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0 | Consider a neutron star with a mass of $\quad 2.0 .10^{30} \mathrm{~kg}$, an average radius of $\quad 1.0 .10^{4} \mathrm{~m}$, and a rotation period of $2.0 .10^{-2} \mathrm{~s}$. a - Calculate the flattening factor, given that the gravitational constant is $6.67 .10^{-11}$ N. m ${ }^{2} . \mathrm{kg}^{-2}$. | $3,7.10^{-4}$ | For equilibrium we have $\mathrm{F}_{\mathrm{c}}=\mathrm{F}_{\mathrm{g}}+\mathrm{N}$ where N is normal to the surface. Resolving into horizontal and vertical components, we find: $$ \begin{gathered} F_{g} \cdot \cos (\phi)=F_{c}+N \cdot \sin (\alpha) \\ \quad F_{g} \cdot \sin (\phi)=N \cdot \cos (\alpha) \end{gather... | IPHO 1990 | value | ||
1 | In the long run (over many years) the rotation of the star slows down, due to energy loss, and this leads to a decrease in the flattening. The star has however a solid crust that floats on a liquid interior. The solid crust resists a continuous adjustment to equilibrium shape. Instead, starquakes occur with sudden chan... | 0.95 | For a point mass of 1 kg on the surface, $$ U_{p o t}=-\frac{G \cdot M}{r} \quad U_{k i n}=\frac{1}{2} \cdot \omega^{2} \cdot r^{2} \cdot \cos ^{2}(\phi) $$ The form of the surface is such that $\mathrm{U}_{\text {pot }}-\mathrm{U}_{\text {kin }}=$ constant. For the equator ( $\Phi=0$, $r=r_{e}$ ) and for the pole ( ... | IPHO 1990 | value | ||
2 | An electron after leaving a device, which accelerated it with the potential difference $U$, falls into a region with an inhomogeneous field $\boldsymbol{B}$ generated with a system of stationary coils $L_{1}, L_{2}, \ldots, L_{n}$. The known currents in the coils are $i_{1}, i_{2}, \ldots, i_{n}$, respectively. What s... | $$ i^{\prime}{ }_{n}=-35.0 i_{n} . $$ | At the beginning one should notice that the kinetic energy of the electron accelerated with the potential difference $U=511 \mathrm{kV}$ equals to its rest energy $E_{0}$. Therefore, at least in the case of the electron, the laws of the classical physics cannot be applied. It is necessary to use relativistic laws. The... | IPHO 1989 | equation | ||
3 | How many times would the resolving power of the above microscope increase or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave pro... | 35 | The resolving power of the microscope (in the meaning mentioned in the text of the problem) is proportional to the wavelength, in our case to the length of the de Broglie wave: $$ \lambda=\frac{h}{p} $$ where $h$ denotes the Planck constant and $p$ is the momentum of the particle. We see that $\lambda$ is inversely p... | IPHO 1989 | value | ||
4 | Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or B$)$ of their saturated vapors obey, to a good approximation, the formula: $$ \ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}, $$ where $p_{o}$ denotes the normal atmo... | 77 | The liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\mathrm{A}$ or B$)$, one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$. Then $\ln \left(p_{i} / p_{0}\right)=0$, and... | IPHO 1989 | value | ||
5 | A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the s... | 72.4 | conservation of energy: $$ \mathrm{M} \cdot \mathrm{~g} \cdot \mathrm{~s}=\frac{1}{2} \cdot \mathrm{I}_{\mathrm{A}} \cdot \omega^{2} $$ where $\omega$ is the angular speed of the wheel and $\mathrm{I}_{\mathrm{A}}$ is the moment of inertia about the axis through A. Note: If we would take the moment of inertia about ... | IPHO 1988 | value | ||
6 | A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the s... | $5,13 \cdot 10^{-3}$ | Kinetic energy of linear motion of the centre of mass of the wheel is $\mathrm{E}_{\mathrm{T}}=\frac{1}{2} \cdot \mathrm{M} \cdot \mathrm{v}^{2}=\frac{1}{2} \cdot \mathrm{M} \cdot \omega^{2} \cdot \mathrm{r}^{2}=\frac{1}{2} \cdot 0,40 \cdot 72,4^{2} \cdot 9 \cdot 10^{-6}=9,76 \cdot 10^{-3} \mathrm{~J}$ Potential energy... | IPHO 1988 | value | ||
7 | A cylindrical wheel of uniform density, having the mass $\mathrm{M}=0,40 \mathrm{~kg}$, the radius $\mathrm{R}=0,060 \mathrm{~m}$ and the thickness $\mathrm{d}=0,010 \mathrm{~m}$ is suspended by means of two light strings of the same length from the ceiling. Each string is wound around the axle of the wheel. Like the s... | 1.96 | Let $\frac{\mathrm{T}}{2}$ be the tension in each string. Torque $\tau$ which causes the rotation is given by $\tau=\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{r}=\mathrm{I}_{\mathrm{A}} \cdot \alpha$ where $\alpha$ is the angular acceleration $\quad \alpha=\frac{\mathrm{M} \cdot \mathrm{g} \cdot \mathrm{r}}{\mathrm{I}_{... | IPHO 1988 | value | ||
8 | A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \vareps... | $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$ | $\mathrm{r}_{0}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2}$ $\mathrm{p}_{0}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$ since $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdo... | IPHO 1988 | expression | ||
