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charle [14.2K]
3 years ago
10

The frequency of which type of eletromagtic wave i just higher than that of visible light

Physics
2 answers:
Gwar [14]3 years ago
7 0

Hello, Your hero is here and that's The Human Spider!


Answer: Electromagnetic waves with wavelengths just shorter than those of visible light; Ultraviolet rays carry out more energy than visible light because it has a higher frequency. Electromagnetic Waves with wavelengths just shorter than those of ultraviolet rays; X-Rays can penetrate through most matter.



Note: If you loved this answer why not mark brainliest.


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Your friendly neighborhood Human Spider!!

yawa3891 [41]3 years ago
3 0

visible light goes from red to violet in increasing frequencies. so just higher than visible light is the <u>ultra-violet</u> light/electromagnetic wave.

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When using a different calorimeter, and mixing 50 ml of hot water at 65 degrees c with 60 ml of water in the calorimete
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<h3>Conseervation of energy</h3>

The heat capacity of the calirometer is determined by applying the principle of conservation of energy.

Heat lost by the hot water = Heat gained by the calirometer

Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c

where;

  • M is mass

mass = density x volume = ρV

Density of water = 1 g/ml

Mass of hot water = 1 x (50) = 50 g

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<h3>Equilibrium temperature</h3>

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<h3>Specific heat capacity of the calirometer</h3>

50 \times 4.184 \times (65 - 30.5) = 60 \times C_c \times (30.5 - 25)\\\\7217.4 = 330C_c\\\\C_c = \frac{7217.4}{330} \\\\C_c = 21.87 \ J/g^0C

Thus, the specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

The complete question is below

When using a different calorimeter, and mixing 50 ml of hot water at

65 degrees C with 60 ml of water in the calorimeter at 25 degrees C, the temperature of the calorimeter increased by 5.5 degrees C.

a. Calculate the heat capacity of this calorimeter?

Learn more about heat capacity here: brainly.com/question/16559442

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