Writing Assignment 2_2(1)

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Life Chiropractic College West *

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Physics

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Apr 3, 2024

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Written Assignment 2-2 Answer the following questions. Do not send in a partially completed assignment. Answer each question as completely as possible. Do not just submit the final answer. Show the main steps in the solution. Use Equation Editor or MathType for typing symbols and equations when needed. Submit your assignment in MS Word docx, doc, or rich format. Submissions in other formats will not be graded and will be treated as non-submissions. 1. If a grandfather clock were running slow, how could we adjust the length of the pendulum to correct the time? Identify the Issue : First, confirm that the clock is indeed running slow by comparing its time with a reliable timekeeping device. Determine the Correction : Calculate the amount of time the clock is running slow. This will help you determine how much you need to adjust the length of the pendulum. Adjust the Pendulum Length : To make the pendulum shorter, move the pendulum bob (the weight at the bottom of the pendulum) upwards. Conversely, to make the pendulum longer, move the pendulum bob downwards. Adjust the Pendulum Length : To make the pendulum shorter, move the pendulum bob (the weight at the bottom of the pendulum) upwards. Conversely, to make the pendulum longer, move the pendulum bob downwards. 2. A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance d = 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 47.5 N/m , determine the mass of the object. To determine the mass of the object attached to the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is: F=kx Where: F is the force exerted by the spring (in Newtons, N). k is the spring constant (in Newtons per meter, N/m). x is the displacement from the equilibrium position (in meters, m). Given that k=47.5 N/m and x=0.050 m (since 5.00 cm is equivalent to 0.050 m), we can calculate the force exerted by the spring: F=(47.5 N/m)×(0.050 m) F=2.375 N Now, since the system is in equilibrium, the force exerted by the spring is balanced by the force due to gravity acting on the object. The force due to gravity can be calculated using Newton's second law: F=mg Where: m is the mass of the object (in kilograms, kg).

g is the acceleration due to gravity, which is approximately 9.81 m/s^2 on the surface of the Earth. Now we can rearrange the equation to solve for the mass m: m=Fg Substituting the values we've found: m=(2.375 N) / (9.81 m/s^2 ) m≈0.242 kg Therefore, the mass of the object attached to the spring is approximately 0.242 kg 3. A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500 kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency  , the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (e) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s . (a) Force Constant of the Spring: We can use Hooke's Law to find the force constant of the spring: F=kx Where: F is the force applied to the spring (in Newtons, N). k is the spring constant (in Newtons per meter, N/m). x is the displacement from the equilibrium position (in meters, m). Given that F=7.50N and x=0.030m (3.00 cm converted to meters), we can rearrange the equation to solve for k: k= xF k=(7.50 N) / (0.030 m) k=250N/m So, the force constant of the spring is 250N/m. (b) Angular Frequency, Frequency, and Period of the Motion: The angular frequency (ω) can be calculated using the formula: ω=km Where: k is the spring constant (in N/m). m is the mass of the particle (in kg). Given k=250N/m and m=0.500kg, we can calculate ω: ω= √ (250 N/m) / (0.500 kg) ω= √ 500 ω≈22.36 rad/s The frequency (f) is related to the angular frequency by f=ω/2 π . Thus: f= (22.36 rad/s) / (2 π ) f≈3.56 Hz The period (T) of the motion is the reciprocal of the frequency: T=1/f . Therefore: T=1/3.56 Hz T≈0.281s

So, the angular frequency is ω≈22.36 rad/s, the frequency is f≈3.56 Hz and the period is T≈0.281 s. (c) Total energy of the system: The potential energy of the spring at the equilibrium position (x=0.03 m) is: PE=1/2kx^2 PE=1/2(k)(0.03)^2 Given that x=0.03 m, we need to find k, the force constant of the spring. We can use Hooke's law: F=k/x Given that F=7.50 N and x=0.03 m: k=F/x k=7.50 N/0.03 m k=250 N/m Now, we can find the potential energy: PE=1/2(250)(0.03)^2 PE=1/2×250×0.0009 PE=112.5×10^−6 J So, the total energy of the system is approximately 112.5×10^−6 J. (d) Amplitude of the motion: Given as A=0.05 m. (e) Maximum velocity and maximum acceleration of the particle: Using the given formula: V_max= √(k(0.03)^2)/m) V_max= √(250×(0.03)^2/0.500) V_max= √(2.25/0.500) V_max= √4.5 V_max≈2.12 m/s’ A_max=kx/m A_max=(250×0.05/0.500) A_max=(12.5/0.500) A_max=25 m/s^2 (f) Displacement xx of the particle from the equilibrium position at t=0.500 s: Given the equation x=Acos(ωt), we need to find ω first. From the given displacement and amplitude, we can find ω using x=Acos(ωt): 0.05=0.03cos(ω×0) Cos^−1(0.05/0.03)=ω×0 ω=cos^−1(0.05/0.03) ω≈1.23 rad/s Now, at t=0.500 s: x=0.05cos(1.23×0.500) x=0.05cos(0.615) x≈0.05×0.796 x≈0.0398 m So, the displacement xx of the particle from the equilibrium position at t=0.500 s is approximately 0.0398m.