9 | A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \vareps... | $m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$ | $\left|\overrightarrow{\mathrm{v}}_{\mathrm{e}}\right| \ldots .$. speed of the external electron before the capture $\left|\vec{V}_{i}\right| \ldots \ldots$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ before capturing $\left|\vec{V}_{f}\right|$...... speed of $A^{(\mathrm{Z}-1)+}$ after capturing $\mathrm{E}_{\mathrm{n}}=... | IPHO 1988 | equation | ||
10 | A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \vareps... | $-E_{R} \cdot Z^{2}$ | Determination of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$ : potential energy $=-\frac{Z \cdot e^{2}}{4 \cdot \pi \cdot \varepsilon_{0} \cdot r_{0}}=-\frac{Z \cdot q^{2}}{r_{0}}$ kinetic energy $=\frac{p^{2}}{2 \cdot m}$ If the motion of the electrons is confined within the $x-y$-plane, principles of uncertainty in ... | IPHO 1988 | value | ||
11 | A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \vareps... | $-2 \cdot E_{R} \cdot\left(Z-\frac{1}{4}\right)^{2}$ | In the case of $\mathrm{A}^{(\mathrm{Z}-1)+}$ ion captures a second electron potential energy of both electrons $=-2 \cdot \frac{Z \cdot q^{2}}{r_{0}}$ kinetic energy of the two electrons $=2 \cdot \frac{\mathrm{p}^{2}}{2 \cdot \mathrm{~m}}=\frac{\hbar^{2}}{\mathrm{~m}_{\mathrm{e}} \cdot \mathrm{r}_{0}^{2}}$ potential ... | IPHO 1988 | value | ||
12 | A gas consists of positive ions of some element (at high temperature) and electrons. The positive ion belongs to an atom of unknown mass number Z. It is known that this ion has only one electron in the shell (orbit). Let this ion be represented by the symbol $A^{(Z-1)+}$ Constants: electric field constant $$ \vareps... | 4.1 | The ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ is at rest when it captures the second electron also at rest before capturing. From the information provided in the problem, the frequency of the photon emitted is given by $v=\frac{\omega}{2 \cdot \pi}=\frac{2,057 \cdot 10^{17}}{2 \cdot \pi} \mathrm{~Hz}$ The energy equation can ... | IPHO 1988 | value | ||
13 | Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ... | 279 | Temperature $\mathrm{T}_{1}$ where the cloud ceiling forms $$ \mathrm{T}_{1}=\mathrm{T}_{0} \cdot\left(\frac{\mathrm{p}_{1}}{\mathrm{p}_{0}}\right)^{1-\frac{1}{\mathrm{x}}}=279 \mathrm{~K} $$ | IPHO 1987 | value | ||
14 | Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ... | 1410 | Height $\mathrm{h}_{1}$ of the cloud ceiling: $$ \begin{aligned} & \mathrm{p}_{0}-\mathrm{p}_{1}=\frac{\rho_{0}+\rho_{1}}{2} \cdot \mathrm{~g} \cdot \mathrm{~h}_{1}, \text { with } \rho_{1}=\rho_{0} \cdot \frac{\mathrm{p}_{1}}{\mathrm{p}_{0}} \cdot \frac{\mathrm{~T}_{0}}{\mathrm{~T}_{1}} \\ & \mathrm{~h}_{1}=1410 \mat... | IPHO 1987 | value | ||
15 | Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ... | 271 | Temperature $T_{2}$ at the ridge of the mountain. The temperature difference when the air is ascending from the cloud ceiling to the mountain ridge is caused by two processes: - adiabatic cooling to temperature $\mathrm{T}_{\mathrm{x}}$, - heating by $\Delta \mathrm{T}$ by condensation. $$ \begin{aligned} & \mathrm{... | IPHO 1987 | value | ||
16 | Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ... | 35 | Height of precipitated water column $$ \mathrm{h}=35 \mathrm{~mm} $$ | IPHO 1987 | value | ||
17 | Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100 kPa are measured at meteorological stations $\mathrm{M}_{0}$ and $\mathrm{M}_{3}$ and a pressure of 70 kPa at station $\mathrm{M}_{2}$. The temperature of the air at $\mathrm{M}_{0}$ is $20^{\circ... | 300 | Temperature $\mathrm{T}_{3}$ behind the mountain $$ \mathrm{T}_{3}=\mathrm{T}_{2} \cdot\left(\frac{\mathrm{p}_{3}}{\mathrm{p}_{2}}\right)^{1-\frac{1}{\mathrm{x}}}=300 \mathrm{~K} $$ The air has become warmer and dryer. The temperature gain is caused by condensation of vapour. | IPHO 1987 | value | ||
18 | $A$ beam of electrons emitted by a point source $P$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small ( $2 \cdot \alpha_{0} \ll 1$ ). The injection of the electrons occurs on the mean ... | $1.48 \cdot 10^{-2}$ | Determination of B : The vector of the velocity of any electron is divided into components parallel with and perpendicular to the magnetic field $\overrightarrow{\mathrm{B}}$ : $$ \vec{v}=\vec{v}_{\|}+\vec{v}_{\perp} $$ The Lorentz force $\overrightarrow{\mathrm{F}}=-\mathrm{e} \cdot(\overrightarrow{\mathrm{v}} \tim... | IPHO 1987 | value | ||
19 | $A$ beam of electrons emitted by a point source $P$ enters the magnetic field $\vec{B}$ of a toroidal coil (toroid) in the direction of the lines of force. The angle of the aperture of the beam $2 \cdot \alpha_{0}$ is assumed to be small ( $2 \cdot \alpha_{0} \ll 1$ ). The injection of the electrons occurs on the mean ... | $0.37 \cdot 10^{-2} $ | Determination of $\mathrm{B}_{1}$ : Analogous to eq. (2) $$ m \cdot \frac{v_{0}^{2}}{R}=e \cdot v_{0} \cdot B_{1} $$ must hold. From eq. (7) follows $$ \mathrm{B}_{1}=\frac{1}{\mathrm{R}} \cdot \sqrt{2 \cdot \frac{\mathrm{~m}}{\mathrm{e}} \cdot \mathrm{~V}_{0}}=0.37 \cdot 10^{-2} \frac{\mathrm{Vs}}{\mathrm{~m}^{2}}... | IPHO 1987 | value | ||
