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storchak [24]
3 years ago
5

If there is 3.4 m3 of methane gas in a container with a pressure of 18.9 atm and the container expands until the methane has a p

ressure of 2.2 atm, what is the final volume of the methane? Temperature is constant at 305 K. A. 12.23 m3 B. 141.37 m3 C. 29.21 m3 OD. 2.53 m3​
Physics
1 answer:
Veseljchak [2.6K]3 years ago
7 0

Answer:

C. 29.21 m³

Explanation:

<u>Given the following data;</u>

  • Initial volume, V1 = 3.4 m³
  • Initial pressure, P1 = 18.9 atm
  • Final pressure, P2 = 2.2 atm

To find the final volume, we would use Boyle's law;

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Making V2 the subject of formula, we have;

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

Substituting the values into the formula, we have;

V_{2} = \frac {18.9 * 3.4}{2.2}

V_{2} = \frac {64.26}{2.2}

<em>Final volume, V2 = 29.21 m³</em>

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Answer:308 N/m

Explanation:

Given

mass\left ( m\right )=0.129 kg

wavelength\left ( \lambda \right )=3.86\times 10^7

We know frequency =\frac{c}{\lambda }=\frac{3\times 10^8}{3.86\tmes 10^7}

f=7.772 Hz

As the frequency of radio waves is same as the frequency at which object oscillates

f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}

7.772=\frac{1}{2\pi }\sqrt{\frac{k}{0.129}}

7.772\times 2\times \pi =\sqrt{\frac{k}{0.129}}

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7 0
3 years ago
If a car goes at a speed of 80 kph in 10 seconds, what is the distance
Blababa [14]

Answer:222 meters

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Speed=80kph=(80x5/18)=(80x5)/18=22.2m/s

Time=10 seconds

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3 0
4 years ago
I need help with this please
Firdavs [7]
C is the answer to the question
6 0
3 years ago
¿Por qué en algunas partes el agua tiene mayor presión?
Phantasy [73]
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Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,
Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

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a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

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we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

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Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

 

6 0
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