$m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3$

8 0

3 years ago

This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e

Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

$P = \frac{1}{2} \rho v^3 A C_p$

where $\rho = 1.2 kg/m^3$ is the air density, v = 8.89 m/s is the wind speed, A is the swept area and $C_p = 0.45$ is the efficiency

$10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45$

$10000 = 190A$

$A = 10000 / 190 = 52.7 m^2$

The swept area is a circle with radius r being the blade length

$\pi r^2 = A = 52.7$

$r^2 = 52.7 / \pi = 16.79$

$r = \sqrt{16.79} = 4.1 m$

4 0

3 years ago