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3 years ago

How does gay-lussac's law apply to everyday life?

1 answer:

vladimir2022 [97]3 years ago

5 0

Gay-Lussac's Law shows the direct relationship between pressure and temperature for an ideal gas with constant volume. Mathematically it is
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

This particular has a lot of application in our everyday life. In cooking, for example, we apply this concept when using a pressure cooker. We increase/decrease the temperature to meet the right amount of pressure.

In addition, knowing that pressure increases when temperature does can help you with road safety. Knowing that temperature affects heat directly, we must be careful in making sure that tires are not overheated or else they explode out of too much pressure inside.

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Drag the tiles to the correct boxes to complete the pairs.

Brut [27]

Answer:

5 percent = normal matter

68 percent = dark energy

27 percent = dark matter

Explanation:

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2 years ago

Which is a factor that influences the development of cardiovascular disease?

alexgriva [62]

Answer: Several health conditions, your lifestyle, and your age and family history can increase your risk for heart disease. These are called risk factors. About half of all Americans (47%) have at least 1 of 3 key risk factors for heart disease: high blood pressure, high cholesterol, and smoking

Explanation: Have a great day!

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3 years ago

Read 2 more answers

I need 1, 2 and 3 Please help!

svetoff [14.1K]

1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..

And I hope this helped :)

7 0

3 years ago

11. A car travels at 25 m/s to the North. It has an acceleration of 2 m/s’ to the south

Anna11 [10]

Answer:

delta x= 100m

Explanation:

vi= 25 meters

a= -27meters

t=20s

find delta x

4 0

2 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

[density * volume]_salt water = [density * volume]_fresh water ---> 1

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

h = 0.72 m (ANSWER)

8 0

3 years ago