\[\boxed{\mathbf{1245.ОК\ ГДЗ - домашка\ н}а\ 5}\]
\[\mathbf{Дано:}\]
\[p = 11,4\frac{г}{см^{3}};\]
\[\delta = 4\ мм;\]
\[d_{0} = 13\ мм;\]
\[l = 25\ м.\]
\[\mathbf{Найти:}\]
\[m - ?\]
\[\mathbf{Решение.}\]
\[1)\ p = 11,4\ \frac{г}{см^{3}} =\]
\[= 11,4\ \bullet 10^{3}\ \frac{кг}{м^{3}}.\]
\[2)\ \delta = 4\ мм = 4 \bullet 10^{- 3}\ м.\]
\[3)\ d_{0} = 13\ мм = 13 \bullet 10^{- 3}\ м.\]
\[4)\ S = \pi r_{1}^{2} - \pi t_{0}^{2} =\]
\[= \frac{\pi}{4}\left( d_{1}^{2} - d_{0}^{2} \right) =\]
\[= \frac{\pi}{4}(\left( d_{0} + 2\delta \right)^{2} - d_{0}^{2} =\]
\[= \frac{\pi}{4}\left( d_{0} + 2\delta - d_{0} \right)\left( d_{0} + 2\delta + d_{0} \right) =\]
\[= \frac{\pi}{4} \cdot 2\delta \cdot 2\left( d_{0} + \delta \right) =\]
\[= \pi\delta\left( d_{0} + \delta \right).\]
\[5)\ m = pV = pSl =\]
\[= \pi l\delta p\left( d_{0} + \delta \right);\]
\[m =\]
\[= \pi \cdot 11,4 \cdot 10^{3} \cdot 25 \cdot 4 \cdot 17 \cdot 10^{- 3} \approx\]
\[\approx 60,9\ кг.\]
\[Ответ:\ масса\ трубы\ примерно\ \]
\[60,9\ кг.\]