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3 years ago

does a Neon atom emits specific frequencies of light that correspond to decreases in energy of its electrons.

Physics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

Yes, the frequency of light emitted is a property of the difference between the levels of energy of its electrons.

Explanation:

Neon atom is a noble gas which glows when its electrons de-excites after absorbing energy.

Niels Bohr postulated that the energy level in all atoms are quantized, thus electrons do not exist in-between two levels. When electrons in the Neon atom are excited, this increase in energy causes them to jump to a higher energy levels. On de-excitation, the electrons drops to their initial level releasing the absorbed energy in the form of a photon.

The photon emitted has a frequency that is directly proportional to the energy change in the electron.

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The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is 54.55 %.

Learn more here:

brainly.com/question/20793607

3 0

3 years ago

You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.

dem82 [27]

Angle (θ) = 60°
Force (F) = 20 N
Distance (s) = 200 m
Therefore, work done
= Fs Cos θ
= (20 × 200 × Cos 60°) J
= (20 × 200 × 1/2) J
= (20 × 100) J
= 2000 J

Answer:

2000 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago

Read 2 more answers

Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 23.8 ∘ and 65.0 ∘, r

GalinKa [24]

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4 0

4 years ago

A student is told to use 20.0 g of sodium chloride to make an aqueous solution that has a concentration of 10.0 g/L (grams of so

Stells [14]

Answer:

she should add solute to the solvent

Explanation:

Given data :

Mass of the sodium chloride, = 20.0 g

Concentration of the solution = 10 g/L

Volume of 20.0 g of sodium chloride = 7.50 mL

Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L

thus, for 20 g of sodium chloride volume of the solution is 2 L or 2000 mL

also,

Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)

thus,

2000 mL = 7.5 mL + volume of solvent (water)

volume of water = (2000 - 7.5) mL

volume of water = 1992.5 mL

volume of water = 199.25 L ≈ 199 L

6 0

3 years ago

Read 2 more answers

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -

slega [8]

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

$v=\omega\sqrt{A^2-x^2}$

$\omega=\sqrt{\dfrac{k}{m}}$

$\omega=\sqrt{\dfrac{300}{2}}$

$\omega=12.24\ rad/s$

So, $\dfrac{v^2}{\omega^2}+x^2=A^2$

$\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2$

A = 0.326 m

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.

4 0

3 years ago