20 | When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image>Determine how $\Phi$ depends on $\omega$, L and C ( $\omega$ is the angular frequency of the sine wave). | $\varphi=2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right) \text { with } 0 \leq \omega \leq \frac{2}{\sqrt{\mathrm{~L} \cdot \mathrm{C}}}$ | $$ \text { Current law: } \quad \mathrm{I}_{\mathrm{L}_{\mathrm{n}-1}}+\mathrm{I}_{\mathrm{C}_{\mathrm{n}}}-\mathrm{I}_{\mathrm{L}_{\mathrm{n}}}=0 $$ Voltage law: $\quad \mathrm{V}_{\mathrm{C}_{\mathrm{n}-1}}+\mathrm{V}_{\mathrm{L}_{\mathrm{n}-1}}-\mathrm{V}_{\mathrm{C}_{\mathrm{n}}}=0$ Capacitive voltage drop: $\mat... | IPHO 1987 | expression | ||
21 | When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image> Determine the velocity of propagation of the waves if the length of each unit is $\ell$. | $\mathrm{v}=\frac{\ell}{\Delta \mathrm{t}}=\frac{\omega \cdot \ell}{\varphi} \quad \text { or } \quad \mathrm{v}=\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\right)}$ | The distance $\ell$ is covered in the time $\Delta \mathrm{t}$ thus the propagation velocity is $$ \mathrm{v}=\frac{\ell}{\Delta \mathrm{t}}=\frac{\omega \cdot \ell}{\varphi} \quad \text { or } \quad \mathrm{v}=\frac{\omega \cdot \ell}{2 \cdot \arcsin \left(\frac{\omega \cdot \sqrt{\mathrm{~L} \cdot \mathrm{C}}}{2}\ri... | IPHO 1987 | expression | ||
22 | When sine waves propagate in an infinite LC-grid (see the figure below) the phase of the acvoltage across two successive capacitors differs by $\Phi$. <image>State under what conditions the propagation velocity of the waves is almost independent of $\omega$. Determine the velocity in this case. | $\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2} \ll 1$ and $ \mathrm{v}_{0}=\frac{\ell}{\sqrt{\mathrm{L} \cdot \mathrm{C}}} $ | Slightly dependent means arc $\sin \left(\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2}\right) \sim \omega$, since v is constant in that case. This is true only for small values of $\omega$. That means $\frac{\omega \cdot \sqrt{\mathrm{L} \cdot \mathrm{C}}}{2} \ll 1$ and therefore $$ \mathrm{v}_{0}=\frac{\e... | IPHO 1987 | expression | ||
23 | Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse sei... | $ t=\frac{2 R \sin \theta}{v}, \quad \text { for } \theta \leq \cos ^{-1}\left(\frac{R_{C}}{R}\right) $ | Figure2.1 $$ \mathrm{EX}=2 R \sin \theta \quad \therefore t=\frac{2 R \sin \theta}{v} $$ where $v=v_{P}$ for P waves and $v=v_{S}$ for S waves. This is valid providing $X$ is at an angular separation less than or equal to $X^{\prime}$, the tangential ray to the liquid core. X ' has an angular separation given by, fr... | IPHO 1986 | equation | ||
24 | Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse sei... | $\theta=\left[90-\sin ^{-1}\left(\frac{v_{C P}}{v_{P}} \sin i\right)+i-\sin ^{-1}\left(\frac{R_{C}}{R} \sin i\right)\right]$ | $\frac{R_{C}}{R}=0.5447 \quad$ and $\quad \frac{v_{C P}}{v_{P}}=0.831 .3$ Figure 2.2 From Figure 2.2 $\theta=\hat{A O C}+E \hat{O A} \Rightarrow \theta=(90-r)+(1-\alpha)$ (ii) Continued Snell's Law gives: $\frac{\sin i}{\sin r}=\frac{v_{P}}{v_{C P}}$. From the triangle EAO , sine rule gives $\frac{R_{C}}{\sin x}=\f... | IPHO 1986 | expression | ||
25 | Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_{c}$. The core region within radius $R_{c}$ contained a liquid. Figure 2.1 <image> Figure 2.1 The velocities of longitudinal and transverse sei... | $114^{\circ}$ | Using the result $t=\frac{2 r \sin \theta}{v}$ the time delay $\Delta t$ is given by $\Delta t=2 R \sin \theta\left[\frac{1}{v_{S}}-\frac{1}{v_{P}}\right]$ Substituting the given data $131=2(6370)\left[\frac{1}{6.31}-\frac{1}{10.85}\right] \sin \theta$ Therefore the angular separation of E and X is $2 \theta=17.84^{\... | IPHO 1986 | value | ||
26 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and... | $m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right) $ | $m \frac{d^{2} u_{2}}{d t^{2}}=k\left(u_{3}-u_{2}\right)+k\left(u_{1}-u_{2}\right)$ | IPHO 1986 | equation | ||
27 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and... | $m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right)$ | $m \frac{d^{2} u_{1}}{d t^{2}}=k\left(u_{2}-u_{1}\right)+k\left(u_{3}-u_{1}\right)$ | IPHO 1986 | equation | ||
28 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image>If each mass is displaced from equilibrium by small displacements $u_{1}, u_{2}$ and... | $m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)$ | $m \frac{d^{2} u_{3}}{d t^{2}}=k\left(u_{1}-u_{3}\right)+k\left(u_{2}-u_{3}\right)$ | IPHO 1986 | equation | ||
