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Paraphin [41]
2 years ago
9

The low-frequency speaker of a stereo set produces 10.0 W of acoustical power. If the speaker projects sound uniformly in all di

rections, at what distance from the speaker is the intensity level 80.0 dB
Physics
1 answer:
Paha777 [63]2 years ago
6 0

Answer:

the required distance is 89.125 m

Explanation:

Given the data in the question;

we know that, sound intensity B in decibels of sound is;

β(dB) = 10log₁₀( I / I_0 )

where intensity I = power / area carried by wave

I_0 = 10⁻¹² W/m² { minimum threshold intensity }

Now,

intensity I = power / area carried by wave = P/A = P/4πr² { spherical  }

given that; β = 80.0 dB and P = 10 W

so

β(dB) = 10log₁₀( I / I_0 )

we substitute

80 = 10log₁₀( P / 4πr²× I_0)

80 = 10log₁₀( 10 / 4πr²× 10⁻¹² )

8 = log₁₀(10) - log₁₀( 4πr²× 10⁻¹² )  

8 = 1 - log₁₀( 4πr²× 10⁻¹² )

8 - 1 = -log₁₀( 4πr²× 10⁻¹² )

7 = -log₁₀( 1.2566 × 10⁻¹¹ × r² )

7 = -[ log₁₀( 1.25 × 10⁻¹¹) + log₁₀( r² ) ]

7 = -[ -10.9 + log₁₀( r² ) ]

7 = 10.9 - log₁₀( r² )

-log₁₀( r² ) = 7 - 10.9

-log₁₀( r² )  = - 3.9

log₁₀( r² ) = 3.9

2log₁₀r = 3.9

log₁₀r  = 3.9 /2

log₁₀r = 1.95

r = 89.125 m

Therefore, the required distance is 89.125 m

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harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

\displaystyle P=\frac{W}{t}

Since W=F.d

\displaystyle P=\frac{F.d}{t}

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

\displaystyle P=\frac{(75). (4)}{10}=30\ Watt

The second option has a lower power output

7 0
3 years ago
In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u

Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

Object distance is 9 cm

v=-4\times 9=-36\ cm

Image distance is -36 cm (other side of object)

7 0
3 years ago
Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
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Answer:

Answer is

A. I = 6.3×10^8 A

B. Yes

C. No

Refer below.

Explanation:

Refer to the picture for brief explanation.

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