Pho.rs: Fuel tank and Gearbox

A1 ^0.60 Write the expression that can be used to calculate the volume of fuel $V$ in the tank, given the voltmeter reading $U_1$ and the values of $R_0$, $\lambda$, $\mathcal{E}$ and $a$.

1 It's noticed that the same current flows through all elements	0.20
2 \[U_1 = \mathcal{E}\frac{\lambda h}{\lambda a + R_0}\]	0.20
3 \[V = a^3 \frac{U_1}{\mathcal{E}} \left( 1 + \frac{R_0}{\lambda a} \right)\]	0.20

A2 ^1.00 Write the expression that can be used to calculate the volume of fuel $V$ in the tank, given the voltmeter reading $U_2$ and the values of $R_1/R_2$, $\mathcal{E}$ and $a$.

1 The readings of the voltmeter $U_2$ is written using the difference of the potentials	0.40
2 \[ U_2 = \mathcal{E} \left( \frac{h}{a} - \frac{R_1}{R_1 + R_2} \right) \]	0.40
3 \[ V= a^3 \left( \frac{U_2}{\mathcal{E}} + \frac{R_1/R_2}{R_1/R_2 + 1} \right)\]	0.20

A3 ^1.60 Determine the volume of the fuel tank $V_0$. Determine ratio $R_1/R_2$ that should be used in the fuel gauge for it to give correct readings.

1 It's written that a deviated value of $R_1/R_2$ doesn't change the slope	0.40
2 It's determined that $V_0=76~\text{dm}^3$	0.40
3 A method of determinig $R_1/R_2$ is proposed, e.g. readings when $t=0$	0.40
4 \[R_1/R_2 = 23/15 \approx 1.53\]	0.40

B1 ^0.60 Sketch a graph of $\eta$ vs $\omega$.

1 The graph is a parabola with downward branches	0.20
2 The parabola goes from $(0,0)$ to $(\omega_0,0)$	0.20
3 The graph is a horizontal line $\eta=0$	0.20

B2 ^0.50 Show that if the engine's axis rotates with the angular velocity $\omega$ and an external torque $M$ acts on it, then the mechanical power of the engine equals $\omega M$.

1 It's written that $dW = M d \theta$ or analogous equation

0.50

B3 ^1.00 Determine torque $M$ of the engine. Express the answer in terms of $P_0$, $\alpha$, $\omega$ and $\omega_0$.

1 The substitution is made \[ M\omega = \eta P_0\]	0.50
2 \[ M = \frac{\alpha P_0 }{\omega_0} \left( 1 - \frac{\omega}{\omega_0} \right)\]	0.50

B4 ^1.20 Determine the angular velocity $\omega_W$ of the wheel and the torque $M_W$ acting on it from its axis $W$. Express the answer in terms of $k$, $\omega$ and $M$.

1 \[\omega_W = \omega k\]	0.30
2 \[ \omega M = \omega_W M_W \]	0.50
3 \[ M_W = \frac{M}{k}\]	0.40

B5 ^2.40 Determine the terminal velocity $v$ of the car with the engine running. Express the answer in terms of $P_0$, $\alpha$, $k$, $r$, $\omega_0$ and $\beta$. Sketch the graph of $v$ vs $k$.

1 Relation between powers \[ \eta P_0 = \beta v^2 \]or the 2nd Newton's law \[ \frac{M_W}{r} = \beta v\]is written	0.60
2 The substitution $v = k \omega r$ is made	0.20
3 \[v = \omega_0 r \frac{1}{\frac{1}{k} + k \frac{\beta r^2 \omega_0^2}{\alpha P_0}}\]	0.60
4 The graph is linear for low values of $k$	0.50
5 The graph is hyperbolic for high values of $k$	0.50

B6 ^1.10 For what value of $k=k_\text{max}$ does the car velocity reach its maximum? Express the answer in terms of $\alpha$, $P_0$, $\beta$, $r$ and $\omega_0$.

1 To determine $k_\text{max}$ the derivative is proposed	0.20
2 The derivative is calculated: \[ \left( \frac{a}{k} + bk \right)' = -\frac{a}{k^2} + b\]	0.40
3 \[k_\text{max} = \sqrt{\frac{\alpha P_0}{\beta r^2 \omega_0^2}}\]	0.50

Fuel tank and Gearbox

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