29 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> in terms of its displacement and those of the adjacent masses when the particles ar... | $$ \begin{array}{ll} m \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+k\left(u_{n-1}-u_{n}\right) & n=1,2 \ldots \ldots N \\ \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+\omega_{o}^{2}\left(u_{n-1}-u_{n}\right) & \end{array} $$ | Equation of motion of the n'th particle: $$ \begin{array}{ll} m \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+k\left(u_{n-1}-u_{n}\right) & n=1,2 \ldots \ldots N \\ \frac{d^{2} u_{n}}{d t^{2}}=k\left(u_{1+n}-u_{n}\right)+\omega_{o}^{2}\left(u_{n-1}-u_{n}\right) & \end{array} $$ Substituting $u_{n}(t)=u_{n}(... | IPHO 1986 | equation | ||
30 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> Determine the ratio $$ u_{n} / u_{n+1} $$ for large $N$ in the case of low freque... | 1 | For s'th mode $$ \frac{u_{n}}{u_{n+1}}=\frac{\frac{u_{n}}{u_{n+1}}=\frac{\sin \left(2 n s \frac{\pi}{N}\right)}{\sin \left(2(n+1) s \frac{\pi}{N}\right)}}{\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)+\cos \left(2 n s \frac{\pi}{N}\right) \sin \left(2 s \frac{\pi}{N}\right)} $$ (a) For sma... | IPHO 1986 | value | ||
31 | Three particles, each of mass $m$, are in equilibrium and joined by unstretched massless springs, each with Hooke's Law spring constant $k$. They are constrained to move in a circular path as indicated in Figure 3.1. Figure 3.1 <image> Determine the ratio $$ u_{n} / u_{n+1} $$ for large $N$ in the case of $\omega=\o... | -1 | For s'th mode $$ \frac{u_{n}}{u_{n+1}}=\frac{\frac{u_{n}}{u_{n+1}}=\frac{\sin \left(2 n s \frac{\pi}{N}\right)}{\sin \left(2(n+1) s \frac{\pi}{N}\right)}}{\sin \left(2 n s \frac{\pi}{N}\right) \cos \left(2 s \frac{\pi}{N}\right)+\cos \left(2 n s \frac{\pi}{N}\right) \sin \left(2 s \frac{\pi}{N}\right)} $$ The highest ... | IPHO 1986 | value | ||
32 | A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensit... | $$ a=\frac{\lambda}{4 \sin \frac{1}{2} \varphi}, \quad \mathcal{G}_{A}=\frac{1}{2} \varphi, \quad \mathcal{G}_{B}=-\frac{1}{2} \varphi \quad \text { and } \quad \delta=\frac{1}{2} \pi-2 \pi N $$ | a) Let the electrical signals supplied to rods 1 and 2 be $E_{1}=E_{0} \cos \omega t$ and $E_{2}=E_{0} \cos (\omega t+\delta)$, respectively. The condition for a maximum signal in direction $\mathfrak{9}_{A}$ (Fig. 4) is: $$ \frac{2 \pi a}{\lambda} \sin 9_{A}-\delta=2 \pi N $$ and the condition for a minimum signal i... | IPHO 1985 | equation | ||
33 | A young radio amateur maintains a radio link with two girls living in two towns. He positions an aerial array such that when the girl living in town A receives a maximum signal, the girl living in town B receives no signal and vice versa. The array is built from two vertical rod aerials transmitting with equal intensit... | 114.5 | b) The wavelength $\lambda=c / v=11.1 \mathrm{~m}$, and the angle between directions A and $\mathrm{B}, \varphi=157^{\circ}-72^{\circ}=85^{\circ}$. The minimum distance between the rods is $a=4.1 \mathrm{~m}$, while the direction of the symmetry line of the rods is $72^{\circ}+42.5^{\circ}=114.5^{\circ}$ measured from ... | IPHO 1985 | value | ||
34 | In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be... | 4.06 | a) First the electron velocity is calculated from the current I: $$ I=j S=n e_{0} v b c, \quad v=\frac{I}{n e_{0} b c}=25 \mathrm{~m} / \mathrm{s} $$ The components of the electric field are obtained from the electron velocity. The component in the direction of the current is $$ E_{\|}=\frac{v}{\mu}=3.2 \mathrm{~V} ... | IPHO 1985 | value | ||
35 | In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be... | 25 | b) The potential difference is $$ U_{H}=E_{\perp} b=25 \mathrm{mV} $$ | IPHO 1985 | value | ||
36 | In a long bar having the shape of a rectangular parallelepiped with sides $a$, $b$, and $c(a \gg b \gg c)$, made from the semiconductor InSb flows a current $I$ parallel to the edge $a$. The bar is in an external magnetic field $B$ which is parallel to the edge $c$. The magnetic field produced by the current $I$ can be... | $$ \bar{U}_{H}=\frac{I_{0} B_{0}}{2 n e_{0} c} \cos \delta $$ | c) The potential difference $U_{H}$ is now time dependent: $$ U_{H}=\frac{I B b}{n e_{0} b c}=\frac{I_{0} B_{0}}{n e_{0} c} \sin \omega t \sin (\omega t+\delta) $$ The DC component of $U_{H}$ is $$ \bar{U}_{H}=\frac{I_{0} B_{0}}{2 n e_{0} c} \cos \delta $$ | IPHO 1985 | expression | ||
37 | A particle moves along the positive axis $O x$ (one-dimensional situation) under a force having a projection $F_{x}=F_{0}$ on $O x$, as represented, as function of $x$, in the figure 1.1. In the origin of the $O x$ axis is placed a perfectly reflecting wall. A friction force, with a constant modulus $F_{f}=1,00 \mathrm... | 20 | a. It is possible to make a model of the situation in the problem, considering the Ox axis vertically oriented having the wall in its' lower part. The conservative force $F_{x}$ could be the weight of the particle. One may present the motion of the particle as the vertical motion of a small elastic ball elastically col... | IPHO 1984 | value | ||
