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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\(^3\)

Induced emf (electromotive force) refers to the voltage generated across a conductor or circuit due to a changing magnetic field, as explained by Faraday's laws of induction. This voltage drives the flow of induced current when the circuit is complete.

The ampere is a fundamental SI base unit, representing electric current, and is one of the seven base units used in physics

A. Reduce the friction of fluid. The pointed front of a boat or airplane is designed to minimize drag and reduce the frictional resistance as it moves through water or air, enhancing efficiency and improving performance.

Equivalent \(\, emf \, E_{eq} = 1.5 \, \text{V} \, (\text{same as single cell})\). Equivalent \(\, internal \, resistance \, r_{eq} = \frac{1}{2} = 0.5 \, \Omega \, (\text{parallel})\). Total \(\, resistance \, R_{total} = r_{eq} + 2 = 2.5 \, \Omega\). Current \(\, I = \frac{E_{eq}}{R_{total}} = \frac{1.5}{2.5} = 0.6 \, \text{A}\).

LASER stands for Light Amplification by Stimulated Emission of Radiation. It works when an incoming photon makes an excited electron release another identical photon, producing a coherent, highly focused beam of light.

The velocity-time graph has three stages: 0 - 10s: constant \(12 \, \text{m/s}\) 10 - 14s: deceleration from \(12 \, \text{m/s}\) to \(0 \, \text{m/s}\) (last stage) The area of the triangle gives the distance in the last stage: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (14 - 10) \, \text{s} \times 12 \, \text{m/s}\) \(= \frac{1}{2} \times 4 \times 12 = 24 \, \text{m}\).

1a. Tables of values M(g) W\(_1\)(g) W\(_2\)(g) W\(_3\)(g) U = (W\(_1\) - W\(_2\)(g) V = (W\(_1\) - W\(_3\)(g) 5.0 5.00 4.20 4.70 0.80 0.30 10.0 10.00 7.90 8.60 2.10 1.40 15.0 15.00 12.75 13.50 2.25 1.50 20.0 17.10 17.10 16.40 2.90 3.60 25.0 22.10 22.10 21.40 2.90 3.60 S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{3 - 0.8}{2.8 - 1.4}\) = \(\frac{11}{7}\) = 1.57. Precaution (i) I carefully avoided parallax error while reading the scale of the spring balance to ensure accurate measurements. (ii) I checked and accounted for any zero error on the spring balance before taking measurements. (iii) I handled the liquid carefully to prevent any splashing, which could lead to loss of material or inaccurate readings. (iv) I ensured that the object did not touch the bottom or sides of the beaker to avoid interference with the force readings. (v) I thoroughly cleaned the object (mass) before dipping it into the liquid to prevent contamination or alteration of the liquid's properties. bii) Law of Floatation: The law of floatation states that an object will float in a fluid if its weight is equal to the weight of the fluid it displaces. For a floating object, the weight of the object is balanced by the upward buoyant force.

This process disrupts the alignment of magnetic domains, effectively demagnetizing the bar.

2a. Table of observations D(cm) D\(^2\)(cm\^2\) X\(_1\)(cm) L = (X\(_1 - X_2\)) L\(^2\)(cm\(^2\)) 100 10000 80.50 62.20 3868.8 90 8100 70.20 51.20 2621.4 80 6400 59.10 39.10 1528.8 70 4900 46.70 24.70 610.09 60 3600 29.00 21.40 457.96 S = \(\frac{y_2 - y_1}{x_2 - x_1}\) = \(\frac{6250 - 3250}{100 - 60}\) = \(\frac{3000}{40}\) = 75 K = \(\frac{\text{s}}{4}\) = \(\frac{75}{4}\) = 18.75 Precautions: (i) I carefully avoided parallax error by ensuring my eyes were directly in line with the markings on the metre rule while taking measurements to improve accuracy. (ii) I adjusted the setup and focused precisely to ensure a sharp and well-defined image was formed on the screen, minimizing errors due to blurriness or misalignment. (iii) I made sure that the experimental apparatus was stable and free from vibrations to avoid any disturbance during measurements. (iv) I ensured that the metre rule was properly calibrated and positioned perpendicular to the setup to prevent measurement inaccuracies due to angle errors. bi. A real image is formed when light rays converge and meet at a point, allowing it to be projected onto a screen. It is typically inverted relative to the object, as seen in concave mirrors or convex lenses when the object is beyond the focal point. On the other hand, a virtual image is formed when light rays appear to diverge from a point but do not actually meet. This type of image cannot be projected onto a screen and is always upright, as seen in plane mirrors or convex mirrors. bii. See diagram above.

A lone pair of electrons is a pair of valence electrons in an atom that are not shared with another atom in a chemical bond, also known as a non-bonding pair. These electrons reside in the outermost electron shell, occupy space, and significantly influence a molecule's shape, polarity, and reactivity, often depicted as dots in Lewis structures. To determine the number of lone pairs, we analyze the Lewis structure of each molecule: CH\(_4\)(Methane): Carbon forms four single bonds with hydrogen atoms. There are zero lone pairs in the molecule. NH\(_3\)(Ammonia): Nitrogen forms three single bonds with hydrogen atoms and has one lone pair of electrons. H\(_2\)O(Water): Oxygen forms two single bonds with hydrogen atoms and has two lone pairs of electrons. CO\(_2\)(Carbon Dioxide): Carbon forms two double bonds with oxygen atoms. Each oxygen atom has two lone pairs, totaling four lone pairs of electrons in the molecule. :. The molecule with the highest number of lone pairs of electrons is CO\(_2\) - Option D