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6x\(^2\) - 2x + 1 \(\frac{6x^{2+1}}{2+1} - \frac{2x^{1+1}}{1+1}+ 1x = \frac{6x^3}{3} - \frac{2x^2}{2} + x\) \(2x^3 - x^2 + x + c\)

Efficiency = \(\frac{\text{useful energy output from machine}}{\text{energy input into machine}}\) = \(\frac{E3}{E2}\)

Work is done whenever the given conditions are satisfied. (ii) there is a displacement of the body caused by the applied force along the direction of the applied force. No displacement takes place in this case. So, work done by a centripetal force is always zero. The work done by centripetal force in the case of a uniform circular motion is 0 joules.

Given Data: Efficiency = 40% Load = 120N, V.R = 6 (Number of Pulleys) Efficiency = \(\frac{M.A}{V.R} * 100\) 40 = \(\frac{M.A}{6} * 100\) 6 * 40 = 100 M.A M.A = \(\frac{240}{100}\) → 2.4 Where M.A = \(\frac{Load}{Effort}\) 2.4 = \(\frac{120}{Effort}\) : Effort = \(\frac{120}{2.4}\) → 50N

H = \(\frac{U^2sin^2 \theta}{2g}\) H = \(\frac{200^2(sin 45^o)^2}{20}\) = 999.8m or 1000m

(a) Strain is the ratio of the extension to original length of an object. (b) Strain = \(\frac{e}{l_o} = \frac{2l_o - l_o}{l_o}\) = 1

His movements 1km due east and then 1 km due north make a Right-Angled Triangle. Using Pythagoras' theorem; The Resultant becomes the Hypotenuse [h] Hyp\(^2\) = Adj\(^2\) + \Opp(^2\) Hyp\(^2\) = 1\(^2\) + 1\(^2\) Hyp\(^2\) = 2 → Hyp = \sqrt2 : Resultant = \sqrt2 DIRECTION TAN Ø = \(\frac{OPP}{ADJ}\) TAN Ø = \(\frac{1}{1}\) Ø = TAN\(^{-1}\) [1] Ø= 45^\circ Displacement is \sqrt2km N 45^\circE

The steam engine uses the Heat supplied from steam to supply kinetic energy into the generator which converts it to electricity to light the lamp, the lamp gives out light and heat as it works

Given Data: Mass of Ag = 36.0g, Z which is electrochemical equivalent of Ag = 0.0012coulomb t = 5mins → 5 * 60 = 300secs. Using M =Z * I * t make I the subject of the formula I = \(\frac{M}{Z * t}\) → \(\frac{36}{0.0012 * 300}\) Current (I) = 100A

Power delivered to head lamp = \(\frac{V^2}{R}\) = 40W R = \(\frac{V^2}{40}\) = \(\frac{(12 \times 12)}{40}\) = 3.6\Omega

S/N d(cm) I(A) I\(^{1}\)(A\(^{-1}\)) Log I\(^{-1}\) 1 25 0.65 1.54 0.19 2 35 0.55 1.82 0.26 3 50 0.44 2.50 0.40 4 65 0.30 3.33 0.52 5 80 0.23 4.35 0.64 vii. Scale: 2cm to 10 units on x-axis 2cm to 0.1 unit on y-axis viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\) = \(\frac{80 -25}{0.64 - 0.20}\) = \(\frac{55}{0.44}\) = 125 (ix) Precautions (a) Key opened in between reading or keys opened when readings were not taken (b) Tight connections ensured (c) Avoided parallax error when taking reading on ammeter or metre rule (d) Repeated readings shown (e) Avoided zero error on ammeter or metre rule (f) Clean terminals ensured. (b)i. 1\(^{-1}\) = \(\frac{1}{1.5}\) = 0.67 Log(1\(^{-1}\)) =-0.174 ii. Factors affecting the resistance of a wire - Length of wire - Cross-sectional area/ thickness/ diameter/ radius of wire - Resistivity or nature of wire - Temperature.

At resonance in a series R-L-C circuit, the voltage across the capacitor equals that across the inductor. Given V\(_L\) = 40V, the voltage across the capacitor (V\(_C\)​) is also 40V.

Cathode rays are streams of electrons observed in vacuum tubes.

\(\lambda\) max = \(\frac{2.9 \times 10^{-3}}{330}\) = 8.8 x 10\(^{-6}\)m v = f\(\lambda\) or f = \(\frac{v}{\lambda}\) where v = velocity of light = 3.0 x10\(^{-8}\) m/s \(\lambda\) = 8.8 x 10\(^{-6}\) f = \(\frac{v}{\lambda}\) = \(\frac{3.0 \times 10^8}{8.8 \times 10^{-6}}\) = 3.4 x 10\(^{13}\) Hz

V = -6cm M = 0.6cm Recall m = \(\frac{V}{U}\) U = \(\frac{V}{M}\) U = \(\frac{6}{0.6}\) U = 10 \(\frac{1}{U}\) + \(\frac{1}{V}\) = \(\frac{1}{F}\) \(\frac{1}{10}\) + \(\frac{1}{-6}\) = \(\frac{1}{F}\) \(\frac{-4}{60}\) = \(\frac{1}{F}\) F = \(\frac{60}{-4}\) = -15cm