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stellarik [79]
3 years ago
5

All of the celestial bodies of the solar system were said to have formed from -a nebular cloud of dust and gas -the protostar -d

ark matter -nuclear fusion
Physics
2 answers:
Kitty [74]3 years ago
5 0
The answer is a. It cant
be any other one.
Bumek [7]3 years ago
4 0
A : A nebular cloud of dust and gas is the correct answer.
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As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, b
katovenus [111]

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

<h3>What do you mean by sound radiates?</h3>

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level,  \beta= 101 dB

Area, A = 22\;m^{2}

Time, \triangle t = 20\;min=1200\;s

Intensity, I=1\times 10^{-12}\;W/m^{2}

r=1\;km=1000\;m

(a)

We know that, Sound level is,

\beta=10\times log(\frac{I}{I_{o} } )

Solving the above equation for sound intensity,

I=I_{o} \times 10^{\frac{\beta}{10} }

I=1 \times 10^{-12}  \times 10^{\frac{101}{10} }

I=0.0126\;W/m^{2}

Therefore, The sound energy is,

E=P\times \triangle t

Substitute P=I \times A in the above equation,

E=I \times A \times \triangle t

E=0.0126 \times 22 \times 1200

E=332.6\;J

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

A_{hemisphere}  = \frac{1}{2} \times 4 r^{2} \pi

Substitute the known value in the above equation ,

A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi

A_{hemisphere} = 6283185\;m^{2}

Sound Intensity is,

I = \frac{P}{A_{hemisphere}}

Substitute P=I \times A in the above equation,

I = \frac{I \times A}{A_{hemisphere}}

Substitute the known value in the above equation,

I = \frac{0.0126 \times 22}{6283185}

I = 4.4 \times 10^{-8}\;W/m^{2}

Sound level is,

\beta=10\times log(\frac{I}{I_{o} } )

Substitute the known value in the above equation,

\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )

\beta=46.4\;dB

To learn more about sound radiates, Visit:

brainly.com/question/20360072

#SPJ4

8 0
1 year ago
A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre
disa [49]

Answer:

I =  26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x  π x 49

ω = 307.72 rad/s

As we know that voltage given as

V= Vo sinω t

V_o=\sqrt2\ V_{rms}

V_o=\sqrt2\ \times 82\

Vo=115.96 V

V=115.96 sinω t

The current given as

I=C\dfrac{dV}{dt}

I=0.74\times \dfrac{dV}{dt}\ mA

\dfrac{dV}{dt}=115.96\omega cos\omega t

I=0.74\times 115.96\times 307.22 cos\omega t\ mA

I = 26362.67 cosω t mA

I =  26.36 cosω t A

This is the current at time ant time t.

7 0
3 years ago
The scientific unit used to measure distance is the?
alexgriva [62]

Answer:

Meter.

Explanation:

4 0
3 years ago
What is the formula to find the second charge in Coulomb's Law?
kicyunya [14]

Answer:

1.F is the electrostatic force between charges (in Newtons),

2.q₁ is the magnitude of the first charge (in Coulombs),

3.q₂ is the magnitude of the second charge (in Coulombs),

4.r is the shortest distance between the charges (in m),

5.ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C² .

5 0
2 years ago
Read 2 more answers
A complete path through which charge can flow is an electric what?
grigory [225]
<span>A complete path through which charge can flow is an "Electric Circuit"

Hope this helps!</span>
5 0
3 years ago
Read 2 more answers
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