38 | Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\fr... | 500 | a. The ray with the wavelength $\lambda_{0}$ pass trough the prisms system without refraction on $A C$ face at any angle of incidence if : $n_{1}\left(\lambda_{0}\right)=n_{2}\left(\lambda_{0}\right)$ Because the dependence of refraction indexes of prisms on wavelength has the form : $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{... | IPHO 1983 | value | ||
39 | Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\fr... | 30,7 | c. In the figure 1.2 is presented the path of ray with wavelength $\lambda_{0}$ at minimum deviation (the angle between the direction of incidence of ray and the direction of emerging ray is minimal). Figure 3.2 In this situation $n_{1}\left(\lambda_{0}\right)=n_{2}\left(\lambda_{0}\right)=\frac{\sin \frac{\delta_{\t... | IPHO 1983 | value | ||
40 | Two dispersive prisms having apex angles $\hat{A}_{1}=60^{\circ}$ and $\hat{A}_{2}=30^{\circ}$ are glued as in the figure ( $\hat{C}=90^{\circ}$ ). The dependences of refraction indexes of the prisms on the wavelength are given by the relations $n_{1}(\lambda)=a_{1}+\frac{b_{1}}{\lambda^{2}}$; $n_{2}(\lambda)=a_{2}+\fr... | 1.2 | d. Using the figure 1.3 the refraction law on the $A D$ face is $\sin i_{1}=n_{1} \cdot \sin r_{1}$ The refraction law on the $A C$ face is $n_{1} \cdot \sin r_{1}{ }^{\prime}=n_{2} \cdot \sin r_{2}$ Figure 3.3 As it can be seen in the figure 1.3 $$ r_{2}=A_{2} $$ and $i_{1}=30^{\circ}$ Also, $r_{1}+r_{1}{ }^{\prim... | IPHO 1983 | value | ||
41 | A photon of wavelength $\lambda_{i}$ is scattered by a moving, free electron. As a result the electron stops and the resulting photon of wavelength $\lambda_{0}$ scattered at an angle $\theta=60^{\circ}$ with respect to the direction of the incident photon, is again scattered by a second free electron at rest. In this ... | $1,24 \times 10^{-10} $ | The purpose of the problem is to calculate the values of the speed, momentum and wavelength of the first electron. To characterize the photons the following notation are used: Table 4.1 | | initial <br> photon | photon - <br> after the <br> first scattering | final <br> photon | | :--- | :--- | :--- | :--- | | momen... | IPHO 1983 | value | ||
42 | Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant in... | 150 | a. As is very well known in the study of AC circuits using the formalism of complex numbers, a complex inductive reactance $\overline{X_{L}}=L \cdot \omega \cdot j,(j=\sqrt{-1})$ is attached to the inductance $L$ - part of a circuit supplied with an alternative current having the pulsation $\omega$. Similar, a complex ... | IPHO 1983 | value | ||
43 | Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant in... | $\frac{10^{5}}{2 \pi}$ | b. The fact that immediately after the source is detached it is a current in the coils, allow as to admit that currents dependents on time will continue to flow through the coils. Figure 2.1 The capacitors will be charged with charges variable in time. The variation of the charges of the capacitors will results in cu... | IPHO 1983 | value | ||
44 | Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant in... | -0.3 | c. If the momentary tension on circuit is like in (2.43), one may write $\left\{\begin{array}{l}u(0)=A \cdot \sin (\delta)=u_{0} \\ \sin (\delta)=\frac{u_{0}}{A}\end{array}\right.$ From the currents (2.47) is possible to write $\left\{\begin{array}{l}i_{01}=\frac{1}{L_{1} \cdot \omega} \cdot A \cdot \cos (\delta)+M \\ ... | IPHO 1983 | value | ||
45 | Let's consider the electric circuit in the figure, for which $L_{1}=10 \mathrm{mH}$, $L_{2}=20 \mathrm{mH}, C_{1}=10 \mathrm{nF}, C_{2}=5 \mathrm{nF}$ and $R=100 \mathrm{k} \Omega$. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant in... | -0.3 | d. The amplitude of the current through the inductance $L_{1}$ is $$ \max \left(\tilde{i}_{1}\right)=\max \left(\frac{4 \sqrt{26}}{100} \cdot \cos \left(10^{5} \cdot t+\operatorname{arctg}(1 / 5)\right) A\right)=\frac{4 \sqrt{26}}{100} A \approx 0,2 A $$ representing the answer at the question $\mathbf{d}$. | IPHO 1983 | value | ||
46 | Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\... | 68.38 | a) Floating condition: The total mass of the balloon, consisting of the mass of the envelope $\mathrm{m}_{\mathrm{H}}$ and the mass of the air quantity of temperature $\vartheta_{2}$ must equal the mass of the displaced air quantity with temperature $\vartheta_{1}=20^{\circ} \mathrm{C}$. $$ \begin{aligned} & V_{B} \c... | IPHO 1982 | value | ||
47 | Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\... | 1.21 | b) The force $F_{B}$ acting on the rope is the difference between the buoyant force $F_{A}$ and the weight force $\mathrm{F}_{\mathrm{G}}$ : $$ \mathrm{F}_{\mathrm{B}}=\mathrm{V}_{\mathrm{B}} \cdot \rho_{1} \cdot \mathrm{~g}-\left(\mathrm{V}_{\mathrm{B}} \cdot \rho_{3}+\mathrm{m}_{\mathrm{H}}\right) \cdot \mathrm{g} $... | IPHO 1982 | value | ||
48 | Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\... | 843 | c) The balloon rises to the height $h$, where the density of the external air $\rho_{h}$ has the same value as the effective density $\rho_{\text {eff }}$, which is evaluated from the mass of the air of temperature $\vartheta_{3}=110^{\circ} \mathrm{C}$ (inside the balloon) and the mass of the envelope $\mathrm{m}_{\ma... | IPHO 1982 | value | ||
