Question Library

One of the process in which nutrient is fixed into the soil is electrical discharge which helps to add nutrient and nitrate to the soil.

Since the salary increases by #36 after every 6 months : 2 years and 8 months imply an increase of five times only: 36 * 5 → #180 His salary then = initial salary + increment = 540 + 180 = #720 Answer is C

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

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

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

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.

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

Given: P = 4 atm , V = 10 dm\(^3\) , n = ? R = 0.082 (constant) T = 273 + 273 = 546K Using the ideal gas law, PV = nRT n = \(\frac{PV}{RT}\) n = \(\frac{{4}\times{10}}{{0.082}\times{546}}\) n = \(\frac{40}{44.772}\) n = 0.893 mole

Tap water is likely to contain less Ca\(^{2+}\) and Mg\(^{2+}\). The reason is because looking through all the given options, spring water, river water and sea water are natural water that has not undergone any treatment. Conversely, it is believed that before tap water is distributed for consumption, it must have been treated, thereby reducing the Ca\(^{2+}\) and Mg\(^{2+}\) content in the water.

Let the number of people that offer both Mathematics and Physics = y Then, \((32 - y) + y + (24 - y) + 4 = 40\) \(60 - y = 40 \implies y = 20\) \(\therefore\) 20 students offer both Mathematics and Physics.