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marshall27 [118]
3 years ago
5

Which is a correct reason to physically connect objects with a bond wire when transferring flammable liquids? A. To create a pat

hway for the liquid to flow easily B. To eliminate a difference in the static charge potential C. To lessen the likelihood of a spill
Physics
2 answers:
Alona [7]3 years ago
8 0

Answer

Correct Option is A

Explanation:

Correct Option is A that is to create a pathway for the liquid to flow easily

Option B

This option is not correct because objects are not conductive.  

Option C

This option is also not correct because better options are there to lessen the spill

Hope this answer will help you


RideAnS [48]3 years ago
5 0
<h2>When Transferring Flammable Liquids - Option A</h2>

To create a pathway for the liquid to flow easily is a correct reason to physically connect objects with a bond wire when transferring flammable liquids. Physically connect the two conductive objects together with a bond wire to eliminate a difference in static charge potential among them. The bond wire is attached between the containers in the process of flammable filling liquid actions except for a metallic path is found.


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2 years ago
A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.
Shtirlitz [24]

Answer:

Wavelength = 0.7083 meters

Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

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The red shift of visible light waves that is observed by astronomers on Earth is used to determine the
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3 years ago
An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the
gregori [183]

Answer:

\lambda = 672 nm

so this is nearly red colour light

Explanation:

As we know that the interference of light from reflected light then the path difference is given as

\Delta x = 2\mu t + \frac{\lambda}{2}

now we know that for constructive interference of light the path difference is given as

\Delta x = n\lambda

so we will have

2\mu t + \frac{\lambda}{2} = N\lambda

so we will have

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8 0
3 years ago
Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys
max2010maxim [7]

Answer:

The spring was compressed the following amount:

\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount \Delta x), and the total final energy is the addition of the kinetic energies of both masses:

E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2

E_i=E_f\\

\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }

8 0
2 years ago
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