49 | Consider a hot-air balloon with fixed volume $\mathrm{V}_{\mathrm{B}}=1.1 \mathrm{~m}^{3}$. The mass of the balloonenvelope, whose volume is to be neglected in comparison to $\mathrm{V}_{\mathrm{B}}$, is $\mathrm{m}_{\mathrm{H}}=0.187 \mathrm{~kg}$. The balloon shall be started, where the external air temperature is $\... | harmonic oscillations | d) For small height differences ( 10 m in comparison to 843 m ) the exponential pressure drop (or density drop respectively) with height can be approximated by a linear function of height. Therefore the driving force is proportional to the elongation out of the equilibrium position. This is the condition in which harm... | IPHO 1982 | expression | ||
50 | A space rocket with mass $M=12 \mathrm{t}$ is moving around the Moon along the circular orbit at the height of $h=100 \mathrm{~km}$. The engine is activated for a short time to pass at the lunar landing orbit. The velocity of the ejected gases $u=10^{4} \mathrm{~m} / \mathrm{s}$. The Moon radius $R_{M}=1,7 \cdot 10^{3}... | 29 | 1) During the rocket moving along the circular orbit its centripetal acceleration is created by moon gravity force: $$ G \frac{M M_{M}}{R^{2}}=\frac{M v_{0}^{2}}{R}, $$ where $R=R_{M}+h$ is the primary orbit radius, $v_{0}$-the rocket velocity on the circular orbit: $$ v_{0}=\sqrt{G \frac{M_{M}}{R}} $$ Since $g_{M}... | IPHO 1979 | |||
51 | A space rocket with mass $M=12 \mathrm{t}$ is moving around the Moon along the circular orbit at the height of $h=100 \mathrm{~km}$. The engine is activated for a short time to pass at the lunar landing orbit. The velocity of the ejected gases $u=10^{4} \mathrm{~m} / \mathrm{s}$. The Moon radius $R_{M}=1,7 \cdot 10^{3}... | 116 | 2) In the second case the vector $\vec{v}_{2}$ is directed perpendicular to the vector $\vec{v}_{0}$ thus giving $$ \vec{v}_{A}=\vec{v}_{0}+\Delta \vec{v}_{2}, \quad v_{A}=\sqrt{v_{0}^{2}+\Delta v_{2}^{2}} $$ Based on the energy conservation law in this case the equation can be written as $$ \frac{M\left(v_{0}^{2}+\... | IPHO 1979 | value | ||
52 | .During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal ... | $2.6 \cdot 10^{-7}$ | 1) The beam divergence angle $\delta \varphi$ caused by diffraction defines the accuracy of the telescope optical axis installation: $$ \delta \varphi \approx \lambda / D \approx 2.6 \cdot 10^{-7} \text { rad. } \approx 0.05^{\prime \prime} . $$ | IPHO 1979 | value | ||
53 | .During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal ... | $10^{-12}$ | 2) The part $K_{1}$ of the light energy of a laser, directed to a reflector, may be found by the ratio of the area of $S_{1}$ reflector ( $S_{1}=\pi d^{2} / 4$ ) versus the area $S_{2}$ of the light spot on the Moon ( $S_{2}=\pi r^{2}$, where $r=L \delta \varphi \approx L \lambda / D, L-$ the distance from the Earth to... | IPHO 1979 | value | ||
54 | .During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal ... | 12 | 3) The pupil of a naked eye receives as less a part of the light flux compared to a telescope, as the area of the pupil $S_{e}$ is less than the area of the telescope mirror $S_{t}$ : $$ K_{e}=K_{0} \frac{S_{e}}{S_{t}}=K_{0} \frac{d_{e}^{2}}{D^{2}} \approx 3.7 \cdot 10^{-18} $$ So the number of photons $N$ getting in... | IPHO 1979 | value | ||
55 | .During the Soviet-French experiment on the optical location of the Moon the light pulse of a ruby laser ( $\lambda=0,69 \mu \mathrm{~m}$ ) was directed to the Moon's surface by the telescope with a diameter of the mirror $D=2,6 \mathrm{~m}$. The reflector on the Moon's surface reflected the light backward as an ideal ... | $2 \cdot 10^{6}$ | 4) In the absence of a reflector $\alpha=10 \%$ of the laser energy, that got onto the Moon, are dispersed by the lunar surface within a solid angle $\Omega_{1}=2 \pi$ steradian. The solid angle in which one can see the telescope mirror from the Moon, constitutes $$ \Omega_{2}=\mathrm{S}_{\mathrm{t}} / \mathrm{L}^{2}... | IPHO 1979 | value | ||
56 | A hollow sphere of radius $R=0.5 \mathrm{~m}$ rotates about a vertical axis through its centre with an angular velocity of $\omega=5 \mathrm{~s}^{-1}$. Inside the sphere a small block is moving together with the sphere at the height of R/2 (Fig. 6). ( $g=10 \mathrm{~m} / \mathrm{s}^{2}$.)a) What should be at least the ... | 0.2259 | a) The block moves along a horizontal circle of radius $R \sin \alpha$. The net force acting on the block is pointed to the centre of this circle (Fig. 7). The vector sum of the normal force exerted by the wall $N$, the frictional force $S$ and the weight $m g$ is equal to the resultant: $m \omega^{2} R \sin \alpha$. ... | IPHO 1976 | value | ||
