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poizon [28]
3 years ago
11

A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source,

the intensity of the sound is 5.0 × 10−3 W/m2. What is the total sound power P emitted by the source?
Physics
1 answer:
Allushta [10]3 years ago
3 0

To solve this problem we will apply the concept related to Intensity, that is, the acoustic power transferred by a sound wave per unit of normal area to the direction of propagation:

I = \frac{P}{A} \rightarrow P= AI

Where,

P = Power

A = Acoustic Area

Our values are given as,

r= 12 m

I = 4.3*10^{-3} W/m^2

The Area then would be

A = 4\pi r^2

A = 4\pi (12)^2

A = 1809.55m^2

Replacing the values we have that

P = (4.3*10^{-3} W/m 2)(1809.55m^2 )

P = 7.781 W

The total sound power P emitted by the source is 7.781W

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Which of the following accurately describe some aspect of gravitational waves? Select all that apply.
steposvetlana [31]
<h2>Answers:</h2>

-The first direct detection of gravitational waves came in 2015

-The existence of gravitational waves is predicted by Einstein's general theory of relativity

-Gravitational waves carry energy away from their sources of emission

<h2>Explanation:</h2>

Gravitational waves were discovered (theoretically) by Albert Einstein in 1916 and "observed" for the first time in direct form in 2015 (although the results were published in 2016).  

These gravitational waves are fluctuations or disturbances of space-time produced by a massive accelerated body, modifying the distances and the dimensions of objects in an imperceptible way.  

In this context, an excellent example is the system of two neutron stars that orbit high speeds, producing a deformation that propagates like a wave,<u> in the same way as when a stone is thrown into the water</u>. So, in this sense, gravitational waves carry energy away from their sources .

Therefore, the correct options are D, E and F.

5 0
3 years ago
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the
Eduardwww [97]

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3  m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g  

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0
3 years ago
A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
GuDViN [60]

Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=k \epsilon_0 \frac{A}{d}

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

4 0
3 years ago
A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the
Lelechka [254]

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0
3 years ago
A pelican flying along a horizontal path drops
Ludmilka [50]

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

8 0
3 years ago
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