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erastova [34]
3 years ago
9

Sound source A has a deciBel rating of 50 dB. Sound source B is 1000 times more intense. What is the deciBel rating of B

Physics
1 answer:
masha68 [24]3 years ago
3 0

The decibel system of sound intensity operates by a logarithmic scale, meaning that sound intensity increases exponentially in relation to the decibel rating.

For decibels, the equation between intensity and the dB equivalent is:

dB = 10log(i),

where “i” is the intensity of the sound. The ten in front of the log means that an increase in ten dB results in a tenfold increase in sound intensity; for example, a 30 dB sound is ten times softer than a 40 dB sound.

In this case, a sound with a dB of 80 would be 1000 times more intense than a 50 dB sound, so the decibel rating of B is 80.

Hope this helps!

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Instructions:Select all the correct answers.
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The correct answer is 1 and 4.

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

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Hi!


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~CoCo

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Fundamental frequency
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