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2 years ago

The interference of two sound waves of slightly different frequencies produces

2 answers:

7 0

The interference produces 'beats' ... two new sounds
in addition to the original ones.

The frequencies of the beats are

-- the sum of the two original frequencies
and
-- the difference of the two original frequencies .

Technically, there are actually even more new sounds than that.
There are:
-- beats between the beats, and
-- beats between each beat and each of the original two sounds.
But their amplitudes are so low that they can be detected only with
sensitive equipment, and they can't be heard in the presence of
the two original sounds and the two 'first order' beats.

Leviafan [203]2 years ago

4 0

Beats.

When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats. The beat frequency is equal to the absolute value of the difference in frequency of the two waves.

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A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and

kicyunya [14]

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

3 0

2 years ago

an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei

wel

If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

1: √2 : 1

<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

Learn more about the radii of their circular paths here:

brainly.com/question/16816166

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6 0

1 year ago

Provides most of the energy used in the world today.

lara [203]

fossil fuels is used the most often in the world.

6 0

2 years ago

A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0

3 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

1 year ago