Question

Представьте решение:

Answer 1

$14)Q=cm(t_{2} -t_{1})\\\\Q=cmt_{2}-cmt_{1}\\\\cmt_{1} =cmt_{2}-Q\\\\t_{1}=\frac{cmt_{2}-Q }{cm}\\\\\\15)V=\frac{1}{3}\pi R^{2}H\\\\3V=\pi R^{2}H\\\\H=\frac{3V}{\pi R^{2}}\\\\\\16)p=\frac{nmV^{2}}{3}\\\\3p=nmV^{2}\\\\m=\frac{3p}{nV^{2}}$

Answer 2

$14)\ \ \ Q=cm\, (t_2-t_1)\ \ \ \to \ \ \ t_2-t_1=\dfrac{Q}{cm}\ \ \ \to \ \ \ t_2-\dfrac{Q}{cm}=t_1\ \ ,\\\\t_1=\dfrac{cmt_2-Q}{cm}\\\\\\15)\ \ \ V=\dfrac{1}{3}\, \pi R^2H\ \Big|\cdot 3\ \ \ \ \to \ \ \ \ 3V=\pi R^2H\ \ ,\ \ H=\dfrac{3V}{\pi R^2}\\\\\\16)\ \ p=\dfrac{nmV^2}{3}\ \Big|\cdot 3\ \ \ \to \ \ \ 3p=nmV^2\ \ \ ,\ \ \ m=\dfrac{3p}{nV^2}$