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kati45 [8]
3 years ago
8

. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe

what had to happen to the pressure (if temperature remained constant). Include which law supports this observation.
Physics
2 answers:
mash [69]3 years ago
8 0
<h2>Answer:</h2>

<u>The pressure increased to 450 Kilo pascals.</u>

<h3>Explanation:</h3>

This the condition of Boyle's law:

P1V1 = P2V2

By putting values in above formula:

= 300 * 75 = P2 * 50

= P = 300 * 75/50

= p = 450 Kilo pascals

In this question description, two variables volume and pressure are involved while the temperature remains constant. So it is the condition of <em><u>Boyle's law which states that the Pressure and volume of a gas are inversely proportional if the temperature of the gas remains constant.</u></em>

julia-pushkina [17]3 years ago
3 0
By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P<span>₂*50

</span>P<span>₂*50= 300*75
</span>
P<span>₂ = 300*75/50 = 450
</span>
P<span>₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.</span>
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Where does the heat come from that drives this convection current in the mantle
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Earth's interior (Core)

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3 0
3 years ago
I need help with this question how to solve it for Brass and Cooper
Ksenya-84 [330]

Take into account that density and relative density are given by:

\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}

Cooper:

\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}

3 0
1 year ago
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound
Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

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a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

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IgorLugansk [536]

Answer:

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8 0
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