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The general relationship is $\frac{9}{Y} = \frac{3}{\eta} + \frac{1}{K}$.
- Rearranging for $K$: $\frac{1}{K} = \frac{9}{Y} - \frac{3}{\eta} = \frac{9\eta - 3Y}{Y\eta}$
- Thus, $K = \frac{Y\eta}{9\eta - 3Y}$[cite: 492].
We integrate $\frac{dP}{dx} = 100 - 12\sqrt{x}$ from $x = 0$ to $x = 25$:
$P = \int_0^{25}(100 - 12\sqrt{x})\,dx = \left[100x - 12 \cdot \frac{2}{3}x^{3/2}\right]_0^{25}$
$= \left[100x - 8x^{3/2}\right]_0^{25} = 100(25) - 8(25)^{3/2} = 2500 - 8(125) = 2500 - 1000 = 1500$
New production $= 2000 + 1500 = \mathbf{3500}$
For identical trajectories at the same angle, the ratio $\dfrac{u^2}{g}$ must be equal (since all trajectory dimensions scale with $u^2/g$):
$\dfrac{25}{9.8} = \dfrac{9}{g'} \Rightarrow g' = \dfrac{9 \times 9.8}{25} = \dfrac{88.2}{25} \approx \mathbf{3.5\text{ m/s}^2}$
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