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(g) Velocity and acceleration of the particle when t=0.500t=0.500 s: Using the formulas: v=−Aωsin(ωt) v=−0.05×1.23×sin(1.23×0.500) v=−0.05×1.23×sin(0.615) v≈−0.05×1.23×0.544 v≈−0.0336 m/s a=−Aω^ 2 cos(ωt) a=−0.05×(1.23)^ 2 ×cos(1.23×0.500) a=−0.05×(1.23)^ 2 ×cos(0.615) a≈−0.05×1.5129×cos(0.615) a≈−0.0756×0.796 a≈−0.0602 m/s^2 So, the velocity of the particle when t=0.500 s is approximately −0.0336 m/s and the acceleration is approximately −0.0602m/s² 4. A simple pendulum is 5.00 m long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s 2 ? (b) What is its period if the elevator is accelerating downward at 5.00 m/s 2 ? (c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s 2 ? To solve this problem, we'll use the formula for the period (T) of a simple pendulum undergoing simple harmonic motion: T=2π √(L/g) Where: T is the period of the pendulum. L is the length of the pendulum. g is the acceleration due to gravity. Given that the length of the pendulum (L) is 5.00 m, we can calculate the period (T) for each scenario. (a) Elevator accelerating upward at 5.00 m/s^2 : When the elevator is accelerating upward, the effective acceleration due to gravity for the pendulum will be g+a, where g=9.81 m/s^2 is the acceleration due to gravity and a=5.00 m/s^2 is the acceleration of the elevator. Tup=2π√(L/g+a) Tup=2π√(5.00 m/9.81 m/s^2+5.00 m/s^2) Tup≈3.41 s (b) Elevator accelerating downward at 5.00 m/s^2 : When the elevator is accelerating downward, the effective acceleration due to gravity for the pendulum will be g−a. Tdown=2π√(L/g−a) Tdown=2π√(5.00 m/9.81 m/s^2−5.00 m/s^2 Tdown≈6.49 s (c) Truck accelerating horizontally at 5.00 m/s^2 :

Horizontal acceleration doesn't affect the period of a simple pendulum because gravity is acting perpendicular to the direction of motion. So, the period remains the same as if there were no acceleration. Thorizontal=2π√(L/g) Thorizontal=2π√(5.00 m/9.81 m/s) Thorizontal≈7.09 s So, the periods for the three scenarios are approximately 3.41 s (elevator accelerating upward), 6.49 s (elevator accelerating downward), and 7.09 s (truck accelerating horizontally). 5. A 5.00 g bullet is moving, as shown in the figure, with an initial speed of 400 m/s is fired into and passes through a 1.00 kg block, as in the figure. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 900 N/m . If the block moves 5.00 cm to the right after impact, find (a) the speed at which the bullet emerges from the block and (b) the mechanical energy lost in the collision.When dealing with sound intensities and decibel levels, a convenient approximation (accurate to 2 significant figures) is: For every doubling of the intensity, the decibel level increases by 3.0. Suppose the sound level at some location is 85 dB . Find the decibel levels if the sound intensity is increased by factors of (a) 2.0, (b) 4.0, (c) 8.0, and (d) 16. (a) Speed at which the bullet emerges from the block: To find the speed at which the bullet emerges from the block, we'll use the principle of conservation of momentum. Since momentum is conserved in the collision, we can write: M_bv_b,_i = (m_b+m_s)v_f where: m_b is the mass of the bullet (5.00 g, or 5.00×10^−3 kg), v_b,_i is the initial velocity of the bullet (400 m/s), m_s is the mass of the block (1.00 kg), v_f is the final velocity of the bullet-block system. Solving for v_f: V_f=m_bv_b,_i / m_b+m_s V_f=(5.00×10^−3 kg)(400 m/s)/(5.00×10^−3 kg+1.00 kg) V_f≈3.96 m/s So, the speed at which the bullet emerges from the block is approximately 3.96 m/s (b) Mechanical energy lost in the collision: The mechanical energy lost in the collision can be calculated by considering the change in kinetic energy of the bullet-block system. Initially, the bullet possesses kinetic energy, which is transferred to the block-spring system. The mechanical energy lost is equal to the initial kinetic energy of the bullet minus the final kinetic energy of the bullet-block system. Elost= ½ m_bv^2_b,_i – ½ (m_ b +m_ s )v ^2_f Substituting the given values Elost=1/2(5.00×10^−3 kg)(400 m/s)^2−1/2(5.00×10^−3 kg+1.00 kg)(3.96 m/s)^2 Elost≈397 J So, the mechanical energy lost in the collision is approximately 397 J The relationship between sound intensity (measured in watts per square meter, W/m²) and decibel level (dB) is given by: dB=10log _ 10 (I/I_0) where:

