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torisob [31]
3 years ago
5

A 7 L sample of gas has a pressure of 1.1 atm at a temperature of 285 K. If the pressure decreases to 0.6 atm, causing the volum

e to increase to 10 L, what is the new temperature? Round your answer to the nearest tenth.
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

A 7 L sample of gas has a pressure of 1.1 atm at a temperature of 285 K. If the pressure decreases to 0.6 atm, causing the volume to increase to 10 L, what is the new temperature? Round your answer to the nearest tenth.

Answer: 222.1K


Alecsey [184]3 years ago
4 0
A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs. 
Read more on Brainly.com - brainly.com/question/1581851#readmore
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If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:

Momentum before and after the collision is conserved.

Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
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p = m * v = (50000kg + 55000kg) * v

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What is the difference between mass and density?
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Answer:

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3 0
3 years ago
A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p
Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d =  2.0 mm = 2 x 10⁻³ m

magnitude of the electric field =  200 kN/C

Capacitance of the capacitor is calculated as follows;

C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

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