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nadya68 [22]
3 years ago
14

What are some ways that use physics in your everyday life?

Chemistry
1 answer:
Gwar [14]3 years ago
3 0

Answer:

Roasting s'mores over a fire.

Ironing the wrinkles out of my shirts.

Wet clothes are dried with the hot air of the dryer.

Small speakers in your headphones use electricity and moving magnets to create sound waves.

Explanation:

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How many millions of years ago did the earliest marine reptiles evolve from land reptiles?
Nezavi [6.7K]
Roughly 240 million years ago
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PLS HELPP DUE TODAY NEED DONE
Zarrin [17]

Answer:

Explanation:

Each coil increases it by a multiple of 100.

=> 50 | 3 | <u><em>15,000</em></u>

=> 100 | 3 | <u><em>30,000</em></u>

=> 150 | 3 | <u><em>45,000</em></u>

3 0
2 years ago
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A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

6 0
3 years ago
Which ones react with water?
slavikrds [6]

Answer:

I think its the last one

Explanation:

its supposed to be H 2 O 2.

8 0
2 years ago
BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.
Yakvenalex [24]

Answer:

T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

T_2=\frac{T_1P_2}{P_1}

Thus, we plug in the given data (temperature in Kelvins) to obtain:

T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C

Best regards!

3 0
3 years ago
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