This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Right Blessing, let's go. Part b(i): Wavelength limits in water Step 1: Identify the given values. Wavelength range in air: _air, min = 400 nm, _air, max = 700 nm. Refractive index of water: n = 1.33. Step 2: Use the relationship between wavelength in air, wavelength in water, and refractive index. The formula is n = _air_water, which means _water = _airn. Step 3: Calculate the minimum and maximum wavelengths in water. For the minimum wavelength: _water, min = 400 nm1.33 ≈ 300.75 nm For the maximum wavelength: _water, max = 700 nm1.33 ≈ 526.32 nm The limits of the wavelength in water are: 300.75 nm to 526.32 nm Part b(ii): Speed of a transverse wave on a steel wire Step 1: Identify the given values and convert units. Length of wire: L = 64 cm = 0.64 m. Mass of wire: m = 5 g = 0.005 kg. Tension force: F = 8 N. Step 2: Calculate the linear mass density () of the wire. = (m)/(L) = 0.005 kg0.64 m ≈ 0.0078125 kg/m Step 3: Calculate the speed of the transverse wave (v). The formula for the speed of a transverse wave on a string is v = sqrt((F)/()). v = sqrt(8 N)0.0078125 kg/m v = sqrt(1024) m/s v = 32 m/s The speed of the transverse wave is: 32 m/s Part 18a(i): Definitions of pitch, resonance, and Doppler effect • Pitch: The characteristic of a sound that determines whether it is high or low, primarily determined by the frequency of the sound wave. Higher frequency corresponds to higher pitch. • Resonance: A phenomenon that occurs when an oscillating system is driven at its natural frequency, resulting in a large amplitude of oscillation. This happens when the frequency of the applied force matches the natural frequency of the system. • Doppler effect: The apparent change in frequency or wavelength of a wave (sound or light) observed by an observer moving relative to its source. The frequency appears higher when the source and observer are approaching each other, and lower when they are moving apart. Part 18a(ii): Depth of the sea and wavelength of ultrasound Step 1: Identify the given values and convert units. Frequency of ultrasound: f = 50 kHz = 50 × 10^3 Hz. Time for round trip: t = 0.80 s. Speed of sound in sea water: v = 1500 m/s. Step 2: Calculate the depth of the sea. The signal travels to the ocean bed and back, so the distance covered is 2 × depth. Distance = speed × time. 2 × depth = v × t depth = (v × t)/(2) = 1500 m/s × 0.80 s2 depth = 1200 m2 = 600 m Step 3: Calculate the wavelength of the signal in water. The formula relating speed, frequency, and wavelength is v = f. = (v)/(f) = 1500 m/s50 × 10^3 Hz = (1500)/(50000) m = 0.03 m The depth of the sea is 600 m and the wavelength of the signal is 0.03 m. Part b(i): Frequency heard by the observer on the other train Step 1: Identify the given values and convert units. Speed of trains (source and observer): v_s = v_o = 90 kmh^-1. Convert to m/s: 90 kmh^-1 = 90 × 1000 m3600 s = 25 m/s. Source frequency: f_s = 520 Hz. Speed of sound in air: v = 350 m/s. Step 2: Apply the Doppler effect formula for approaching source and observer. When the source is approaching the observer, the denominator is (v - v_s). When the observer is approaching the source, the numerator is (v + v_o). The formula for the observed frequency f_o is: f_o = f_s ((v + v_o)/(v - v_s)) f_o = 520 Hz (350 m/s + 25 m/s350 m/s - 25 m/s) f_o = 520 Hz (375 m/s325 m/s) f_o = 520 Hz × 1.1538 f_o ≈ 600 Hz The frequency heard by the observer on the other train is: 600 Hz Part b(ii): Atmospheric temperature Step 1: Identify the given values. Speed of sound at temperature T: v_T = 336 m/s. Speed of sound at 0^: v_0 = 332 m/s. Step 2: Use the approximate formula for the speed of sound in air as a function of temperature. The speed of sound in air increases by approximately 0.6 m/s for every 1^ rise in temperature above 0^. The formula is v_T = v_0 + 0.6T, where T is the temperature in ^. 336 m/s = 332 m/s + (0.6 m/s/^) × T Step 3: Solve for T. 336 - 332 = 0.6T 4 = 0.6T T = (4)/(0.6) T ≈ 6.67^ The atmospheric temperature is: 6.67^ That's all for this set! Send me the next one 📸