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The alpha rays are the least penetrating, followed by beta rays, and gamma rays are the most penetrating. in descending order \(\gamma, \beta, \alpha\)

\beta Particles are equivalent to electrons but arise from radioactive decay of unstable atoms. They are emitted with a continuous range of energies up to a maximum that is characteristic of each radionuclide. Electrons have a greater range and penetrating power, but much less ionizing potential compared with \alpha particles. Beta particles are deflected by a magnetic field in an opposite direction to alpha particles confirming they must hold a charge opposite to alpha particles.

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

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

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

\(X_L = \frac{V}{I} = \frac{24}{1.5} = 16\Omega\) L = \(\frac{XL}{2 \pi f} = \frac{16}{2 \pi 50}\) = 0.05H

In an A.C circuit, the voltage(V\(_{rms}\) is completely in phase with the current for a resistor and lags the current by 90º for a capacitor(R - C)

C = \(C_1 + C_2 = 2 + 2 = 4 \mu F\) but Q = CV = 4 x 10\(^{-6}\) x 120 = 4.8 x 10\(^{-4}\) \(\therefore\) Each of the capacitor has a charge of \(\frac{4.8 \times 10^{-4}}{2}\) = 2.4 x 10\(^{-4}\) C

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

I = \(\frac{f}{n} = \frac{1}{6 - 2}\) I = \(\frac{v}{R + r} = \frac{2}{3} + \frac{1}{6} = 0.63\)A

The determinant of the brightness of a bulb is the power (watts) of the circuit. Recall that Power (P) = \(\frac{V^2}{R}\). Since the resistance is to be used as a divisor, it means the smaller the resistance, the greater the power.

S/N OP(\(_{cm}\)) I(\(_{A}\)) V(\(_{v}\)) 1 20.0 0.13 1.08 2 30.0 0.14 0.97 3 40.0 0.15 0.85 4 50.0 0.16 0.74 5 60.0 0.17 0.60 Slope, s = \(\frac{y_{2} - y_{1}}{x_{2} - x _{1}} = \frac{0.85 - 0.30}{0.20 - 0.5}\) \(\frac{0.55}{0.05}\) = 11 When I = o V = 3.15v Precautions; i - ensured that the key was open whenever the circuit was not in use so as to avoid the cells being run down. - ensured there was no zero error in the ammeter and the voltmeter by making necessary adjustments before taking my readings. - ensured that the jockey did not scrap the potentiometer wire. - ensured that the electrical wires were tightly connected to the terminals in order to avoid current fluctuations. (b)i. Lead-acid accumulator Leclanche cell (i). It has the capacity to supply large current for a long time (ii). It can be charged It has no such capacity It cannot be recharged ii. e.m.f = 2V r = 1\(\Omega\) (internal resistance) R = 9\(\Omega\) (external resistance) Calculate the potential drop across the cell. By formula; E = I(R +r) 2 = I(9 + 1) 2 = 101 I = \(\frac{2}{10}\) = 0.2A Therefore, the p.d drop across the cell is Ir. = 0.2 x 1 = 0.2V

n = \(\frac{sin i}{sin r}\) 1.5 = \(\frac{sin \theta}{sin(190 - \theta)} = \frac{sin \theta}{cos theta} = \tan \theta\) Hence, 0 = tan\(^{-1}\) 1.5 = 56.31\(^o\)

(1) If F = uniform speed of body \(\mu = \frac{F}{R} = \frac{Mg sin \theta}{Mg cos \theta} = tan \theta\)

(a) An electrode is a conductor through which electrons flow into and out of the electric cell. (b) I = \(\frac{M}{zt} = \frac{200}{8.33 \times 10^{-4} \times 60 \times 2}\)