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alina1380 [7]
3 years ago
10

I NEED HELP ASAP! Compare the spectrum of the unknown gas collected at the bulb company

Physics
1 answer:
Maru [420]3 years ago
4 0

The leaking gas is Neon.

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<em>Your question is not complete, kindly check the image uploaded for the remaining part of the question.</em>

The gases commonly used in fluorescent light bulbs include Hydrogen, Helium, Neon, Argon, Krypton, Xenon.

Fluorescent lamps emit light of wavelength between 400 nm to 700 nm.

In the given chart the suspected gases in the bulb include;

Hydrogen (H), maximum wavelength (nm) = 410, 434, 86, 656

Helium (He), maximum wavelength, (nm) = 438, 443, 7, 471, 42, 501, 50, 587, 66.

Neon, (Ne), maximum wavelength, (nm) = 421, 423, 425, 436,

The leaking gas is determined as follows;

  • For hydrogen the maximum wavelength fluctuated between 86 and 656, which caused the average of the maximum wavelength to be 396.5 nm. The average of the maximum wavelength is below the minimum wavelength of the Fluorescent lamp.
  • For Helium, the maximum wavelength fluctuated between 7 and 587, which caused the average of the maximum wavelength to be 282.11 nm. The average of the maximum wavelength is below the minimum wavelength of the Fluorescent lamp.
  • For Neon (Ne), the maximum wavelength is fairly constant and the average of the maximum wavelength is 426.25 nm. This value is above the minimum wavelength of the fluorescent lamps.

Thus, the leaking gas is Neon.

Learn more here: brainly.com/question/7388951

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If an object in space is giving off a frequency of 10^13 wavelength of 10^-6 what will scientist be looking for?
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Answer:

The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s

Explanation:

If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.

The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;

v = f¶

Given f = 10^13Hz and ¶ = 10^-6m,

v = 10¹³ × 10^-6

v = 10^7 m/s

The value of the velocity of the object in space that the scientist will be looking for is 10^7m/s

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
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Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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