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Total balls = 2 + 6 + 5 = 13 Pr(black) = \(\frac{5}{13}\)

a. Total number of cards = 52 There are 4 fives and 2 red nines A card is drawn at random Pr(Either a five or a red nine) = Pr(five) + Pr(red nine) = \(\frac{4}{52}\) + \(\frac{2}{52}\) = \(\frac{6}{52}\) = \(\frac{3}{26}\) b. see diagram above ∠Q = 90^\circ \Delta PQR is a right-angled triangle. PR\(^2\) = PQ\(^2\) + QR\(^2\) PR\(^2\) = 5\(^2\) + 3\(^2\) = 25 + 9 PR\(^2\) = 34 = > PR = \(\sqrt{34}\) = 5.831 Let angle P = \theta Therefore, \(\frac{5.831}{sin 90^\circ}\) = \(\frac{3}{sin \theta}\) 5.831 sin \theta = 3 Therefore, Sin \theta = \(\frac{3}{5.831}\) = 0.5145 \(\theta\) = Sin\(^{-1}\)(0.5145) = 30.96º Bearing of R from P = 150^\circ - \theta = 150^\circ - 30.96^\circ = 119^\circ (to the nearest degree)

Answer: D. 2.06 kg (Steam at 100º C: heat lost = m(L\(_v\) + c_w(100 - 15)) = 5400 \times 10\(^3\) J; m ≈ \(\frac{5.4e6}{(2.26e6 + 4200 \times 85)}\) ≈ 2.06 kg.)

Answer: C Time = \(\frac{(mc\DeltaT)}{(VI)}\) = \(\frac{588000}{1200}\) = 490 s.

Specifically, the heat required to change ice to water at the same temperature is known as the latent heat of fusion (of ice).

c = 450 Jkg\(^{-1}\)K\(^{-1}\); h = 20 m; \theta = ?; g = 10 ms\(^{-2}\) mc\(\Delta\)\(\theta\) = mgh c x \(\Delta\)\(\theta\) = g x h 450 x \(\Delta\)\(\theta\) = 20 x 10 \(\Delta\)\(\theta\) = \(\frac{200}{450}\) = \(\frac{20}{45}\) \(\therefore\) \(\Delta\)\(\theta\) = \(\frac{4}{9}\)ºC

Heat loss by hot water = Heat gained by water mc(t - 50) = 2mc(50 - 30) t - 50 = 2(20) t - 50 = 40 t = 40 + 50 \(\therefore\) t = 90^\circC

m\(_s\) = 2 kg; c\(_s\) = 460 Jkg\(^{-1}\)K\(^{-1}\); m\(_w\) = ?; c\(_w\) = 4200 Jkg\(^{-1}\)K\(^{-1}\); \theta\(_1\) = 280^\circC; \theta\(_2\) = 30^\circC; \theta\(_3\) = 70^\circC Heat loss = Heat gained m\(_s\) c\(_s\) (\theta\(_1\) - \theta\(_3\)) = m\(_w\) c\(_w\) (\theta\(_3\) - \theta\(_2\)) 2 \times 460 \times (280 - 70) = m\(_w\) \times 4200 \times (70 - 30) 2 \times 460 \times 210 = m\(_w\) \times 4200 \times 40 193,200 = 168,000 m_w m\(_w\) = \(\frac{193200}{168000}\) ≈ 1.15kg.

The heat required to convert ice at 0^\circC to water at the same temperature is given by the formula: Q = m \times L, where m is the mass of the ice, and L is the specific latent heat of ice. Substituting the given values, we get Q = 20 g \times 336 Jg\(^{-1}\) = 6.72 \times 10\(^3\) J.

\Delta\theta\(_P\) = 2 \times \Delta\theta\(_Q\) m\(_P\) = m\(_Q\) c\(_Q\) = c\(_P\) = ? H\(_P\) = H\(_Q\) (given) i.e. m\(_P\) \times c\(_P\) \times \Delta\theta\(_P\) = m\(_Q\) \times c\(_Q\) \times \Delta\theta\(_Q\) Since \Delta\theta\(_P\) = 2 \times \Delta\theta\(_Q\) and m\(_P\) = m\(_Q\) ∴ m\(_Q\) \times c\(_P\) \times 2 \times \Delta\theta\(_Q\) = m\(_Q\) \times c\(_Q\) \times \Delta\theta\(_Q\) c\(_P\) \times 2 = c\(_Q\) \(\frac{c_Q}{c_P}\) = \(\frac{2}{1}\) \(\therefore\) c\(_Q\): c\(_P\) = 2: 1

The general ordering of the electromagnetic spectrum from shortest to longest wavelength is as follows: 1. Gamma rays 2. X-rays 3. Ultraviolet (UV) rays 4. Visible light 5. Infrared (IR) radiation 6. Microwaves 7. Radio waves Therefore, X-rays are the correct answer.

Standing waves are formed when waves reflect back from a boundary and interfere with the incoming waves of the same frequency and amplitude, creating points that appear to stand still (nodes) and points of maximum oscillation (antinodes).

Water waves are mechanical waves that require a material medium, such as water, to propagate. They involve the vibration of water particles. In contrast, light waves are electromagnetic waves that do not need a medium for propagation; they can travel through a vacuum. For example, sunlight travels through space without the need for any material medium.

Answer: A. 1:4 (Pressure equal: \(\frac{F_1}{(\pi r_1^2)}\) = \(\frac{F_2}{(\pi r_2^2)}\); \(\frac{_r2}{r_1}\) = \(\sqrt{(F_2}{F_1)}\) = \(\sqrt{16}\) = 4.)

Efficiency = \(\frac{\text{MA}{\text{VR}}\) x 100 But V.R = \(\frac{1}{sin\theta}\) = \(\frac{1}{sin 30}\) = 2 M.A = \(\frac{\text{L}}{\text{E}}\) Efficiency = \(\frac{\text{MA}{\text{VR}}\) x 100 0.75 = \(\frac{\text{M.A}}{2}\) 0.75 x 2 = M.A M.A = 1.5 = \(\frac{7.5 x 10}{\text{E}}\) 1.5E = 75 E = \(\frac{75}{1.5}\) = 50N