C

9000036110

Level: 
C
A tower is observed from two different places \(A\) and \(B\). The direct distance between \(A\) and \(B\) is \(65\, \mathrm{m}\). If we denote the bottom of the tower by \(C\), we get a triangle \(ABC\) in which the measure of \(\measuredangle CAB \) is \(71^{\circ }\) and the measure of \(\measuredangle ABC \) is \( 34^{\circ }\). From the point \(A\) the angle of elevation of the top of the tower is \(40^{\circ }18'\). Find the height of the tower. Suppose that all \(A\), \(B\) and \(C\) are in the same height above sea level and round your answers to nearest meters.
\(32\, \mathrm{m}\)
\(30\, \mathrm{m}\)
\(35\, \mathrm{m}\)
\(38\, \mathrm{m}\)

9000038701

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) The friction \(F_{t}\) is given by the formula \(F_{t} = fF_{n}\), where \(f\) is the coefficient of the friction.What is the influence of the increasing angle \(\alpha \) on the forces acting on the box?
\(F_{1}\) becomes bigger and \(F_{t}\) becomes smaller
both \(F_{1}\) and \(F_{t}\) become smaller
\(F_{1}\) becomes bigger, \(F_{t}\) does not change
\(F_{1}\) becomes smaller, \(F_{t}\) does not change
both \(F_{1}\) and \(F_{t}\) become bigger
\(F_{1}\) becomes smaller and \(F_{t}\) becomes bigger

9000038702

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) Find \(F_{1}\).
\(F_{1} = F_{G}\sin \alpha \)
\(F_{1} = \frac{F_{G}} {\sin \alpha } \)
\(F_{1} = F_{G}\mathop{\mathrm{tg}}\nolimits \alpha \)
\(F_{1} = \frac{F_{G}} {\mathop{\mathrm{tg}}\nolimits \alpha } \)
\(F_{1} = F_{G}\cos \alpha \)
\(F_{1} = \frac{F_{G}} {\cos \alpha } \)

9000038703

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) Find \(F_{p}\).
\(F_{p} = F_{G}\cos \alpha \)
\(F_{p} = \frac{F_{G}} {\cos \alpha } \)
\(F_{p} = F_{G}\mathop{\mathrm{tg}}\nolimits \alpha \)
\(F_{p} = \frac{F_{G}} {\mathop{\mathrm{tg}}\nolimits \alpha } \)
\(F_{p} = F_{G}\sin \alpha \)
\(F_{p} = \frac{F_{G}} {\sin \alpha } \)

9000038704

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) For \(F_{1} = 20\, \mathrm{N}\) and \(F_{n} = 55\, \mathrm{N}\) find the corresponding \(\alpha \).
\(\alpha \doteq 20^{\circ }\)
\(\alpha \doteq 21^{\circ }\)
\(\alpha \doteq 69^{\circ }\)
\(\alpha \doteq 70^{\circ }\)
\(\alpha \doteq 30^{\circ }\)
\(\alpha \doteq 29^{\circ }\)

9000038705

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha = 45^{\circ }\). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) The friction \(F_{t}\) is given by the formula \(F_{t} = fF_{n}\). The coefficient of the friction \(f = 0.5\). Consider the standard acceleration of gravity \(g = 10\, \mathrm{m\, s^{-2}}\). Find the acceleration of the box.
\(a = \frac{5\sqrt{2}} {2} \, \mathrm{m\, s^{-2}}\)
\(a = 5\sqrt{2}\, \mathrm{m\, s^{-2}}\)
\(a = 5\sqrt{3}\, \mathrm{m\, s^{-2}}\)
\(a = 0\, \mathrm{m\, s^{-2}}\)
\(a = 5\, \mathrm{m\, s^{-2}}\)
\(a = \frac{5\sqrt{3}} {2} \, \mathrm{m\, s^{-2}}\)

9000038706

Level: 
C
The box is on the slope as in the picture. The angle of the slope is \(\alpha \). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) The friction \(F_{t}\) is given by the formula \(F_{t} = fF_{n}\). The coefficient of the friction is \(f = 0.47\). Consider the standard acceleration of gravity \(g = 10\, \mathrm{m\, s^{-2}}\). Find the angle \(\alpha \) which ensures that the box moves on the slope with zero acceleration.
\(\alpha \doteq 25^{\circ }\)
\(\alpha \doteq 15^{\circ }\)
\(\alpha \doteq 20^{\circ }\)
\(\alpha \doteq 65^{\circ }\)
\(\alpha \doteq 28^{\circ }\)
\(\alpha \doteq 62^{\circ }\)

9000038707

Level: 
C
The box is on the slope as in the picture. The length of the slope is \(l = 2\, \mathrm{m}\) and the height is \(h = 1.2\, \mathrm{m}\). The forces acting on the box are the force of gravity \(\vec{F_{G}}\) and the friction \(\vec{F_{t}}\). The force of gravity can be replaced by two components \(\vec{F_{1}}\) and \(\vec{F_{n}}\). (The force \(\vec{F_{1}}\) is parallel to the slope and \(\vec{F_{n}}\) is perpendicular to the slope.) The friction \(F_{t}\) is given by the formula \(F_{t} = fF_{n}\) where \(f\) is the coefficient of the friction. Consider the standard acceleration of gravity \(g = 10\, \mathrm{m\, s^{-2}}\). Find the minimal value for the coefficient of the friction \(f\) to ensure that the box does not move with an acceleration.
\(f = 0.75\)
\(f = 0.6\)
\(f = 0.65\)
\(f = 0.7\)
\(f = 0.55\)
\(f = 0.8\)

9000035810

Level: 
C
Given the complex number \(z = -2 + 2\mathrm{i}\), find all the roots of \(\root{3}\of{z}\).
\(\begin{aligned}[t] &w_{0} = \root{6}\of{8}\left (\cos \frac{\pi } {4} + \mathrm{i}\sin \frac{\pi } {4}\right ) & \\&w_{1} = \root{6}\of{8}\left (\cos \frac{11\pi } {12} + \mathrm{i}\sin \frac{11\pi } {12}\right ) \\&w_{2} = \root{6}\of{8}\left (\cos \frac{19\pi } {12} + \mathrm{i}\sin \frac{19\pi } {12}\right ) \\ \end{aligned}\)
\(\begin{aligned}[t] &w_{0} = 2\left (\cos \frac{\pi } {4} + \mathrm{i}\sin \frac{\pi } {4}\right ) & \\&w_{1} = 2\left (\cos \frac{11\pi } {12} + \mathrm{i}\sin \frac{11\pi } {12}\right ) \\&w_{2} = 2\left (\cos \frac{19\pi } {12} + \mathrm{i}\sin \frac{19\pi } {12}\right ) \\ \end{aligned}\)
\(\root{3}\of{-2} + \root{3}\of{2}\)
\(\begin{aligned}[t] &w_{0} = 2\left (\cos \frac{\pi } {3} + \mathrm{i}\sin \frac{\pi } {3}\right )& \\&w_{1} = 2\left (\cos \pi +\mathrm{i}\sin \pi \right ) \\&w_{2} = 2\left (\cos \frac{5\pi } {3} + \mathrm{i}\sin \frac{5\pi } {3}\right ) \\ \end{aligned}\)