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solucionario tema 4 oxford, Ejercicios de Física

Solucionario primeros temas del libro de IB physics de Oxford

Tipo: Ejercicios

2019/2020

Subido el 27/10/2020

victor-moreno-23
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5 documentos

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¡Descarga solucionario tema 4 oxford y más Ejercicios en PDF de Física solo en Docsity! 1 E N D - O F -T O P I C Q U E S T I O N S © Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute Solutions for Topic 4 – Oscillations and waves 1. a) (i) point at minima of curve (acceleration maximum when displacement is minimum) (ii) either point of intercept with time axis (maximum speed at zero displacement) b) weight of the bob is opposed by the component of tension along the vertical axis i.e. T sinθ = mg 2. a) displacement graph shifted by – π _ 2 b) velocity graph is the gradient of the displacement graph; velocity is maximum when displacement is zero 3. a) magnitude of acceleration is proportional to the displacement from a fixed point; direction of acceleration is towards that fixed point b) (i) A = 0.5 × d (ii) and (iii) x A t T 4. a) a wave in which the positions of maximum and minimum amplitude travel through the medium b) 4.0 mm; 2.4 cm; 3.3 Hz; 7.9 cm s–1 5. a) transverse: direction of energy transfer perpendicular to direction of travel longitudinal: direction of energy transfer parallel to direction of travel b) frequency = (time for one period)–1 = (0.135)–1 = 7.4 Hz; amplitude = maximum displacement = 8 mm c) c = fλ; λ = 0.15 _ 7.4 = 0.020 m = 2.0 cm d) d x 6. a) (i) ray: line showing direction in which wave transfers energy; (ii) wave speed is the distance wave has travelled per unit time; wave energy is the sum of the kinetic energy (maximum when speed is maximum) and potential energy (maximum when speed is zero) b) (i) f = 1 _ 6 × 10–3 = 167 Hz (ii) at t = 1.0 ms, x A = 1.7 mm, x B = 0.7 mm, so total displacement = 1.7 + 0.7 = 2.4 mm at t = 8.0 ms; x A = 1.7 mm; x B = – 0.7 mm, so total displacement = 1.7 – 0.7 = 1.0 mm 7. a) transverse: direction of energy transfer perpendicular to direction of travel b) (i) frequency is 1 ÷ time period (time between successive crests) = 1 ÷ 0.13 = 7.7 Hz (ii) amplitude is maximum displacement = 8 mm longitudinal: direction of energy transfer parallel to direction of travel c) c = fλ; λ = 0.12 _ 7.7 = 0.016 m = 1.6 cm 839213_Solutions_Ch04.indd 1 12/17/14 4:09 PM
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