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Volume $V = \frac{\pi d^2 h}{4} = \frac{\pi \times 12.6^2 \times 34.2}{4} \approx 4262.2 \text{ cm}^3$.
Since inputs have 3 significant figures, $V \approx 4260 \text{ cm}^3$.
Error calculation: $\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta h}{h}$
$\Delta V = 4260 \left( 2 \frac{0.1}{12.6} + \frac{0.1}{34.2} \right) \approx 4260 (0.01587 + 0.00292) \approx 80 \text{ cm}^3$.
Final value: $4260 \pm 80 \text{ cm}^3$.
For two Carnot engines in series where the work output is equal ($W_A = W_B$):
$Q_1 - Q_2 = Q_2 - Q_3$
Since $Q \propto T$ for a Carnot cycle, we can write:
$T_1 - T_2 = T_2 - T_3$
Rearranging for $T_2$:
$2T_2 = T_1 + T_3 \implies T_2 = \frac{600 + 400}{2} = 500 \text{ K}$.
A body starts from rest and moves under constant acceleration for 20 s. It covers distance $S_1$ in the first 10 s and $S_2$ in the next 10 s. Then $S_2$ equals:
Using $s = \dfrac{1}{2}at^2$ (from rest):
$S_1 = \dfrac{1}{2}a(10)^2 = 50a$
Total in 20 s: $S_{\text{total}} = \dfrac{1}{2}a(20)^2 = 200a$
$S_2 = 200a - 50a = 150a = 3 \times 50a = \mathbf{3S_1}$
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