Sapewa

2022-01-16

Air enclosed in a sphere has density $\rho =1.4k\frac{g}{{m}^{3}}$ . What will the density be if the radius of the sphere is halved, compressing the air within?

braodagxj

Beginner2022-01-17Added 38 answers

Mass = density $\times$ volume

$m=\rho V$

Since mass is constant

this implies

$\rho V$ is constant

$\rho}_{1}{V}_{1}={\rho}_{2}{V}_{2$

$\rho}_{1}\left(\frac{4}{3}\right)\pi {R}^{3}={\rho}_{2}\left(\frac{4}{3}\right)\pi {\left(\frac{R}{2}\right)}^{3$

$\rho}_{1}=\frac{{\rho}_{2}}{8$

$\rho}_{2}=8{\rho}_{1}=8\times 1.4=11.2\text{}k\frac{g}{{m}^{3}$

Since mass is constant

this implies

RizerMix

Expert2022-01-20Added 656 answers

Let m be the mass of the enclosed air and $V=\frac{4}{3}\pi {r}^{3}$ be the volume of the sphere. Then,$\rho =\frac{m}{V}$ Modified volume ${V}^{\prime}=\frac{4}{3}\pi (r/2{)}^{3}=\frac{1}{8}\times \frac{4}{3}\pi {r}^{3}=\frac{V}{8}$ New density ${\rho}^{\prime}=\frac{m}{{V}^{\prime}}=8\frac{m}{V}=8\rho =8\times 1.4=11.2\text{}kg/{m}^{3}$ Result: $11.2\text{}kg/{m}^{3}$

alenahelenash

Expert2022-01-23Added 556 answers

I tried to solve this example, but it didn’t work for me, I want to know the answer too.Thanks

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