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Answers to Chapter 10 Problems

10.2.1 0.33

10.2.1 Solution

10.2.2 0.023 N m; 0.0020 J

10.2.2 Solution

10.2.3 (b) 5250 J

10.2.3 Solution

10.2.4 0.79 N; 5.0 × 10⁷ N m^–2

10.2.4 Solution

10.2.5 L_steel = 79.95 cm; L_copper = 79.69 cm

10.2.5 Solution

10.2.6 4 x 10^–18 m N

10.2.6 Solution

10.3.1 (a) 11.26 kg (b) 1.78 m (c) 196 J

10.3.1 Solution

10.3.2 0.91

10.3.2 Solution

10.3.3 (a) 4.0 × 10⁴ Pa (b) 5.0 × 10⁴ Pa (c) 1.6 × 10⁴ Pa

10.3.3 Solution

10.4.1 0.28 mm

10.4.1 Solution

10.5.1 \( \rho_0 g \over \alpha \); 1.3 x 10^–4 m^–1; the model ignores many significant factors such as air circulation, water vapour, thermal effects, etc.

10.5.1 Solution

10.6.1 \( {1 \over 2 \pi}{\sqrt{\rho_0 g \over \rho h}} \)

10.6.1 Solution

10.6.2 800 kg m^–3; 239 g; 232 g

10.6.2 Solution

10.6.3 For He: ��_a&�Բ��;–&�Բ��;��_He = 1.29 – 0.18 =1.11 kg m^–3. For H, ��_a&�Բ��;–&�Բ��;��_H = 1.29 – 0.09 = 1.20 kg m^–3. The lifting ability of He is 0.92 that of H. R = 6.0 m

10.6.3 Solution

10.6.4 (i) 47.5 mm (ii) 4.39 × 10^–⁴ m³

10.6.4 Solution

10.6.5 0.52 mm

10.6.5 Solution

10.6.6 450 m³

10.6.6 Solution

10.6.7 (a) 0.31 (b) 465 g (c) 510 g

10.6.7 Solution

10.7.1 45.1 W

10.7.1 Solution

10.7.2 126 m s⁻¹

10.7.2 Solution

10.7.3 13.9 N m⁻²

10.7.3 Solution

10.7.4 640 s = 10 m 40 s

10.7.4 Solution

10.7.5 3.6 m s⁻¹; 0.28 litre per second

10.7.5 Solution

10.8.1 1.88 x 10⁻⁵ N s m⁻²

10.8.1 Solution

10.8.2 1.1 × 10^–6 m³ s^–1

10.8.2 Solution

10.9.1

10.9.1 Solution

10.9.2 ≈ 0.7 \( \mu \)m

10.9.2 Solution

10.10.1 734 mm Hg

10.10.1 Solution

10.10.2 one half

10.10.2 Solution

10.11.1 0.69 mN

10.11.1 Solution

10.11.2 15 kJ

10.11.2 Solution

10.12.1 (a) 1.75 × 10²⁹ atoms m^–3 (b) 3.3 × 10²⁸ molecules m^–3 (c) 2.27 × 10²⁸ atoms m^–3

(a) 0.083 mol g^–1 (b) 0.056 mol g^–1 (c) 0.017 mol g^–1

10.12.1 Solution

10.13.1

10.13.1 Solution

Understanding Physics

Mansfield and O'Sullivan, Understanding Physics, 3rd ed., John Wiley & Sons, Chichester (2020),

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