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V125BC [204]
3 years ago
9

Which would most likely cause the cylinder head temperature and engine oil temperature gauges to exceed their normal operating r

anges?
Physics
1 answer:
dusya [7]3 years ago
3 0

Answer: Using fuel that has a lower-than-specified fuel rating.

Explanation:

Most likely, what causes the cylinder head temperature and engine oil temperature gauges to exceed their normal operating ranges is using fuel that has a lower-than-specified fuel rating. This can lead to detonation of the engine which is the tendency for the fuel to pre-ignite or auto-ignite in an engine's combustion chamber.The cylinder head and the engine oil are part of the automobile systems that helps in fuel combustion.

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What portion of the electromagnetic spectrum is used for mass spectrometry
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The energy of the electromagnetic spectrum is not used in mass spectrography to make measurements.

<h3>What is mass spectrometry?</h3>

In physics and chemistry, mass spectrometry refers to statistical analytical techniques that allow scientists to determine the mass distribution of various types of molecules based on their mass on a substance.

The energy of the electromagnetic spectrum is not used in mass spectrography to make measurements. The process of mass spectrometry is primarily based on the interaction of molecules with a beam of electrons (rather than photons) and the subsequent measurement.

Hence the energy of the electromagnetic spectrum is not used in mass spectrography to make measurements.

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What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
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The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

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