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 \(\alpha \).
\(\alpha \doteq 20^{\circ }\)
\(\alpha \doteq 21^{\circ }\)
\(\alpha \doteq 69^{\circ }\)
\(\alpha \doteq 70^{\circ }\)
\(\alpha \doteq 30^{\circ }\)
\(\alpha \doteq 29^{\circ }\)

9000036102

Level: 
C
Three forces act on the same body in the same point and the total force on the body is zero (the forces cancel). The first two forces are \(8\, \mathrm{N}\) and \(10\, \mathrm{N}\) and the angle between these forces is \(55^{\circ }\). Find the third force.
\(16\, \mathrm{N}\)
\(15\, \mathrm{N}\)
\(17\, \mathrm{N}\)
\(18\, \mathrm{N}\)

9000036103

Level: 
C
Three forces \(F_{1}\), \(F_{2}\) and \(F_{3}\) act on the same body in the same point and the total force on the body is zero (the forces cancel). The first two forces are \(F_{1} = 8\, \mathrm{N}\) and \(F_{2} = 10\, \mathrm{N}\) and the angle between \(F_{1}\) and \(F_{2}\) is \(55^{\circ }\). Find the angle between \(F_{3}\) and \(F_{1}\). Round your answer to the nearest degrees.
\(149^{\circ }\)
\(125^{\circ }\)
\(55^{\circ }\)
\(30^{\circ }\)

9000036106

Level: 
C
Two straight roads go off from the crossing. The angle between directions of the roads is \(52^{\circ }18'\). A significant tree is on the first road in the distance \(250\, \mathrm{m}\) from the crossing. A rock with a beautiful view is on the second road in the distance \(380\, \mathrm{m}\) from the crossing. Find the direct distance (length of a line segment) from the rock to the tree and round your answer to nearest meters.
\(301\, \mathrm{m}\)
\(411\, \mathrm{m}\)
\(568\, \mathrm{m}\)
\(629\, \mathrm{m}\)

9000035602

Level: 
C
Find the values of the parameter \(m\in \mathbb{C}\) which guarantee that the following quadratic equation has a double solution. \[ mx^{2} - 2x - 1 + \mathrm{i} = 0 \]
\(m = -\frac{1} {2} -\frac{1} {2}\mathrm{i}\)
\(m = -1\)
\(m = -1 + \mathrm{i}\)
\(m = -\frac{1} {2} + \frac{1} {2}\mathrm{i}\)