dB is the decibel level, I is the sound intensity, I0 is the reference intensity (usually the threshold of human hearing, which is 10^−12W/m²). To calculate the change in decibel level for a given change in sound intensity, we can use the fact that for every doubling of intensity, the decibel level increases by 3.0 dB. Given that the initial decibel level is 85 dB, let's calculate the decibel levels for different increases in sound intensity: 6. When dealing with sound intensities and decibel levels, a convenient approximation (accurate to 2 significant figures) is: For every doubling of the intensity, the decibel level increases by 3.0. Suppose the sound level at some location is 85 dB . Find the decibel levels if the sound intensity is increased by factors of (a) 2.0, (b) 4.0, (c) 8.0, and (d) 16. (a) Increase by a factor of 2.0: Since doubling the intensity corresponds to an increase of 3.0 dB, increasing by a factor of 2.0 means doubling the intensity once. dB_new=85 dB+3.0 dB=88.0 dB (b) Increase by a factor of 4.0: Doubling the intensity twice corresponds to an increase of 3.0 dB+3.0 dB=6.0 dB dB_new=85 dB+6.0 dB=91.0 dB (c) Increase by a factor of 8.0: Doubling the intensity three times corresponds to an increase of 3.0 dB+3.0 dB+3.0 dB=9.0 dB dB_new=85 dB+9.0 dB=94.0 dB (d) Increase by a factor of 16: Doubling the intensity four times corresponds to an increase of 3.0 dB+3.0 dB+3.0 dB+3.0 dB=12.0 dB dB_new=85 dB+12.0 dB=97.0 dB So, the decibel levels for increases in sound intensity by factors of (a) 2.0, (b) 4.0, (c) 8.0, and (d) 16 are 88.0 dB, 91.0 dB, 94.0 dB, and 97.0 dB, respectively. 7. A stone is dropped from rest into a well. The sound of the splash is heard exactly 2.00 s later. Find the depth of the well if the air temperature is 10.0 °C . To find the depth of the well, we can use the kinematic equations for motion under constant acceleration. The stone is initially at rest, so its initial velocity v_0= 0. The time it takes for the stone to reach the bottom of the well is the same as the time it takes for the sound of the splash to travel back up the well to the top. Let's denote: d as the depth of the well, t as the time it takes for the stone to reach the bottom of the well, v as the velocity of sound in air. The time it takes for the stone to fall is the same as the time it takes for the sound to travel back up: t=2.00 s

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The velocity of sound in air can be calculated using the formula: v=331.5+0.6T where T is the air temperature in Celsius. Given that the air temperature is 10.0°C v=331.5+0.6×10.0 v=331.5+6.0 v=337.5m/s Now, let's use the fact that the time taken for the stone to fall is equal to the time taken for the sound to travel back up: t=2d/v Solving for d: d=1/2 vt d=1/2×337.5 m/s×2.00 s d=337.5 m So, the depth of the well is 337.5 m. 8. Expectant parents are thrilled to hear their unborn baby’s heartbeat, revealed by an ultrasonic motion detector. Suppose the fetus’s ventricular wall moves in simple harmonic motion with amplitude 1.80 mm and frequency 115 beats per minute . The motion detector in contact with the mother’s abdomen produces sound at precisely 2 MHz , which travels through tissue at 1.50 km/s . (a) Find the maximum linear speed of the heart wall. (b) Find the maximum frequency at which sound arrives at the wall of the baby’s heart. (c) Find the maximum frequency at which reflected sound is received by the motion detector. (By electronically “listening” for echoes at a frequency different from the broadcast frequency, the motion detector can produce beeps of audible sound in synchrony with the fetal heartbeat.) (a) Maximum linear speed of the heart wall: The maximum linear speed of an object undergoing simple harmonic motion can be found using the formula: V_max=ωA where: v_max is the maximum linear speed, ω is the angular frequency, A is the amplitude of the motion. Given that the frequency of the motion is f=115 beats per minute, we need to convert it to Hz: f=115 beats/1 min×1 min/60 s =1.92 Hz Now, we can calculate the angular frequency: ω=2πf ω=2π×1.92 Hz ω≈12.05 rad/s Given that the amplitude A=1.80 mm, we need to convert it to meters: A=1.80 mm×1 m/1000 mm=0.00180 m Now, we can find the maximum linear speed: vmax=ωA vmax=(12.05 rad/s)(0.00180 m) vmax≈0.0217 m/s So, the maximum linear speed of the heart wall is approximately 0.0217 m/s.

