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Triss [41]
2 years ago
12

In a ruby laser, an electron jumps from a higher energy level to a lower one. if the energy difference between the two levels is

1.8 ev, what is the wavelength of the emitted photon?
Physics
1 answer:
Simora [160]2 years ago
4 0

In a ruby laser, an electron jumps from a higher energy level to a lower one. if the energy difference between the two levels is 1.8 EV, 649 is the wavelength of the emitted photon

Only pulse mode is employed with the ruby laser. Water does not readily absorb radiation, but pigments like melanin and hemoglobin do so to a large extent. The anterior ocular structures are easily penetrated by the ruby laser. Vascular and colored retinal lesions are photocoagulation with it.

Ruby served as the active medium in Maiman's 1960 development of the solid-state laser known as the Ruby laser. Aluminum oxide crystals are found in rubies. where chromium ions are used to partially replace aluminum ions. Chromium ions are the substance that makes the ruby laser active.

To learn more about ruby laser please visit
brainly.com/question/28168316
#SPJ4

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Pls helpp!! Write a fictional story that takes place in a world without gravity.
kirza4 [7]

Answer: Once upon a time there was a world without gravity. Billy Bob was sitting at home and his dad went to get some milk. But when he stepped out he floated away. Billy Bob never saw his dad again. The end.

Explanation:

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A child is sitting in a sled at the top of a snowy hill. Which statement is true regarding the position of the child?
sergejj [24]

Answer:

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3 years ago
The water behind Grand Coulee Dam is 1000 m wide and 200 m deep. Find the hydrostatic force on the back of the dam. (Hint: the t
Vika [28.1K]

Answer:

The hydro static force on the back of the dam is 1.96\times10^{11}\ N

Explanation:

Given that,

Width b= 1000 m

Depth d= 200 m

We need to calculate the average pressure

Using formula of  average pressure

P_{avg}=\rho\times g\times d_{avg}

Put the value into the formula

P_{avg}=1000\times9.8\times100

P_{avg}=980000\ Pa

We need to calculate  the hydro static force on the back of the dam

Using formula of force

F = P_{avg}\times A

Put the value into the formula

F = 980000\times1000\times200

F=1.96\times10^{11}\ N

Hence, The hydro static force on the back of the dam is 1.96\times10^{11}\ N

7 0
3 years ago
2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an
Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

\delta = \frac{Pl}{AE}

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

\delta = \frac{Pl}{AE}

\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}

\delta = 0.1973mm

Therefore the elongaton of the rod in a 200mm gage length is 0.1973mm

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

\upsilon = \frac{l_{a}}{l_{s}}

Where,

\upsilon =Poission's ratio

l_{a}= Lateral strain

l_{s}= Linear strain

\upsilon = \frac{\delta/d}{\delta/l}

0.3 = \frac{\delta/22}{0.1973/200}

0.3(\frac{0.1973}{200}) = \frac{\delta}{22}

\delta = 6.5109*10^{-3}mm

Therefore the change in diameter of the rod is 6.5109*10^{-3}mm

5 0
3 years ago
_______ are mountain ranges in the oceans, where the crust is young and there is almost no sediment
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The answer is mid-ocean ridge
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