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Sliva [168]
1 year ago
12

the threshold of hearing is defined as the minimum discernible intensity of the sound. it is approximately 10−12w/m2 . find the

distance d from the car at which the sound from the stereo can still be discerned. assume that the windows are rolled down and that each speaker actually produces 0.06 w of sound, as suggested in the last follow-up comment.
Physics
1 answer:
vivado [14]1 year ago
3 0

The distance is 97720.5 m

From the question, we have

P = 0.06 W × 2 = 0.12 W

d = ?

Sound intensity, I = P/4πd²

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

The distance is 97720.5 m

Sound intensity :

The power carried by sound waves per unit area in the direction  perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

To learn more about Sound intensity visit: brainly.com/question/12899113

#SPJ4

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Physicist Max Planck showed how objects like stars give off different colors based on their temperature. What color are the hott
jenyasd209 [6]

Answer:

the brightest found are Blue - White with

Explanation:

The energy emission of objects increases with their temperature, specifically Wien described the process in an expression

 

          \lambda_{maximum} T = 2,898 10⁻³

With this expression we can find the temperature of the stars by the color they emit.

Specifically the Sun has a color of 550 nm which corresponds to 5400K

bright stars have a BLUE color corresponding to 7500K

the brightest found are Blue - White with a temperature of 20000K

7 0
3 years ago
A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-
kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}

Here,

\Delta L= is the distance moved by the mirror M

\lambda is the wavelenght of the light used.

\Delta L= 0.017m

\lambda = 656.45*10^{-9}m

\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}

\Delta m = 51793.72

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

3 0
2 years ago
The woman in the first photo lifts a plant 2 m
Serjik [45]

Answer:

65x2=130

that's what she does

8 0
3 years ago
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
2 years ago
If a car go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/
Taya2010 [7]

Answer:

v = 87.57 m/s

Explanation:

Given,

The initial velocity of the car, u = 0

The final velocity of the car, v = 60 mi/hr

The time period of car, t = 8 s

                                         = 0.00222 hr

The acceleration of the car is given by the formula,

                                       a = (v -u) / t

                                           = 60 / 0.00222

                                            = 27027 mi/hr²

If the car has initial velocity, u = 50 mi/hr

The time period of the car, t = 5.0 s

                                         = 0.00139 hr

Using first equations of motion

                                      <em> v = u + at</em>

                                          = 50 + (0.00139 x 27027)

                                          = 87.57 mi/hr

Hence, the final velocity of the car, v = 87.57 mi/hr

3 0
3 years ago
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