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mariarad [96]
3 years ago
7

If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev

el. A. 1.6 W
B. 0.06 W
C. 97 dB
D. 223 dB
E. 179 dB
Physics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

(c) 97 dB sound intensity level

Explanation:

We have given the intensity of the loud car horn I=0.005w/m^2

We know that I_O=10^{-12}w/m^2

Now the sound intensity level is given by \beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

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Answer:

600,000,000 degree C

Explanation:

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A car of mass 1500 kg is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s.
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Answer:

Explanation:

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d. μ = Fc/N = Fc/mg = 12000 / 1500(9.8) = 0.8163... ≈ 0.82

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What holds the moon in place, orbiting around the Earth
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a horse moves a sleigh 1.00 kilometers by applying a horizontal 2000 Newton force on its harness for 45 minutes. what is the pow
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--  If 2,000 newtons of force were applied through a distance of 1,000 meters,
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An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro
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Answer:

a) Q_{in} = 13.742\,kW, b) \Delta S = 370.15\,\frac{kJ}{K}

Explanation:

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Q_{in} +U_{sys,1} - U_{sys,2} = 0

Q_{in} = U_{sys,2} - U_{sys,1}

Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})

Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)

Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

3 0
3 years ago
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