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\(X_c=\frac{1}{2\pi{fc}}\)=\(\frac{1}{2\times\pi\times\frac{200}{\pi}\times 25\times 10^{-6}}=\frac{10^6}{10^4}\)=100 Ω

m=zIt, \(I=\frac{m}{zt}= \frac{6.6\times10^{-4}}{3.3 \times10^{-7}\times 1800}=1.1111\\ \)error=1.1111-1.0=0.111A.

E=\(mc^2 = 2\times10^{-6} \times10^{-3}\times(3\times10^{8})^2=1.8\times10^{8}\)

\(E=\lambda v=\frac{hc}{\lambda}\Rightarrow\lambda=\frac{hc}{E}={6.6 \times10^{-34}\times3\times10^8}{1.5\times10^{-19}}\\=1.32\times10^{-6}\)

(a) If a stable element is bombarded by \(\alpha\) - particles the nucleus will become unstable and decay into other elements with the release of energy. Example; \(^{14}_{7}N + ^4_2He \to ^{17}_8 O + ^1_1H\) + Energy (b) When the nucleus of Uranium is bombarded with a neutron, it splits into two nearly equal parts with a release of energy and neutrons. These excess neutron, if not controlled, further bombard the remaining nuclei, causing a chain reaction and consequently an explosion, provided the critical mass of the Uranium is, exceeded \(^{235}_{92}U + ^1_0U \to ^{236}_{92}U \to ^{141}_{56}Ba + ^{92}_{36}Kr + 3^1_0n\) + Energy (c)(i) Radioactive substances give off identifiable radiation i.e \(\alpha, \beta\) and 8 radiations (ii) They are chemically identical to their non-radioactive counterparts and later can be made radioactive by bombardment with nuclear particles (iii) Radioactive materials decay with specific half-life. (d) Three applications of atomic energy (i) It is used for the production of nuclear weapons for warfare (ii) It is used in the generation of electric power (iii) It is used to detect leakages in underground pipes carrying oil or gases. (e) Two Postulates of Bohr's Model of the Atom: (i) Electrons occupy certain fixed orbits around the nucleus of an atom. When an electron is in one of these orbits it does not radiate any energy (ii) The angular momentum is quantized in the allowed orbit. Two Limitations To Bohr's Model: (i) It could not interpret the details of the spectra of complex atoms (ii) There is no theoretical basis for selecting the allowed orbits postulated.

\( \theta _3 \) shows the angle of deviation

F = m (v-u) = kg x m x S-2 = M1L1T-2

To find the average speed of the particle between \( t = 2 \) seconds and \( t = 5 \) seconds, we use the formula for average speed: \(\text{Average Speed} = \frac{\Delta x}{\Delta t}\) Calculate \( x \) at \( t = 2 \) seconds} Using the equation \( x = 10 + 12t^2 \): \(x(2) = 10 + 12(2^2) = 10 + 12(4) = 10 + 48 = 58 \, \text{m}\) Calculate \( x \) at \( t = 5 \) seconds \(x(5) = 10 + 12(5^2) = 10 + 12(25) = 10 + 300 = 310 \, \text{m}\) Calculate \( \Delta x \) and \( \Delta t \) \(\Delta x = x(5) - x(2) = 310 - 58 = 252 \, \text{m}\) \(\Delta t = 5 - 2 = 3 \, \text{s}\) Calculate the average speed \(\text{Average Speed} = \frac{252 \, \text{m}}{3 \, \text{s}} = 84 \, \text{m/s}\).

ma = mgsin\(\theta\) a = \(\frac{5 {\times} 10 {\times} sin30}{t}\) a = 5ms2

M1U1 + M2U2 = (m1 + m2)v = \(\frac{(0.1 {\times} 15) + (0.4 {\times} 0)}{(0.1 + 0.4)}\) v = 3ms-1

Range = \(\frac{U^2sin2{\theta}}{g}\) \(\frac{R_x}{R_y}\) = \(\frac{U^2_xsin60}{U^2_ysin120}\) sin60o = sin120o \(\frac{R_x}{R_y}\) = \(\frac{1}{1}\) hence, = 1:1

In simple harmonic motion (SHM), the force acting on the body is directly proportional to its displacement from the equilibrium position and is in the opposite direction (according to Hooke's Law). This force results in an acceleration of the body, which is also directly proportional to its displacement but in the opposite direction. When the displacement of the body is maximum (either positive or negative), the force acting on it is also maximum, leading to maximum acceleration. Similarly, when the displacement is zero (at the equilibrium position), the force and acceleration are also zero. Thus, the force acting on the body and its acceleration are in phase, meaning they reach their maximum and minimum values simultaneously during the oscillatory motion.

F = mg + ma F = 2 [10 + \(\frac{(40 - 10)}{4}\)] F = 35 N

g = \(\frac{GM}{R^2}\) \(\frac{G_1M_1}{r^2_1}\) = \(\frac{G_2M_2}{r^2_2}\) \(\frac{G_1M_1}{r^2_1}\) = \(\frac{G_2 {\times}10 M_1}{(2r_1)^2}\) \(\frac{G_1}{G_2}\) = \(\frac{2}{5}\)

Using, v\(^2\) = u\(^2\) - 2gx (20)\(^2\) = u\(^2\) - 2 x 10 x 10 u\(^2\) = 400 + 200 = 600(m/s\(^2\)) initial K.E = \(\frac{1}{2}\)mu\(^2\) = \(\frac{1}{2}\) x 0.1 x 600 = 30 J (100g = 0.1kg)