PHYSICS HOMEWORK

Question

Please complete attached assignments. Nothing special, be direct, concise, simple.

professional_writer · Answer

Mathwiz4u · Answer

1. in preparing to shoot an arrow, an archer pulls a box string back 0.40 by exerting a force that increases uniformly from 0 to 230n. what is the equivalent spring constant of the bow?
use hooke's law to solve problem: f= kx, f= force, k = spring constant and x = displacement
find the force exerted by the archer: f initial + f final divided by 2=0 + 230= 115n, f = 115 n
solve for k by transposing the f=kx =k=f divided by x, k=f/x, k=115n/.40m =287.5 n/m, k=287.5 n/m
solution: equivalent spring constant of the bow is 287.5
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2. when a mass of 0.025 kg is attached to a certain spring, it makes 20 complete vibrations in 4.0 seconds. what is the spring constant of the spring?
formula used to solve problem: 
period of mass spring, t= 2 pi square root of m/k
frequency reciprocal of the period: f = 1/t  
a. find frequency of vibrations:  # of vibrations = 20, t = 0.40     f=20/.40 =50 hz
b. find the period of vibrations, t = 1/f =1/5 = 0.2 seconds
c. use the period formula to solve for spring constant t = 2pi times square root of m/k
transpose equation for k = 4 pi^2 x 0.025/ (0.2) ^2 = 24.7 n/m = 39.4784176 x 0.025=.9869604401/0.2^2 =
24.674011 n/m
solution: spring constant of spring = 24.7 n/m
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3.  a spring constant of 1.8 x 10^2 n/m is attached to a 1.5 kg mass and then set in motion.
a. what is the period of the mass spring system?  t = 0.574 seconds
b. what is the frequency of the oscillation? 1.74 hz
formulas needed: period of mass spring system = t = 2 x (pi) times square root of m/k [x=times=multiply]
t=2pi x square root 1.5 divided by 1.8 x 10^2 =2pi x square root of 1.5/180=2pi x .0912870929= 0.5735737208
solutions:
period of mass spring system = 0.574 seconds 
frequency of mass spring system = f = 1/t = 1/0.5735737208 = 1.74345505 = 1.74 hz
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4.  if a grandfather clock is running slow, how can you adjust the length of the pendulum to correct the time.
basically, to correct a pendulum clock that running slow, in researching the simple method is to make the pendulum shorter. the rotation or calibration of the pendulum depends on its gravity. the grandfather clock has a mechanism that you can turn at the bottom clockwise to adjust the pendulum length.
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5.  find the length of a pendulum that oscillates with a frequency of 0.16 hz
solve for the frequency (f)= 1/t= 1/0.16 = 6.25 seconds
 find the pendulum length?  t = 2pi x square root of l/g   transpose equation for l
t^2 = 4 x pi^2 x (l/g)                                              note: acceleration due to gravity = 9.8 m/s^2
l= g x (t^2) divided by 4 x (pi)^2    =     9.8m/s^2 x (6.25 s) ^2    divided   by 4 x pi^2   
l = 9.8 m/s^2   x   39.0625s^2 = 382.815 divided by 39.4784176 = 9.69681723 m         note: m/s^2 x s^2 cancel out
l= 9.7 m     solution: length of pendulum = 9.7 meters
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6.  how are the restoring force and the displacement of an object related in simple harmonic motion?
basically, in simple harmonic motion restoring force is directly proportion to the displacement of an object which always acts toward the equilibrium (balance level) also equal action of opposing forces.
reference: hooke's law theory: stronger the restoring force pulling it back towards equilibrium

Mathwiz4u · Answer

1.  a piano emits frequencies that range from a low of about 28 hz to a high of about 4200 hz. find the range of wavelengths in the air attained by this instrument when the speed of sound in air is 340 m/s.
formula for wavelength is velocity = frequency s wavelength
maximum wavelength = velocity divided by frequency    = 340 m/s divided by 28 hz = 12.14 meters
velocity (speed) = 340 m/s divided by 4200 hz = 340m/s/4200 hz = 0.081 meters
solution: range of wavelengths: range from 0.081 meters to 12.14 meters
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2.  the speed of all electromagnetic waves in empty space is 3.00 x 10^8 m/s.
calculate the wavelength of electromagnetic waves emitted at the following frequencies.
a. radio waves at 88.0 m hz = 3.14 meters
b. visible light at 6.0 x 10^8 m hz = 5.00 x 10^-7 meters
c. x rays at 3.0 x 10^12 m hz = 1.00 x 10^-10 meters
formulas:  
c = frequency x wavelength        c = speed of light      f = frequency       wavelength = lambda symbol 
wavelength = c divided by f
solutions: wavelength for the following:
radio waves wavelength = c divided by frequency = 3.00 x 10^8 m/s divided by 88.0 x 10^6 hz = 3.41 meters
visible light wavelength = c divided by frequency = 3.00 x 10^8 m/s divided by 6.0 x 10^14 hz = 5.00 x 10^-7 meters
x rays' wavelength = c divided by frequency = 3.00 x 10^8 m/s divided by 3.0 x 10^18 hz = 1.00 x 10^-10 meters

