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sergij07 [2.7K]

3 years ago

10

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

aves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s.
(A) If you are 3.00 m from speaker A directly to your right and 3.50 m from speaker B directly to your left, will the sound that you hear be louder than the sound you would hear if only one speaker were in use?

1 answer:

WINSTONCH [101]3 years ago

6 0

Answer:

Explanation:

frequency of sound waves = 688 Hz

wavelength = 344 / 688 = .5 m

The problem is based on interference of sound waves

For the observer , path difference of sound waves reaching his ear

= 3.5 - 3.00

.5 m

= wavelength

Path difference is equal to wavelength so there will be constructive interference and hence louder sound will be heard by the listener than normal sound as sound waves interfere constructively.

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2 years ago

A 0.8 g object is placed in a 534 N/C uniform electric field. Upon being released from rest, it moves 12 m in 1.2 s. Determine t

Mice21 [21]

Explanation:

It is given that,

Mass of the object, m = 0.8 g = 0.0008 kg

Electric field, E = 534 N/C

Distance, s = 12 m

Time, t = 1.2 s

We need to find the acceleration of the object. It can be solved as :

m a = q E.......(1)

m = mass of electron

a = acceleration

q = charge on electron

"a" can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}at^2$

$a=\dfrac{2s}{t^2}$

$a=\dfrac{2\times 12\ m}{(1.2\ s)^2}$

a = 16.67 m/s²

Now put the value of a in equation (1) as :

$q=\dfrac{ma}{E}$

$q=\dfrac{0.0008\ kg\times 16.67\ m/s^2}{534\ N/C}$

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2 years ago

An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s

ad-work [718]

Answer:

a

$\theta = 0.0022 rad$

b

$I = 0.000304 I_o$

Explanation:

From the question we are told that

The wavelength of the light is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

The distance of the slit separation is $d = 0.500 \ mm = 5.0 *10^{-4} \ m$

Generally the condition for two slit interference is

$dsin \theta = m \lambda$

Where m is the order which is given from the question as m = 2

=> $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

substituting values

$\theta = 0.0022 rad$

Now on the second question

The distance of separation of the slit is

$d = 0.300 \ mm = 3.0 *10^{-4} \ m$

The intensity at the the angular position in part "a" is mathematically evaluated as

$I = I_o [\frac{sin \beta}{\beta} ]^2$

Where $\beta$ is mathematically evaluated as

$\beta = \frac{\pi * d * sin(\theta )}{\lambda }$

substituting values

$\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }$

$\beta = 0.06581$

So the intensity is

$I = I_o [\frac{sin (0.06581)}{0.06581} ]^2$

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2 years ago

What's the momentum (in kg-m/s) of a cannonball with a mass of 21 moving with a velocity of 25 m/s?

lbvjy [14]

P=mv
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8 0

3 years ago

the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m

olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

$x = \frac {-nmg}{k}$

Substituting into the formula, we have;

$x = \frac {-49*0.002*9.8}{24}$

$x = \frac {-0.9604}{24}$

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

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2 years ago