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Zarrin [17]
3 years ago
7

1. °C= (°F-32) X 5/9. what will be the answer?

Physics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

F=9/5(°C)+32

Explanation:

Lets take an example

  • Take 25°C

\\ \sf\longmapsto \dfrac{9}{5}(25)+32

\\ \sf\longmapsto 9(5)+32

\\ \sf\longmapsto 45+32

\\ \sf\longmapsto 77°F

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Answer:

i: Gamma Rays

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Explanation:

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Exercise or Getting grader Stomata
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2 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

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3 years ago
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Eddi Din [679]

Answer: I think it’s D.

Explanation:

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3 years ago
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What is the momentum of a two-particle system
Harlamova29_29 [7]

Answer:

-67,500 kgm/s

Explanation:

1300 * 20 + 1100 * (-85) = -67,500 kgm/s

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