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Sav [38]
3 years ago
7

Why shouldn't technicians use oxygen or compressed air to pressurize appliances to check for leaks?

Physics
1 answer:
Flauer [41]3 years ago
5 0

The reason as to why technicians shouldn’t use oxygen or compressed air in pressurizing appliances in cases of checking leaks because when this is mixed with the compressor oil or other refrigerants, this will likely cause explosion because of the oxygen or compressed air mixed to it.

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The speed of light inside a medium is (2.0 x 10^8m/s) What is the index of refraction (n) of the medium?
kondor19780726 [428]

Answer:

Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium

1.5 = 3 x 108 / Speed of light in medium

Speed of light in the medium = 3 x 108 /1.5

= 2 x 108 m/s.

Explanation:

3 0
2 years ago
Light of wavelength 500 nm falls normally on 50 slits that are 2.5×10−3mm wide and spaced 5.0×10−3mm apart. How many interferenc
yanalaym [24]

Answer:

3

Explanation:

The solution is in the attached files below

5 0
4 years ago
Calculate the current through the ammeter (look at photo)
AleksAgata [21]

Answer:

6 amps

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2 years ago
What are the charged particles called in thermoshpere?
ira [324]
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Hope this helps! :D
3 0
4 years ago
(Inelastic) Ballistic Pendulum. A bullet that has a mass of 120g is fired horizontally at a block of jelly of mass 15kg that is
sineoko [7]

Answer:

\Delta E_{loss} = 9100.262\,J

Explanation:

The bullet-Ballistic Pendulum system can be modelled after the Principles of Momentum Conservation and Energy Conservation:

Final speed of the ballistic pendulum:

\frac{1}{2}\cdot (0.12\,kg)\cdot v^{2} = (0.12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)

v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.45\,m)}

v \approx 2.971\,\frac{m}{s}

Initial speed of the bullet:

(15\,kg)\cdot \left(0\,\frac{m}{s}\right) + (0.12\,kg)\cdot v = (0.12\,kg)\cdot \left(20\,\frac{m}{s} \right) + (15\,kg)\cdot \left(2.971\,\frac{m}{s} \right)

v = 391.375\,\frac{m}{s}

Energy loss:

\Delta E_{loss} = \frac{1}{2}\cdot (15\,kg)\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(2.971\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (0.12\,kg)\cdot \left[\left(391.375\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right]\Delta E_{loss} = 9100.262\,J

3 0
4 years ago
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