57 | A hollow sphere of radius $R=0.5 \mathrm{~m}$ rotates about a vertical axis through its centre with an angular velocity of $\omega=5 \mathrm{~s}^{-1}$. Inside the sphere a small block is moving together with the sphere at the height of R/2 (Fig. 6). ( $g=10 \mathrm{~m} / \mathrm{s}^{2}$.)b) Find the minimal coefficient... | 0.1792 | .b) If on the other hand $\frac{\omega^{2} R \cos \alpha}{g}>1$ some friction is necessary to prevent the block to slip upwards. $m \omega^{2} R \sin \alpha$ must be equal to the resultant of forces $S, N$ and $m g$. Condition for the minimal coefficient of friction is (Fig. 8): $$ \begin{aligned} \mu_{b} & \geq \frac... | IPHO 1976 | value | ||
58 | The walls of a cylinder of base $1 \mathrm{dm}^{2}$, the piston and the inner dividing wall are perfect heat insulators (Fig. 10). The valve in the dividing wall opens if the pressure on the right side is greater than on the left side. Initially there is 12 g helium in the left side and 2 g helium in the right side. Th... | 3640 | The volume of 4 g helium at $0^{\circ} \mathrm{C}$ temperature and a pressure of 100 kPa is $22.4 \mathrm{dm}^{3}$ (molar volume). It follows that initially the pressure on the left hand side is 600 kPa , on the right hand side 100 kPa . Therefore the valve is closed. An adiabatic compression happens until the pressur... | IPHO 1976 | value | ||
59 | The focal length f of a thick glass lens in air with refractive index $n$, radius curvatures $r_{l}, r_{2}$ and vertex distance $d$ (see figure) is given by: $\quad f=\frac{n r_{1} r_{2}}{(n-1)\left[n\left(r_{2}-r_{1}\right)+d(n-1)\right]}$ <image> Remark: $\quad \mathrm{r}_{\mathrm{i}}>0$ means that the central curva... | 2 | a) The refractive index $n$ is a function of the wavelength $\lambda$, i.e. $n=n(\lambda)$. According to the given formula for the focal length $f$ (see above) which for a given f yields to an equation quadratic in $n$ there are at most two different wavelengths (indices of refraction) for the same focal length. | IPHO 1975 | value | ||
60 | The focal length f of a thick glass lens in air with refractive index $n$, radius curvatures $r_{l}, r_{2}$ and vertex distance $d$ (see figure) is given by: $\quad f=\frac{n r_{1} r_{2}}{(n-1)\left[n\left(r_{2}-r_{1}\right)+d(n-1)\right]}$ <image> Remark: $\quad \mathrm{r}_{\mathrm{i}}>0$ means that the central curva... | $$ n_{l, 2}=-\frac{B}{2 \cdot A} \pm \sqrt{\frac{B^{2}}{4 \cdot A^{2}}-\frac{C}{A}} $$ | b) If the focal length is the same for two different wavelengths, then the equation $$ f\left(\lambda_{1}\right)=f\left(\lambda_{2}\right) \quad \text { or } \quad f\left(n_{1}\right)=f\left(n_{2}\right) $$ holds. Using the given equation for the focal length it follows from equation (1): $$ \frac{n_{1} r_{1} r_{2}}... | IPHO 1975 | expression | ||
61 | A beam of positive ions (charge +e ) of the same and constant mass $m$ spread from point Q in different directions in the plane of paper (see figure ${ }^{2}$ ). The ions were accelerated by a voltage $U$. They are deflected in a uniform magnetic field $B$ that is perpendicular to the plane of paper. The boundaries of ... | $$ R=\frac{1}{B} \sqrt{\frac{2 \cdot m \cdot U}{e}} $$ | a) The kinetic energy of the ion after acceleration by a voltage $U$ is: $$ 1 / 2 m v^{2}=e U $$ From equation (1) the velocity of the ions is calculated: $$ v=\sqrt{\frac{2 \cdot e \cdot U}{m}} $$ On a moving ion (charge $e$ and velocity $v$ ) in a homogenous magnetic field $B$ acts a Lorentz force $F$. Under the ... | IPHO 1975 | expression | ||
62 | A beam of positive ions (charge +e ) of the same and constant mass $m$ spread from point Q in different directions in the plane of paper (see figure ${ }^{2}$ ). The ions were accelerated by a voltage $U$. They are deflected in a uniform magnetic field $B$ that is perpendicular to the plane of paper. The boundaries of ... | $$ r=\frac{a}{\cos \varphi}\left(1-\frac{R}{a} \sin \varphi\right) $$ | d) It is convenient to deduce a general equation for the boundaries of the magnetic field in polar coordinates $(r, \varphi)$ instead of using cartesian coordinates $(x, y)$. The following relation is obtained from the figure: $$ r \cdot \cos \varphi+R \sin \varphi=a $$ The boundaries of the magnetic field are give... | IPHO 1975 | expression | ||
63 | A hydrogen atom in the ground state, moving with velocity $v$, collides with another hydrogen atom in the ground state at rest. Using the Bohr model find the smallest velocity $v_{0}$ of the atom below which the collision must be elastic. At velocity $v_{0}$ the collision may be inelastic and the colliding atoms may e... | $2 \cdot 10^{-2} \%$ | According to the Bohr model the energy levels of the hydrogen atom are given by the formula: $$ E_{n}=-\frac{E_{i}}{n^{2}}, $$ where $n=1,2,3, \ldots$ The ground state corresponds to $n=1$, while the lowest excited state corresponds to $n=2$. Thus, the smallest energy necessary for excitation of the hydrogen atom is:... | IPHO 1974 | value | ||
64 | Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$... | 1.3 | Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} ... | IPHO 1974 | value | ||
65 | Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$... | 1 | Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} ... | IPHO 1974 | value | ||
