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zepelin [54]
2 years ago
11

Help my math pleaseplease ​

Mathematics
2 answers:
Alex Ar [27]2 years ago
6 0

Answer:

Answer below in picture.

I hope it helps.

polet [3.4K]2 years ago
5 0

Answer:

Volume is <u>135 </u><u>cc</u> when the temperature drops to 270K and the pressure is 150 N/cm²

Step-by-step explanation:

<u>Given</u> :

  • \small\purple\bull Volume = 225 cc
  • \small\purple\bull Temperature = 300K
  • \small\purple\bull Pressure = 100 N/cm²

\begin{gathered}\end{gathered}

<u>To</u><u> </u><u>Find</u> :

  • \small\purple\bull Volume when the temperature drops to 270K and the pressure is 150 N/cm²

\begin{gathered}\end{gathered}

<u>Solution</u> :

Mathematically, the relationship between the three variables is given by the expression :

\longrightarrow{\sf{V = \dfrac{kT}{P}}}

  • \pink\star V = Volume
  • \pink\star P = Pressure
  • \pink\star T = Temperature
  • \pink\star k = constant

\rule{200}2

Substituting all the given values in the formula to find k (constant)

\begin{gathered}\qquad\longrightarrow{\sf{225= \dfrac{300k}{100}}}\\\\\qquad\longrightarrow{\sf{300k = 225 \times 100}}\\\\\qquad\longrightarrow{\sf{300k = 22500}}\\\\\qquad\longrightarrow{\sf{k = \dfrac{22500}{300}}}\\\\\qquad\longrightarrow{\sf{k = \dfrac{225 \:  \cancel{00}}{3 \: \cancel{00}}}}\\\\\qquad\longrightarrow{\sf{k = \dfrac{225}{3}}}\\\\\qquad\longrightarrow{\sf{k = 75}}\end{gathered}

Hence, the value of k is 75.

\rule{200}2

Now, finding the volume by substituting the values in the formula :

\begin{gathered}\qquad\longrightarrow{\sf{V = \dfrac{kT}{P}}}\\\\\qquad\longrightarrow{\sf{V = \dfrac{75 \times 270}{150}}}\\\\\qquad\longrightarrow{\sf{V = \dfrac{20250}{150}}}\\\\\qquad\longrightarrow{\sf{V = \dfrac{2025 \: \cancel{0}}{15 \:  \cancel{0}}}}\\\\\qquad\longrightarrow{\sf{V = \dfrac{2025}{15}}}\\\\\qquad\longrightarrow{\sf{V = 135 \: cc }}\end{gathered}

Hence, the volume is 135 cc.

\rule{300}{2.5}

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Illusion [34]

Answer:

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Step-by-step explanation:

Given:

2 kg of taronges and 3 kg mandarins cost 11.5 euros

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