition in the Bohr hydrogen atom occurs in the Infrared region of the electromagnetic spectrum.

<u>Explanation</u>:

The electron present in a hydrogen atom revolves around the nucleus in their orbits.The electron's energy in an orbit is proportional to its distance from the nucleus. Nearer the electron is from nucleus, the less energy it has. Only a limited number of orbits is allowed with certain energies.

When an electron jumps from an orbit of higher energy, it absorbs light and releases it when an electron falls into an orbit of lower energy. This transition from higher energy level to lower energy level takes place in different region .

When  electron in the hydrogen atom undergoes transition between n=4 energy level and n=8 energy levels it produces spectral line within the infrared region with wavelength of  4.05 to 1.46 microns. This is called as brackett series.

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No they say "Watch out it's the fuzz"

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Why is it not advisable to touch a canvas tent from inside when it is raining?​
Sauron [17]

Answer: When you touch wet canvas, surface tension will draw water to your finger. However, the drop left behind where you touched, like any irregular point on an overhead surface, will draw condensation from inside the tent if it is humid.

3 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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3 years ago
Acceleration of a free-falling object in a frictionless environment increases as a function of time.
Digiron [165]

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

Force = Mass × Acceleration

The force of gravity is F_{g} =G \cdot \dfrac{M \cdot m}{r^{2}}

Where;

G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2

m  = The mass of the object

∴ The force acting on an object in free fall, F_g = m × g

Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>

<u></u>

Learn more about object in free fall here:

brainly.com/question/13712424

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6 0
2 years ago
Weekend A<br> Assignment<br> Differentiate between forced and damped oscillation
4vir4ik [10]

Answer:

A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.

Explanation:

Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.

<h2><em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u> </u></em></h2>

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