(b) Maximum frequency at which sound arrives at the wall of the baby’s heart: The maximum frequency at which sound arrives at the wall of the baby’s heart is equal to the frequency of the motion, which is f=115 beats per minute, or approximately 1.92 Hz. (c) Maximum frequency at which reflected sound is received by the motion detector: The frequency of the sound reflected by the motion detector is the same as the frequency of the sound produced by the motion detector, which is 2 MHz or 2×10^6 9. Two ships are moving along a line due east as shown in the figure. The trailing vessel has a speed relative to a land-based observation point of v1 = 64.0 km/h, and the leading ship has a speed of v2 =45.0 km/h relative to that point. The two ships are in a region of the ocean where the current is moving uniformly due west at vcurrent = 10.0 km/h. The trailing ship transmits a sonar signal at a frequency of 1 200.0 Hz through the water. What frequency is monitored by the leading ship? To find the frequency monitored by the leading ship, we need to consider the Doppler effect. The Doppler effect describes the change in frequency of a wave (in this case, a sonar signal) perceived by an observer due to the relative motion between the source of the wave (the trailing ship) and the observer (the leading ship). The formula for the Doppler effect when the source is moving towards the observer is given by: f′=(v+v_source/v+v_observer) f where: f′ is the observed frequency, f is the frequency of the source, v is the speed of sound in the medium (which we can assume to be constant), v_source is the speed of the source (the trailing ship), v_observerv observer is the speed of the observer (the leading ship). Given: f=1200.0Hz (frequency of the sonar signal), v=10.0 km/h (speed of sound in the water), v_source=64.0 km/h (speed of the trailing ship), v_observer=45.0 km/h (speed of the leading ship). Converting all speeds to the same units, such as m/s, is necessary to maintain consistency: v=10.0×1000/3600=10/3 m/s v_source=64.0×10003600=320/9 m/s v_observer=45.0×1000/3600=25/2 m/s Now, we can calculate f′ using the given values and the Doppler effect formula: f′=((10/3+320/9)/(10/3+25/2)×1200.0 f′≈2.88 kHz So, the frequency monitored by the leading ship is approximately 2.88 kHz 10. A typical sound level for a buzzing mosquito is 40 dB , and that of a vacuum cleaner is approximately 70 dB . Approximately how many buzzing mosquitoes will produce a sound intensity equal to that of a vacuum cleaner? To compare the sound intensities of the buzzing mosquitoes and the vacuum cleaner, we need to understand the relationship between sound intensity and sound level (in decibels).

The relationship between sound intensity (I) and sound level (L) in decibels is given by the formula: L=10log_10(I/I_0) where: I is the sound intensity, I_0 is the reference intensity (typically 10^−12 W/m²). We can rewrite this equation to solve for I: I=I_0×10^(L/10) Given: Sound level of a buzzing mosquito (L_mosquito ) is 40 dB, Sound level of a vacuum cleaner (L_vacuum cleaner) is 70 dB. We can use these values to find the sound intensities of both the buzzing mosquito and the vacuum cleaner. For the buzzing mosquito: I_mosquito=10^−12×10^(40/10) For the vacuum cleaner: I_vacuum cleaner=10^−12×10^(70/10) Now, to find out how many buzzing mosquitoes would produce a sound intensity equal to that of a vacuum cleaner, we'll divide the sound intensity of the vacuum cleaner by the sound intensity of a buzzing mosquito: Number of mosquitoes= I_vacuum cleaner/I_mosquito Let's calculate these values. First, let's calculate the sound intensities for the buzzing mosquito and the vacuum cleaner. For the buzzing mosquito: I_mosquito=10^−12×10^(40/10) I_mosquito=10^−12×10^(4) I_mosquito=10^−12×10,000 I_mosquito=10^−8 W/m^2 For the vacuum cleaner: I_vacuum cleaner=10^−12×10^(70/10) I_vacuum cleaner=10^−12×10^(7) I_vacuum cleaner=10^−12×10,000,000 I_vacuum cleaner=0.01 W/m^2 Now, let's find out how many buzzing mosquitoes would produce a sound intensity equal to that of a vacuum cleaner: Number of mosquitoes=I_vacuum cleaner/I_mosquito Number of mosquitoes=0.01 W/m^2/10^−8 W/m^2 Number of mosquitoes=0.01/10^−8 Number of mosquitoes=10^6 So, approximately 1,000,000 buzzing mosquitoes will produce a sound intensity equal to that of a vacuum cleaner.

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Writing Assignment 2_2(1)

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