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Sagot :
Assuming the bubble as an idel gas, we use the following
[tex]PV=\text{nRT}[/tex]Where the pressure at sea level is 101K Pa.
The pressure would be given by
[tex]P=\rho gh[/tex]Whew rho = 1000 kg/m^3, g = 9.8 m/s^2, and h = 18m.
[tex]\begin{gathered} P=\frac{1000\operatorname{km}}{m^3}\cdot\frac{9.8m}{s^2}\cdot18m \\ P_{\text{under}}=176,400Pa+101,000Pa=277,400Pa \\ \end{gathered}[/tex]The pressure under the water is 277,400 Pa.
Then, form a ratio between the initial condition of the ideal gas and the final conditions.
[tex]\begin{gathered} \frac{P_iV_i}{P_fV_f}=\frac{n_iR_iT_i}{n_fR_fT_f}=\frac{r^3_i}{r^3_f} \\ \end{gathered}[/tex]Observe that the radii are not simplified because there's a change. Solve for the final radius.
[tex]\begin{gathered} r^3_f=\frac{P_{under}\cdot r_i3}{P} \\ r_f=\sqrt[3]{\frac{277,400Pa\cdot r^3_i}{101,000Pa}}=r_i\sqrt[5]{2.75}=1\operatorname{cm}\cdot1.4 \\ r_f=1.4\operatorname{cm} \end{gathered}[/tex]Therefore, the answer is c. 1.4 cm.
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