PHYSICS COLLEGE

An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line

Answers

Answer 1

Answer:

Answer:

2.72 m

Explanation:

wavelength of sound λ = velocity / frequency

= 340 / 1200

= .2833 m

Distance of point of first constructive interference

= λ D / d ( D is distance of the screen and d is distance between source of sound.

Here D = 12.5 m

d = 1.3 m

λ D / d= ( .2833 x 12.5) / 1.3

= 2.72 m

Distance of point of first constructive interference = 2.72 m

Answer 2

Answer:

Final answer:

The wavelength of the produced sound is approximately 0.29 m. Constructive interference occurs when the path difference between the two waves is a multiple of this wavelength, allowing you to calculate the distance between the central maximum and first maximum loud position.

Explanation:

For part (a) of the question, we need to calculate the wavelength of the sound wave. The wave speed (v) is given by the multiplication of frequency (f) and wavelength (λ). The speed of sound in air is approximately 343 m/s and given that the frequency produced by the function generator is 1200 Hz, the wavelength can be calculated using the formula λ = v / f = 343 / 1200 ≈ 0.29 m.

For part (b) the distance between the central maximum (loud) position and the first maximum along this line requires understanding of sound wave interference and constructive interference. For constructive interference to occur, the path difference between the two waves needs to be a multiple of the wavelength. Thus, in the first constructive interference position (first maximum loud position), the path difference equals one wavelength (0.29m). Since the student is walking 12.5 m away and parallel to the line between the speakers (which is the hypotenuse of a right triangle stakeout, with one side being 0.65m), we can use Pythagorean theorem to find out the distance.

Learn more about Sound Waves Interference here:

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A train travels 64 kilometers in 5hours and then 93 kilometers in 2hours. What is it’s average speed?

Answers

I believe the answer is about 22.43 kilometers per hour. However I am not 100% sure.

How I got this: first you add 64 and 93, then 5 and 2. That would leave you with the kilometers traveled (157) over the number of hours (7). You’d have to divide, which would leave you with an estimated 22.4285. Round to the nearest hundredth and you get 22.43 . Please correct me if I’m wrong!

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound

Answers

Answer:

about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

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Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =(k*q)/(d)$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = (8.99e9N*m2/C2*(4e-6C))/(8m) + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.