66 | Consider a parallel, transparent plate of thickness $d$-Fig. 1. Its refraction index varies as $$ n=\frac{n_{0}}{1-\frac{x}{R}} . $$ <image> A light beam enters from the air perpendicularly to the plate at the point $\mathrm{A}\left(x_{A}=0\right)$ and emerges from it at the point B at an angle $\alpha$. Data: $$... | 5 | Fig. 2 Consider a light ray passing through a system of parallel plates with different refractive indexes - Fig. 2. From the Snell law we have $$ \frac{\sin \beta_{2}}{\sin \beta_{1}}=\frac{n_{1}}{n_{2}} $$ i.e. $$ n_{2} \sin \beta_{2}=n_{1} \sin \beta_{1} . $$ In the same way we get $$ n_{3} \sin \beta_{3}=n_{2} ... | IPHO 1974 | value | ||
67 | Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and ... | $$ \mathrm{I}_{1}>\mathrm{I}_{2}>\mathrm{I}_{3} $$ | The inertia moments of the three cylinders are: $$ I_{1}=\frac{1}{2} \rho_{1} \pi\left(R^{4}-r^{4}\right) h, \quad I_{2}=\frac{1}{2} \rho_{2} \pi R^{4} h=\frac{1}{2} m R^{2} \quad, \quad I 3=\frac{1}{2} \rho_{2} \pi\left(R^{4}-r^{4}\right) h, $$ Because the three cylinders have the same mass : $$ m=\rho_{1} \pi\left... | IPHO 1972 | equation | ||
68 | Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and ... | $$ \begin{array}{r} \mathrm{F}_{\mathrm{f}}<\mu \mathrm{N}=\mu \mathrm{mgsin} \alpha \\ \operatorname{tg} \alpha<\mu\left(1+\frac{m R^{2}}{I_{1}}\right) \end{array} $$ | For a cylinder rolling over freely on the inclined plane (fig. 1.1) we can write the equations: $$ \begin{aligned} & m g \sin \alpha-F_{f}=m a \\ & N-m g \cos \alpha=0 \\ & F_{f} R=I \varepsilon \end{aligned} $$ where $\varepsilon$ is the angular acceleration. If the cylinder doesn't slide we have the condition: $$ ... | IPHO 1972 | expression | ||
69 | Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and ... | $$ \mathrm{a}_{1}<\mathrm{a}_{2}<\mathrm{a}_{3} $$ | Fig. 1.1 In the case of the cylinders from this problem, the condition necessary so that none of them slides is obtained for maximum I: $$ \operatorname{tg} \alpha\left\langle\mu\left(1+\frac{m R^{2}}{I_{1}}\right)=\mu \frac{4 n-1}{2 n-1}\right. $$ The accelerations of the cylinders are: $$ a_{1}=\frac{2 g \sin \alp... | IPHO 1972 | equation | ||
70 | Three cylinders with the same mass, the same length and the same external radius are initially resting on an inclined plane. The coefficient of sliding friction on the inclined plane, $\mu$, is known and has the same value for all the cylinders. The first cylinder is empty (tube), the second is homogeneous filled, and ... | $m_{l} g \frac{\cos \alpha}{\cos \phi}$ | In the case than all the three cylinders slide: $$ F_{f}=\mu N=\mu m g \cos \alpha $$ and from (7) results: $$ \varepsilon=\frac{R}{I} \mu m g \cos \alpha $$ for the cylinders of the problem: $$ \begin{aligned} \varepsilon_{1}: & \varepsilon_{2}: \varepsilon_{3}=\frac{1}{I_{1}}: \frac{1}{I_{2}}: \frac{1}{I_{3}}=1:... | IPHO 1972 | value | ||
71 | Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B i... | $$ T V^{2 / 3}=\text { const. } $$ | a) We consider argon an ideal mono-atomic gas and the collisions of the atoms with the piston perfect elastic. In such a collision with a fix wall the speed $\vec{v}$ of the particle changes only the direction so that the speed $\vec{v}$ and the speed $\vec{v}^{\prime}$ after collision there are in the same plane with ... | IPHO 1972 | equation | ||
72 | Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B i... | $64,9 \cdot 10^{5}$ | b) The oxygen is in contact with a thermostat and will suffer an isothermal process. The internal energy will be modified only by the adiabatic process suffered by argon gas: $$ \Delta U=v C_{V} \Delta T=m c_{V} \Delta T $$ where $v$ is the number of kilomoles. For argon $C_{V}=\frac{3}{2} R$. For the entire system $... | IPHO 1972 | value | ||
73 | Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B i... | 8,16 | b) The oxygen is in contact with a thermostat and will suffer an isothermal process. The internal energy will be modified only by the adiabatic process suffered by argon gas: $$ \Delta U=v C_{V} \Delta T=m c_{V} \Delta T $$ where $v$ is the number of kilomoles. For argon $C_{V}=\frac{3}{2} R$. For the entire system $... | IPHO 1972 | value | ||
74 | Two cylinders A and B, with equal diameters have inside two pistons with negligible mass connected by a rigid rod. The pistons can move freely. The rod is a short tube with a valve. The valve is initially closed (fig. 2.1). <image> Fig. 2.1 The cylinder A and his piston is adiabatically insulated and the cylinder B i... | $2,23 \cdot 10^{5}$ | c) When the valve is opened the gases intermix and at thermal equilibrium the final pressure will be $p^{\prime}$ and the temperature T . The total number of kilomoles is constant: $$ \begin{aligned} v_{1}+v_{2}=v^{\prime}, \frac{p_{1}^{\prime} V_{1}^{\prime}}{R T_{1}^{\prime}}+\frac{p_{2}^{\prime} V_{2}^{\prime}}{R T... | IPHO 1972 | value |
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