Đáp án:
\(\begin{array}{l}
a.\\
v = 9,9m/s\\
b.\\
{T_{gio}} = 720p\\
{\omega _{gio}} = \frac{\pi }{{360}}rad/p\\
{v_{gio}} = \frac{\pi }{{120}}cm/p\\
{T_{phut}} = 60p\\
{\omega _{phut}} = \frac{\pi }{{30}}rad/p\\
{v_{phut}} = \frac{{2\pi }}{{15}}cm/p\\
\frac{{{v_{gio}}}}{{{v_{phut}}}} = \frac{1}{{16}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
t = \sqrt {\frac{{2h}}{g}} = \sqrt {\frac{{2.4,9}}{{10}}} = 0,99s\\
v = gt = 10.0,99 = 9,9m/s\\
b.\\
{T_{gio}} = 12h = 720p\\
{\omega _{gio}} = \frac{{2\pi }}{{{T_{gio}}}} = \frac{{2\pi }}{{720}} = \frac{\pi }{{360}}rad/p\\
{v_{gio}} = {\omega _{gio}}{r_{gio}} = \frac{\pi }{{360}}.3 = \frac{\pi }{{120}}cm/p\\
{T_{phut}} = 60p\\
{\omega _{phut}} = \frac{{2\pi }}{{{T_{phut}}}} = \frac{{2\pi }}{{60}} = \frac{\pi }{{30}}rad/p\\
{v_{phut}} = {\omega _{phut}}{r_{phut}} = \frac{\pi }{{30}}.4 = \frac{{2\pi }}{{15}}cm/p\\
\frac{{{v_{gio}}}}{{{v_{phut}}}} = \frac{{\frac{\pi }{{120}}}}{{\frac{{2\pi }}{{15}}}} = \frac{1}{{16}}
\end{array}\)