Triangles

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.
\(f = 0.75\)
\(f = 0.6\)
\(f = 0.65\)
\(f = 0.7\)
\(f = 0.55\)
\(f = 0.8\)

9000124501

Level: 
C
Similar triangles can be used to estimate the distance from a distant object of a given width. Consider a door of the width \(85\, \mathrm{cm}\). A man stands in an unknown distance from the door and holds a thin pencil vertically in his arm in the distance \(35\, \mathrm{cm}\) from his face. If he closes the left eye, the right eye, the pencil and the left side of the door are aligned in one line. In a similar way, his left eye, the pencil and the right hand side of the door are also aligned in one line, which is apparent when closing the right eye. Assuming the distance \(6\, \mathrm{cm}\) between his eyes, estimate the distance from the man to the door. Give your answer in meters and round to one decimal place.
\(5.3\, \mathrm{m}\)
\(5.0\, \mathrm{m}\)
\(0.5\, \mathrm{m}\)
\(4.5\, \mathrm{m}\)

9000124503

Level: 
C
A tall radio mast is attached by several cables. The length of each cable is \(30\, \mathrm{m}\) and all cables are attached \(2\, \mathrm{m}\) under the top of the mast. The second end of the cable is anchored to the ground. The cable is in the height \(6\, \mathrm{m}\) if measured directly above the point which is in the distance \(8\, \mathrm{m}\) from the point where the cable is anchored to the ground. Find the height of the mast.
\(20\, \mathrm{m}\)
\(24\, \mathrm{m}\)
\(22.5\, \mathrm{m}\)
\(24.5\, \mathrm{m}\)

9000124504

Level: 
C
A force due to gravity on a body is \(1\: 800\, \mathrm{N}\). This body has to be lifted to the height \(50\, \mathrm{cm}\) using a slope. The maximal force which can be used to lift the body is \(600\, \mathrm{N}\). Neglect the friction and find the minimal length of the slope required to accomplish this task.Hint: The force due to gravity can be decomposed into two directions. The normal force \(F_{1}\) is compensated by the reaction of the slope. The force \(F_{2}\) parallel to the slope is required to overcome if we wish to lift the body (see the picture).
\(\frac{3} {2}\, \mathrm{m}\)
\(\frac{2} {3}\, \mathrm{m}\)
\(\frac{1} {6}\, \mathrm{m}\)
\(\frac{20} {9} \, \mathrm{m}\)

9000124505

Level: 
C
The picture shows the virtual image \(y'\) of the object \(y\) as created by a concave lens. The points \(F\) and \(F'\) are focal points of the lens. The distance from the lens to each of the focal points is \(20\, \mathrm{cm}\). The object \(y\) is \(25\, \, \mathrm{cm}\) height and it is in the distance \(50\, \mathrm{cm}\) from the lens. Find the height of the virtual image \(y'\).
\(\frac{50} {7} \, \mathrm{cm}\)
\(10\, \mathrm{cm}\)
\(\frac{50} {3} \, \mathrm{cm}\)
\(\frac{175} {2} \, \mathrm{cm}\)