本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A $1,500\,\mathrm{Watt}$ motor is used to pump water a vertical height of $2.0\,\mathrm{meters}$ out of a flooded basement through a cylindrical pipe. The water is ejected though the end of the pipe at a speed of $2.5\,\mathrm{m/s}$. Ignoring friction and assuming that all of the energy of the motor goes to the water, which of the following is the closest to the radius of the pipe? The density of water is $\rho = 1000\,\mathrm{kg/m^{3}}$.
A. $1/3\,\mathrm{cm}$
B. $1\,\mathrm{cm}$
C. $3\,\mathrm{cm}$
D. $10\,\mathrm{cm}$
E. $30\,\mathrm{cm}$
参考答案: D
本题详细解析:
Power $P$ is the rate of energy transfer, $P = dE/dt$.
The energy $dE$ given to a mass $dm$ of water is the sum of its change in kinetic and potential energy: $dE = dKE + dPE_g = \frac{1}{2}(dm)v^2 + (dm)gh$.
$P = \frac{dE}{dt} = \frac{dm}{dt} (\frac{1}{2}v^2 + gh)$.
The mass flow rate $\frac{dm}{dt}$ is $\rho \times \text{Volume flow rate} = \rho (A v) = \rho (\pi r^2) v$.
Substitute this into the power equation:
$P = (\rho \pi r^2 v) (\frac{1}{2}v^2 + gh)$.
Now, solve for $r^2$:
$r^2 = \frac{P}{\rho \pi v (\frac{1}{2}v^2 + gh)}$.
Use $P = 1500\,\mathrm{W}$, $\rho = 1000\,\mathrm{kg/m^3}$, $v = 2.5\,\mathrm{m/s}$, $h = 2.0\,\mathrm{m}$, and $g = 10\,\mathrm{m/s^2}$.
$gh = (10)(2.0) = 20\,\mathrm{J/kg}$.
$\frac{1}{2}v^2 = \frac{1}{2}(2.5)^2 = 3.125\,\mathrm{J/kg}$.
$r^2 = \frac{1500}{(1000) \pi (2.5) (3.125 + 20)} = \frac{1500}{2500 \pi (23.125)} = \frac{0.6}{\pi (23.125)}$.
$r^2 \approx \frac{0.6}{3.1416 \times 23.125} \approx \frac{0.6}{72.648} \approx 0.008259\,\mathrm{m^2}$.
$r = \sqrt{0.008259} \approx 0.0909\,\mathrm{m}$.
$r \approx 9.1\,\mathrm{cm}$, which is closest to $10\,\mathrm{cm}$.
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