Mathwiz4u · Answer

3. what is the fundamental frequency of a 0.20 meters long organ pipe that is closed at one end, when the speed of sound in the pipe is 352 m/s?
valuable information:  length(l) of organ pipe = 0.20 meters      speed of sound in pipe = velocity (v)
fundamental frequency = (f) = f = v divided by 4 x l = 352 m/s divided by 4 x 0.20 meters = 352/0.8 = 440 hz
solution: fundamental frequency of the pipe = 440hz
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4. a flute is essentially the pipe open at both ends. the length of a flute is 0.660 meters. what are the first three harmonics of a flute when all keys are closed? the speed of sound in the flute is 340 m/s.
vital information:  length of flute = l = 0.660 meters           speed of sound of the flute = velocity (v) = 340 m/s
wavelength = 2l divided by n where n = 1,2 and 3     frequency of harmonic = f = velocity divided by wavelength 
wavelength (1) = 2 x l divided by 1 = 2 x 0.660 meters/1 = 1.32 meters
frequency (1) = velocity divided by wavelength = 340 m/s divided by 1.32 meters = 257.5757576 hz = 257.58 hz
wavelength (2) = 2 x l divided by 2 = 2 x 0.660 meters/2 = 0.66 meters
frequency (2) = velocity divided by wavelength = 340 m/s divided by 0.66 meters = 515.1515152 hz = 515.15 hz
wavelength (3) = 2 x l divided by 3 = 2 x 0.660 meters/3 = 0.44 meters
frequency (3) = velocity divided by wavelength = 340 m/s divided by 0.44 meters = 772.7272727 hz = 772.73 hz
solutions: the first (3) harmonics of the flute:
 (1) harmonics =257.58 hz, (2) harmonics = 515.15 and (3) harmonics = 772.73 hz
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5. on a piano, the note middle c has a fundamental frequency of 264 hz.
a. what is the second harmonic of this note?  528 hz
b. if the piano wire is 0.770 meters long, what is the speed of the waves on the wire?
information:  middle c fundamental frequency (formula = f1) = 264 hz
frequency of second harmonic (n=2) f2 = 2 x f1 =2 x 264 hz = 528 hz
f2 = 528 hz   wavelength2 = 0.770 meters       velocity = speed of the waves = f2 x wavelength
 velocity = speed of the waves = f2 x wavelength = 528hz x 0.770 meters = 406.56 m/s
solution:  second harmonic frequency = 528 hz and wave speed = 406.56 m/s

Mathwiz4u · Answer

1. at a maximum level of loudness, the power output of a 75-piece orchestra radiated as sound is 70.0 w. what is the intensity of these sound to a listener who is sitting 25.0 m from orchestra?
power output of the orchestra: p = 70 w   distance from the orchestra: r = 25.0 meters
intensity of sound = i = p/a where p is the power   formula intensity (i) = p/4 x pi x r^2
a = 4 x pi x r^2 = 4 s pi x (25.0 m) ^2 = 4 x pi x (625 m^2) = 2500 x pi m^2
intensity of sound i = p/a = 70.0w divided by 2500 x pi m^2 = 70.0 w divided by 7853.981634 m^2 = .0089126768 w/m^2
i = 0.0089 w/m^2     
solution:  intensity of the sound waves at 25.0 from the orchestra = 0.0089 w/m^2
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2.  if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 meters, how much sound power does that person generate?
intensity = 4.6 x 10^-7 w/m^2            distance: r = 2.0 meters   i = p/a        p = i x a (r=2.0m)
a = 4 x pi (2.0 m) ^2 = 4 x pi (4.0 m^2) = 4 x pi x (4.0 m^2) = 16 x pi m^2 = 50.26548246 = 50.27 m^2
i = 4.6 x 10^-7 w/m^2     and      a = 16 pi m^2
p = i x a = (4.6 x 10^-7 w/m^2) x (16 x pi m^2) = 4.6 x 16 = 73.6 x pi = 231.22 x 10^-7 = 2.3122 x 10^-5 w
solution: sound power generated by the person = 2.3 x 10^-5 w
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3. the energy output of a tuba is 1.05 joules of energy every 3.0 seconds. at what distance is the sound intensity of the tuba. sound intensity = 1.20 x 10^-3 w/m^2?
energy output = e = 1.05 joules        time interval: t = 3.00 seconds       sound intensity = i = 1.20 x 10^-3 w/m^2

p = e/t = 1.05 joules divided by 3.00 seconds = 0.35 w 
formula i = p divided by 4 x pi x r^2
r = distance = square root of p divided by 4 x pi x intensity = square root of 0.35 w/4 x pi x (1.20 x 10^-3 w/m^2) =
= square root of .35 /.0150796447 = 4.817685744 = 4.817685744 = 4.82 meters = distance from the tuba
solution:(distance from the tuba) = 4.82 meters
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