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jenyasd209 [6]
3 years ago
12

Write a question about how changing temperature affects gas

Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0

Answer:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

Explanation:

One possible question can be:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:

V\propto T

Or also written as

\frac{V}{T}=const.

By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.

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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

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Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
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