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Sveta_85 [38]
3 years ago
15

The sound intensity level from one solo flute is 70 dB. If 10 flutists standing close together play in unison, what will the sou

nd intensity level be?
Physics
1 answer:
Masja [62]3 years ago
5 0

Answer:

80 dB

Explanation:

I_0 = Threshold intensity = 10^{-12}\ W/m^2

I = Intensity of sound

\beta = Intensity level of sound = 70 dB

Intensity level of sound is given by

\beta=10log\dfrac{I}{I_0}\\\Rightarrow 70=10log\dfrac{I}{I_0}\\\Rightarrow \dfrac{70}{10}=log\dfrac{I}{10^{-12}}\\\Rightarrow 10^{\dfrac{70}{10}}=\dfrac{I}{10^{-12}}\\\Rightarrow \dfrac{I}{10^{-12}}=10^{\dfrac{70}{10}}\\\Rightarrow I=10^{7}\times 10^{-12}\\\Rightarrow I=10^{-5}\ W/m^2

If there are 10 flutes I=10\times 10^{-5}\ W/m^2

\beta=10log\dfrac{10\times 10^{-5}}{10^{-12}}\\\Rightarrow \beta=10log10^8\\\Rightarrow \beta=10\times 8\\\Rightarrow \beta=80\ dB

The sound intensity level is 80 dB

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igomit [66]

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

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3 years ago
Congugar verbos en primera persona del singular
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This is Spanish !!!!
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A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

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At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di
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Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

F_B = qv\times B

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

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So, we can solve for T from (1):

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Replacing (3) in (2):

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The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

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According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

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