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Answer: A An object sinks when its density is greater than that of water.

Brownian motion is the random motion of particles in fluids.

Answer: D To calculate the mass of silver deposited during electrolysis, we use the formula: \(m = \frac{Q \cdot M}{n \cdot F}\) Where: \( I = 0.8 \, \text{A} \), \( t = 25 \, \text{minutes} = 25 \times 60 = 1500 \, \text{s} \), \( Q = I \cdot t = 0.8 \, \text{A} \cdot 1500 \, \text{s} = 1200 \, \text{C} \) \( M = 107.87 \, \text{g/mol} \), \( n = 1 \), \( F \approx 96485 \, \text{C/mol} \) Substituting the values: \(m = \frac{1200 \, \text{C} \cdot 107.87 \, \text{g/mol}}{1 \cdot 96485 \, \text{C/mol}} = \frac{129444}{96500} \approx 1.34 \, \text{g}\)

The thermal capacity (or heat capacity) of a body is the amount of heat required to raise its temperature by 1^\circC (or 1 K). It is calculated as: C = m.c where: ( m ) is the mass of the body, ( c ) is the specific heat capacity, which depends on the nature (material or substance) of the body. Thus, thermal capacity depends on both the mass and the nature of the material.

The ideal gas law is \( PV = nRT \). Making \( P \) the subject: \[ P = \frac{nRT}{V} \] Substitute the values: - \( n = 8.3 \, \mathrm{mol} \), \( R = 8.31 \, \mathrm{J \, K^{-1} \, mol^{-1}} \), \( T = 17^\circ \mathrm{C} = 290 \, \mathrm{K} \) - \( V = 4.5 \, \mathrm{m}^3 \) \( P = \frac{8.3 \times 8.31 \times 290}{4.5} = 4444.93 \, \mathrm{Pa} \)

Specific gravity (SG) is the ratio of the density of the liquid (\(\rho_l\)) to the density of water (\(\rho_w\)): \( \mathrm{SG} = \frac{\rho_l}{\rho_w} \) Rearranging for density: \((\rho_l = \mathrm{SG} \times \rho_w = 0.76 \times 1000 = 760 \, \mathrm{kg \, m^{-3}}\)

\(C = \frac{Q}{\Delta T} = \frac{48000 \, \text{J}}{53 \, \text{ºC}} \approx 905.66 \, \text{J/ºC}\)

\(F = m \cdot g = 1.5 \, \text{kg} \times 10 \, \text{m/s}^2 = 15 \, \text{N}\) \(A = \pi r^2 = \pi (0.003 \, \text{m})^2 = \pi (9 \times 10^{-6}) \approx 2.827 \times 10^{-5} \, \text{m}^2\) \(\text{Stress} = \frac{F}{A} = \frac{15 \, \text{N}}{2.827 \times 10^{-5} \, \text{m}^2} \approx 5.299 \times 10^5 \, \text{Pa}\)

The apparent depth of an object in water (as viewed from air) is less than the real depth due to refraction. The relationship is given by: \(\mu = \frac{\text{Real depth}}{\text{Apparent depth}}\) where \(\mu\) is the refractive index of water (\(\mu = 1.33\)). Rearranging for real depth: \(\text{Real depth} = \mu \times \text{Apparent depth} = 1.33 \times 10 \, \text{cm} = 13.3 \, \text{cm}\)

\(B = \frac{F}{q \cdot v \cdot \sin \theta}\) \(\theta = 45^\circ \implies \sin 45^\circ = \frac{\sqrt{2}}{2}\) Substituting the values: \(B = \frac{6}{(5 \times 10^{-6})(3 \times 10^4) \left(\frac{\sqrt{2}}{2}\right)}\) Simplifying: \(B = \frac{6}{(15 \times 10^{-2}) \left(\frac{\sqrt{2}}{2}\right)} \approx \frac{6}{0.1061} \approx 56.5 \, \text{T}\)

Photovoltaic cells use semiconductor materials, such as silicon, to convert sunlight into electricity through the photovoltaic effect.

D. 2666.7 \(\text{J kg}^{-1} \, \text{K}^{-1}\) Using the formula: \(Q = mc\Delta T\) We can rearrange to find \( c \): \(c = \frac{Q}{m \Delta T} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\) Thus, the specific heat capacity is: \(c = 2666.7 \, \text{J kg}^{-1} \, \text{K}^{-1}\).

B. \( 6.1 \times 10^{-31} \, \text{kg} \) Using the de Broglie wavelength: \(\lambda = \frac{h}{p}\) Rearranging gives: \(p = \frac{h}{\lambda}\) Given: \(\lambda = 4.0 \times 10^{-7} \, \text{mm} = 4.0 \times 10^{-10} \, \text{m}\) Now substituting for \( p \): \(p = \frac{6.63 \times 10^{-34}}{4.0 \times 10^{-10}} = 1.6575 \times 10^{-24} \, \text{kg m/s}\) Now, to find mass \( m \): \(m = \frac{p}{v} = \frac{1.6575 \times 10^{-24}}{2.7 \times 10^{8}} \approx 6.1 \times 10^{-31} \, \text{kg}\).

E = \(\frac{\text{Kq}}{r^2}\) r = \(\sqrt{\frac{\text{Kq}}{\text{E}}}\) r = \(\sqrt{\frac{9.0 \times 10^9 \times 1.2 \times 10^{-7}}{4.8 \times 10^{-4}}}\) r = \(\sqrt{2.25 \times 10^6}\) = 1500m = 1.5km

\(\gamma\) = 3 \(\alpha\) \(\gamma\) = 3 x 24 x 10\(^{-6}\) = 72 x 10\(^{-6}\)k\(^{-1}\) V\(_2\) = V\(_1\)(1 + \(\gamma\) ∆T) V\(_2\) = 3.48 x (1 + 72 x 10\(^{-6}\)k\(^{-1}\) x 306) V\(_2\) = 3.48 x (1 + 0.022032) V\(_2\) ≈ 3.56cm\(^3\)