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3 years ago

Help my math pleaseplease

Mathematics

2 answers:

Alex Ar [27]3 years ago

6 0

Answer:

Answer below in picture.

I hope it helps.

polet [3.4K]3 years ago

5 0

Answer:

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

Step-by-step explanation:

Given :

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

$\begin{gathered}\end{gathered}$

To Find :

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

$\begin{gathered}\end{gathered}$

Solution :

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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klio [65]

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$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$