Qormaata galuumsa Fiiziksii, bara 2015 A.L.I (2023)

Start with the mirror equation: \(\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}\), where \(f=10.0\) cm and the object distance \(d_o=25.0\) cm.

\(\frac{1}{d_i}=\frac{1}{10}-\frac{1}{25}=\frac{5-2}{50}=\frac{3}{50}\), so \(d_i=\frac{50}{3}\approx 16.7\) cm.

The magnification is \(m=-\frac{d_i}{d_o}=-\frac{50/3}{25}=-\frac{2}{3}\approx -0.67\). Its size is 0.67, so the image is smaller than the object, and the minus sign tells us the image is inverted.

Option B (16.7) is a trap: that is the image distance in cm, not the magnification.

Textbook: Physics Grade 10, Unit 6 Electromagnetic wave and geometrical optics, section 6.5 Mirrors and Lenses.

A systematic error is a bias built into the experiment, for example a scale that always reads a little high or a stopwatch that runs slow. It pushes every reading in the same direction, so the results sit always above or always below the true value. Taking more readings and averaging does not remove it.

A random error comes from small unpredictable changes between readings. It scatters the results on both sides of the true value, so repeating the measurement and averaging does reduce it.

Option B states exactly this, so it is correct. Option A is backwards: averaging helps against random error, not systematic error. Option C swaps the two definitions, and option D swaps which one is biased.

A unit vector has length 1 and points in the same direction as the given vector. To find it, divide the vector by its own magnitude: \(\hat{u}=\frac{\vec{A}}{|\vec{A}|}\).

The magnitude of \(\vec{A}=3\hat{i}+4\hat{j}\) is \(|\vec{A}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\).

So \(\hat{u}=\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j}\). Check: \(\sqrt{(3/5)^2+(4/5)^2}=\sqrt{9/25+16/25}=1\), so its length really is 1.

Textbook: Physics Grade 11, Unit 2 Vectors, section 2.1 Vectors and Types of Vectors.

Two vectors are collinear when they lie along the same line. That happens when one is a number times the other, in the same or the exactly opposite direction.

Look at option D: \(-4\hat{i}-6\hat{j}=-2\,(2\hat{i}+3\hat{j})\). The second vector is \(-2\) times the first, so the pair is collinear (they point in opposite directions along one line).

You can also test with the cross product, which is zero for collinear vectors: \((2)(-6)-(3)(-4)=-12+12=0\).

For the other pairs no single multiplier works. In option C, for example, the \(\hat{i}\) parts need a factor of \(-2\) but the \(\hat{j}\) parts need \(+2\), so those vectors are not collinear.

Textbook: Physics Grade 11, Unit 2 Vectors, section 2.4 Product of Vectors.

Remember what each graph tells you for straight line motion:

If the velocity grows by equal amounts in equal time intervals, the acceleration is constant. Then the velocity-time graph is a straight line with positive slope, and the acceleration-time graph is a horizontal line. That is option C.

Option A is wrong because equal displacement in equal time means constant velocity, which gives a sloped straight line, not a horizontal one. Options B and D mix up which quantity comes from a slope and which comes from an area.

Textbook: Physics Grade 11, Unit 3 Motion in one and two dimensions, section 3.3 Graphical Representation of Uniformly Accelerated Motion.

For a satellite in a circular orbit, gravity supplies the centripetal force: \(\frac{GMm}{r^2}=\frac{mv^2}{r}\). Solving gives the orbital speed \(v=\sqrt{\frac{GM}{r}}\), where \(M\) is the mass of the body being orbited and \(r\) is the orbital radius.

The formula shows that a smaller \(r\) gives a larger \(v\). So a satellite closer to the Earth moves faster than one farther away, which is option D.

The other options fail on the same formula. A: a closer satellite has a shorter path and a higher speed, so its period is shorter, not longer. B: the satellite mass \(m\) cancelled out, so a heavy and a light satellite at the same radius move at the same speed. C: at equal radius the speed depends on the central mass \(M\), and the Earth is far heavier than the Moon, so the satellite around the Earth is the faster one.

Textbook: Physics Grade 12, Unit 2 Two-dimensional motion, section 2.5 Newton's law of universal Gravitation.