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Rasek [7]
2 years ago
13

What portion of the electromagnetic spectrum is used for mass spectrometry

Physics
1 answer:
Alex73 [517]2 years ago
3 0

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.

To know more about Mass spectrometry follow

brainly.com/question/17368088

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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10
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Answer:

1822.14 J/kg C

Explanation:

Mass of aluminum  = 0.100 kg

Mass of Water = 0.250 kg

Mass of copper = 0.300 kg

Mass of unknown object = 0.70 kg

Initial temperature of Aluminum = 10 C

Initial  temperature  of water = 10 C

Initial temperature of copper = 30 C

Initial temperature of = 100 C

Final temperature of all substances = 20 C

Change in temperature of each  of them Aluminum, water , copper and unknown material  respectively = 10 C , 10 C , -10 C, -80 C respectively

Specific heat of  each  of them Aluminum, water , copper and unknown material  respectively = 900,4186,387 and c respectively.

Heat gained by aluminum and water = (0.1)(900)(10) + (0.250)(4186)(10) = 11365 j

heat lost by copper and unknown  material  =(0.3)(387)(-10) = -1161

heat lost by unknown  material  =  (0.070)(c)(-80) = -5.6 c

Now solve for c using the calorimetry principle.

Heat lost + Heat gained = 0

⇒ -5.6 c = -11365 - (-1161) = -10204 j

Specific heat of unknown material = - 10204 / -5.6

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Polyurethane elastomer.

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During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu
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Answer:

The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules

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The given spring constant of the of the spring, k = 88.0 N/m

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For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose

The elastic potential energy, P.E., of a compressed spring is given as follows;

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∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²

1/2 × 88.0 N/m × (4.20 m)² = 776.16 J

The work done on the hose = The potential energy given to hose, P.E. = 776.16 J

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c

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