1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Eduardwww [97]
3 years ago
11

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond

to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75m wide and 1.5m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.A) What are the lowest two frequencies that correspond to resonances on the short axis?B) What are the lowest two frequencies that correspond to resonances on the longer axis?
Physics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

Explanation:

For first overtone

Standing waves will be formed lengthwise and breadth-wise in the enclosures  having dimension of .75m  x 1.5 m

A ) For the formation of lowest two frequencies formed by standing waves along the breadth  , fundamental note and first overtone may be considered.  

For fundamental note ,  

the condition is  

wave length λ = 2L = 2 x 0.75 m  

λ = 1.5 m  

frequency n = v / λ

= 343 / 1.5  

= 229  Hz approx

For first overtone

λ = L = 0.75m

frequency n = v / λ

n = 343 / 0.75  

= 457 Hz approx

B)

For the formation of lowest two frequencies formed by standing waves along the length , fundamental note and first overtone may be considered.  

For fundamental note ,  

the condition is  

wave length λ = 2L = 2 x 1.5 m  

λ = 3 m  

frequency n = v / λ

= 343 / 3  

= 114 Hz approx

frequency n = v / λ

n = 343 / 1.5  

= 229 Hz approx

You might be interested in
9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)
konstantin123 [22]
<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>
6 0
3 years ago
Read 2 more answers
a uniform rule balance on a knife edge at the 55cm mark when a mass of 40g is hung from the 95cm mark. find the weight of the ru
BigorU [14]

Answer:

W = 3.1 N

Explanation:

moments about any convenient point will sum to zero.

I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

I will assume the ruler weight makes a positive moment

W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0

                                                5W = 15.68

                                                   W = 3.136

7 0
2 years ago
The speed of light in vacuum is defined to be c = 299,792,458 m/s = 1 μ0ε0 . The permeability constant of vacuum is defined to b
Radda [10]

Explanation:

It is given that,

The speed of light in vacuum is, c = 299,792,458 m/s

The permeability constant of vacuum is, \mu_o=4\pi\times 10^{-7}\ N.s^2/C^2

Let \epsilon_o is the permittivity of free space. The relation between \mu_o,\epsilon_o\ and\ c is given by :

c=\dfrac{1}{\sqrt{\mu_o\epsilon_o}}

\epsilon_o=\dfrac{1}{c^2u_o}

\epsilon_o=\dfrac{1}{(299792458\ m/s)^2\times 4\pi\times 10^{-7}\ N.s^2/C^2}

\epsilon_o=8.85\times 10^{-12}\ C^2/N.m^2

Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
bezimeni [28]

solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150


5 0
3 years ago
Read 2 more answers
The following are connected in parallel circuit at home EXCEPT:
nikdorinn [45]
Light bulbs, think about parallel circuits as something you plug into, so for example you plug in a tv for it to work, same with a refrigerator and Christmas lights
8 0
2 years ago
Other questions:
  • We humans are on track to increase the amount of CO2 in the atmosphere so that the concentration in the future more than double
    12·1 answer
  • A 4kg mass traveling eastwards at 4m.s per second collides with a 3kg mass traveling westward as 8m.s per second..calculate the
    8·1 answer
  • Why vacuum flask is known as thermos flak?
    15·1 answer
  • The Mariana trench is located in the Pacific Ocean at a depth of about 11 000 m below the surface of the water. The density of s
    9·1 answer
  • Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co
    9·1 answer
  • PLEASE HELP ME TIMED TEST!!!
    10·2 answers
  • -
    14·1 answer
  • Which is a TRUE statement about a series circuit
    8·1 answer
  • What are the symptoms of hepatitis 'b'​
    6·1 answer
  • Two 20kg spheres are placed with their
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!