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(a) \(Time = \frac{2}{3} hour = \frac{2}{3} \times 60 = 40 mins\) 4 students will take \(\frac{8 \times 40}{4} = 80 minutes\) \(Time = \frac{1}{3} hour = \frac{1}{3} \times 60 = 20 minutes\) It will take 20 minutes to fill \(\frac{12 \times 20}{80} = 3 tanks\) (b) \(\tan (\frac{\theta}{2}) = \frac{10}{12}\) \(\frac{\theta}{2} = \tan^{-1} (0.8337) = 39.8^\circ\) \(\theta = 39.8^\circ \times 2 = 79.6^\circ\) (ii) Perimeter of minor segment = Length of chord + length of arc But \(r^{2} = 12^{2} + 10^{2}\) = \(144 + 100 = 244\) \(r = \sqrt{244} = 15.62 cm\) Length of arc = \(\frac{79.6}{360} \times 2 \times 3.142 \times 15.62 = 21.71cm\) Perimeter of minor segment = 20 + 21.71 = 41.71cm

(a) \(3y^{2} - 5y + 2 = 0\) \(\frac{3y^{2}}{3} - \frac{5y}{3} = \frac{-2}{3}\) \(y^{2} - \frac{5}{3}y = \frac{-2}{3}\) \(\frac{1}{2} of \frac{-5}{3} = \frac{-5}{6}\) \(y^{2} - \frac{5}{3}y + (-\frac{5}{6})^{2} = \frac{-2}{3} + (\frac{-5}{6})^{2}\) \((y - (\frac{5}{6}))^{2} = \frac{1}{36}\) Taking square root of both sides, \(\sqrt{(y - (\frac{5}{6}))^{2}} = \sqrt{\frac{1}{36}}\) \(y - \frac{5}{6} =\pm {\frac{1}{6}}\) \(y = \frac{5}{6} \pm \frac{1}{6}\) \(y = \frac{5}{6} + \frac{1}{6} = 1.00\) or \(y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3} = 0.67\). (b) \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) \(MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 2n & 4x + 2y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\) \(m + 2n = 2 ... (1)\) \(4m + 2n = 3 ... (2)\) Multiplying (1) by 4, we have \(4m + 8n = 8 ... (3)\) (3) - (1) : \(8n - 2n = 8 - 2 = 6\) \(6n = 6 \implies n = 1\) \(m + 2n = m + 2(1) = m + 2 = 2\) \(m = 2 - 2 = 0\) \(x + 2y = 1 ... (1)\) \(4x + y = 4 ... (2)\) Multiply (1) by 4, \(4x + 8y = 4 ... (3)\) (3) - (2) : \(8y - y = 4 - 4 = 0\) \(\implies y = 0\) \(x + 2y = x + 2(0) = x + 0 = 1\) \(x = 1 - 0 = 1\) \(\therefore N = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

(a) (i) \(3 * \frac{2}{5} = \frac{3 + \frac{2}{5}}{2}\) = \(\frac{\frac{17}{5}}{2}\) = \(\frac{17}{10}\) (ii) \(8 * y = 8\frac{1}{4}\) \(\frac{8 + y}{2} = \frac{33}{4}\) \(66 = 32 + 4y \implies 4y = 66 - 32 = 34\) \(y = \frac{34}{4} = 8.5\) (b) \(\overrightarrow{AP} = \frac{1}{2}(\overrightarrow{AB})\) = \(\frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) = \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\) Midpoint = \(\overrightarrow{AP} + \overrightarrow{PC} = \overrightarrow{AC}\) \(\overrightarrow{PC} = \overrightarrow{AC} - \overrightarrow{AP}\) = \(\begin{pmatrix} 3 \\ -8 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\) = \(\begin{pmatrix} 5 \\ -11 \end{pmatrix}\) \(\overrightarrow{CP} = -1 \times \begin{pmatrix} 5 \\ -11 \end{pmatrix}\) = \(\begin{pmatrix} -5 \\ 11 \end{pmatrix}\)

Impulse = Force \(\times\) time Force has the dimension \(\frac{ML}{T^{2}}\), hence Impulse= \(\frac{ML}{T^{2}} \times T = \frac{ML}{T}\)

Given that \(u = 0ms^{-1}, a = 2ms^{-2}, s = 9m\) We can use the equation, \(v^{2}=u^{2} +2as\) so that we have, \(v^{2} = 0^{2} + 2\times 2\times 9\) \(v^{2} = 36; v= 6.0ms^{-1}\) Note; u = Initial velocity v = final velocity a = acceleration s = distance covered

All other properties of the different states of matter holds except that the molecules of solids don't move freely due to the fact that the particles in the solids are closely packed so it inhibits free motion.

The body while coming down is acted upon by acceleration due to gravity, hence, the vertical force is constant. Also, at maximum height, the vertical velocity is 0 and then increases as the body comes downward.

Resolving 100N into horizontal and vertical 100 sin45 = 0.7071 * 100 = 70.71 100cos45 = 70.71 Vertical forces = 100 - 70.71 → 29.29 Horizontal force = 70.71 + 60 → 130.71 Resultant of vertical and horizontal forces using pythagoras thoerem Resultant force (F) = m * a a = F/m → 133.9/10 a = 13.4

Upthrust = weight of liquid displaced = mass of liquid displaced \(\times\) acceleration due to gravity = \(\frac{200}{1000} \times 10\) = 2N

No expanded working has been attached for this question yet. The verified answer is Option A.

It is a type of beam balance with great precision, and it can pick up the slightest mass placed on it.

No expanded working has been attached for this question yet. The verified answer is Option B.

Convert velocity from \(kmh^{-1}\) to \(ms^{-1}\), by multiplying \(\frac{1000}{3600}\), \(72 \times \frac{1000}{3600}\) = \(20ms^{-1}\) Therefore, distance covered in 0.4s = \(20 \times 0.4 = 8m\)

The formula is \(\frac{v^{2}sin2\theta}{g}\) And the sine graph is highest at 1 with its inverse as 90^\circ. Hence, \(2\theta\) = 90^\circ \(\theta\) = 45^\circ

Relative motion is the calculation of the motion of an object with regard to some other moving object. Hence, relative motion is not a type of motion.