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3 years ago

At a country music festival, a band is playing at the end of a crowded

Physics

1 answer:

Anna11 [10]3 years ago

4 0

Answer:

t=4.86s

Explanation:

To find the wavelength you use the following formula:

$v=\lambda f$

v: speed of sound = 343m/s

f: frequency = 400Hz

λ: wavelength of the sound

By doing λ the subject of the formula and replacing the values of f and v you obtain:

$\lambda=\frac{v}{f}=\frac{343m/s}{400Hz}=0.85m$

Now, to calculate the time that sound takes to reach the last row you use:

$t=\frac{d}{v}$

t: time

d: distance to the last row = 1947m

$t=\frac{1947m}{400Hz}=4.86s$

hence, the time is 4.86s

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3 years ago

In what way could a random mutation provide an organism with an advantage? With a example please

saveliy_v [14]

Answer:

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3 years ago

the maximum intensity levels of a trumpet, trombone, and a bass drum, each at a distance of 3m are 94 dB, 107dB, and 113dB respe

Gwar [14]

Answer:

β = 114 db

Explanation:

The intensity of sound in decibles is

β = 10 log $\frac{I}{I_{o}}$

in most cases Io is the hearing threshold 1 10-12 W / cm²

let's calculate the intensity of each instrument

I / I₀ = 10 (β / 10)

I = I₀ 10 (β / 10)

trumpet

I1 = 1 10⁻¹² 10 (94/10)

I1 = 2.51 10⁻³ / cm²

Thrombus

I2 = 1 10⁻¹² 10 (107/10)

I2 = 5.01 10-2 W / cm²

low

I3 =1 1-12 (113/10) W/cm²

I3 = 1,995 10-1 W / cm²

when we place the three instruments together their sounds reinforce

I_total = I₁ + I₂ + I₃

I_ttoal = 2.51 10-3 + 5.01 10-2 + 1.995 10-1

I_total = 0.00251 + 0.0501 + 0.1995

I_total = 0.25211 W / cm²

let's bring this amount to the SI system

β = 10 log (0.25211 / 1 10⁻¹²)

β = 114 db

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3 years ago

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3 years ago

Read 2 more answers

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago