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3 years ago

Explain how sound is being produced from this 'Stomp' group.

Physics

1 answer:

Ira Lisetskai [31]3 years ago

6 0

Answer:

Sound is produced when an object vibrates, creating a pressure wave. This pressure wave causes particles in the surrounding medium (air, water, or solid) to have vibrational motion. As the particles vibrate, they move nearby particles, transmitting the sound further through the medium. The human ear detects sound waves when vibrating air particles vibrate small parts within the ear.

In many ways, sound waves are similar to light waves. They both originate from a definite source and can be distributed or scattered using various means. Unlike light, sound waves can only travel through a medium, such as air, glass, or metal. This means there’s no sound in space!

Explanation:

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3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La

jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

$f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}$

$f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}$ ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

$f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz$

Hence, the first harmonic has a frequency of 12.90 Hz

4 0

4 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

5 0

3 years ago

How can you determine the amount of work done on an object? (joules, work, power...etc.)

Novay_Z [31]

Work done is equal to force by distance; so you take the force exerted, in newtons, and multiply that by the direction it's moved (from the starting point in a line, not along the path it's taken.)

7 0

3 years ago

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

Trava [24]

Answer:

- path differnce = 2.18*10^-6

- 1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

$path\ difference\ = dsin\theta$ (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

$path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m$

- The maximum number of bright lines are calculated by using the following formula:

$m\lambda = dsin\theta$ (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

$m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769$

The maxium number of bright lines are twice the previous result, that is, 1538 lines

8 0

3 years ago

Read 2 more answers

The distance from the Earth to the Sun is known as an astronomical unit (AU). This distance is also equal to 1.496e+011 m. Neptu

Lerok [7]

Answer:

The distance from the Sun to Neptune is 29,41 AU.

Explanation:

We know, from the sentence, that the orbit of Neptune has an average diameter around 8.80*10⁹km.

Now, we can calculate the radius of this orbit, which is equivalent to the distance from thsi planet to the Sun. Let's recall tha the radius is the half of the diameter.

$R=\frac{8.80\cdot 10^{9}}{2}=4.4\cdot 10^{9}km=4.4\cdot 10^{12}m$

Ok, we know that 1.496*10¹¹m is an AU, therefore we have:

$4.4\cdot 10^{12}m\cdot \frac{1 AU}{1.496\cdot 10^{11}}=29,41 AU$

Finally, the distance R is 29,41 AU.

I hope it helps you! :)

7 0

3 years ago