(a) Maximum displacement 2.431871022 × 10 − 5 m Given: Frequency of the sound wave is f = 440 Hz Sound level is B = 120 dB Formula used: B = 10 log I I o Here, B is the sound level, I is the intensity level, and I o is the reference intensity which is taken to be equal to 10 − 12 W/m 2 . I = 2 π 2 v ρ f 2 A 2 Here, v is the speed of sound, ρ is density, f is frequency, A is the maximum displacement or amplitude, and I is intensity. Δ P M = 2 π ρ v A f Here, Δ P M is the pressure amplitude, v is the speed of sound, ρ is density, f is the frequency, and A is the maximum displacement or amplitude. Velocity of sound in air is v = 343 m/s Density of air is ρ = 1.29 kg/m 3 Calculation: Use the formula to calculate the intensity. 120 dB = 10 log I 10 − 12 W/m 2 12 dB = log I 10 − 12 W/m 2 I = 1 W/m 2 Use the formula to calculate the maximum displacement. 1 W/m 2 = 2 π 2 343 m/s 1 .29 kg/m 3 440 Hz 2 A 2 A = 2.431871022 × 10 − 5 m Conclusion: Maximum displacement is 2.431871022 × 10 − 5 m . (b) Pressure amplitude 29.7479 Pa Given: Frequency of the sound wave is f = 440 Hz Sound level is B = 120 dB Calculation: Use the formula to calculate the pressure amplitude: Δ P M = 2 π 1.29 kg/m 3 343 m/s 2.431871022 × 10 − 5 m 440 Hz Δ P M = 29.7479 Pa Conclusion: Pressure amplitude is 29.7479 Pa

Children playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not use rounded intermediate values-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (d) What If? If the teacher always launches the ball…

It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The "lenses" of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity ₁ = vi. As it passes through the region x = 0 to x=d, the electron experiences acceleration a = ai +a, where a and a, are constants. For the case v, = 1.67 x 107 m/s, ax = 8.51 x 1014 m/s², and a = 1.50 x 10¹5 m/s², determine the following at x = d = 0.0100 m. (a) the position of the electron y, = 2.60e1014 m (b) the…

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