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Sloan [31]
3 years ago
5

What decibel reading corresponds to a pressure amplitude of 0.2 W/m^2?

Physics
1 answer:
horsena [70]3 years ago
4 0

I  = pressure amplitude given = 0.2 W/m²

dB = decibel reading

decibel reading from the pressure amplitude is given as

dB = 10 log₁₀ (I/10⁻¹²)

inserting the values in the above equation

dB = 10 log₁₀ (0.2/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹.10¹²)

dB = 10 log₁₀ (2 x 10¹²⁻¹)

dB = 10 log₁₀ (2 x 10¹¹)

dB = 113.01 db

hence the decibel reading comes out to be 113.01 db


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asambeis [7]

Answer:

$ 0.48

Explanation:

We can calculate this quantity easily using successive products and taking into account the units.

\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48

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3 0
3 years ago
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

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TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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Answer:

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The electromagnetic force is one of the four fundamental forces.

The electromagnetic force is effective at long range distance.  

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Answer:

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