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3 years ago

8

If the intensity level by 15 identical engines in a garage is 100 dB, what is the intensity level generated by each one of these

engines?

1 answer:

goblinko [34]3 years ago

8 0

To develop this problem we will apply the concepts related to Sound Intensity. This can be defined as the power carried by sound waves per unit area in a direction perpendicular to that area. Mathematically it can be written as,

$\Delta L = 10log(n)$

$\Delta L$ = Level Difference

n = Number of equal loud sound sources

n = 15

Therefore replacing we have,

$\Delta L = 10 log (n)\\\Delta L = 10 log (15)\\\Delta L = 10*1.176\\\Delta L = 11.76dB$

Therefore over 100dB the total difference intensity level would be

$L_T = L_0 -\Delta L$

Here,

$L_0 = \text{ Total intensity}$

$\L_T = 100 -11.76$

$_T= 88.24dB$

The intensity level generated by one engine is

$L_T = 88.44dB$

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Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the

Hoochie [10]

You've told us:

-- 130°x = 212°F

and

-- 10°x = 32°F

Thank you. Those are two points on a graph of °x vs °F . With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are (130, 212) and (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation. I'll use the second point (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are ! Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad. Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for x, and the solution will be the same temperature in F too.

or

We can write [ F = 1.5F + 17 ], solve it for F, and the solution will be the same temperature in x too.

F = 1.5F + 17

Subtract F from each side: 0.5F + 17 = 0

Subtract 17 from each side: 0.5F = -17

Multiply each side by 2 : F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for x:

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works ! -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0

3 years ago

A coil of wire that is carrying a current and produces a magnetic field is.

Akimi4 [234]

A solenoid hope this is right

5 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

What is the ratio of a 0.7 in circle in centimeters.

kifflom [539]

I believe it would be 4.4

4 0

3 years ago

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of

Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

$r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$

m is mass of Singly charged uranium-238 ion, $m=3.95\times 10^{-25}\ kg$

q is charge

So,

$r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m$

So, the radius of their circular path is equal to 0.0548 m.

4 0

3 years ago