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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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shusha [124]

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<h3>What is the definition of destructive interference?</h3>

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5 0
3 years ago
A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on
zepelin [54]

Answer:

D=200\ km

Explanation:

distance on terrain, d_t=30\ km

  • distance on the road, d_r=70\ km
  • speed on terrain, v_t=30\ km.hr^{-1}
  • speed  on road, v_r=130\ km.hr^{-1}

<u>time taken on the terrain,</u>

t_t=\frac{d_t}{v_t}

t_t=\frac{30}{30}

t_t=1\ hr

<u>time taken to cover the distance on the road:</u>

t_r=\frac{d_r}{v_r}

t_r=\frac{70}{130}

t_r=\frac{7}{13}\ hr

<u>Now the distance covered on terrain in the total time:</u>

D= v_r\times (t_r+t_t)

D= 130\times (\frac{7}{13}+1)

D=130\times \frac{20}{13}\

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