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stira [4]
3 years ago
14

Which of these would best define sound intensity? A) how fast the wave is traveling B) the frequency of the sound wave the power

of the sound wave per unit area the mass of the wave divided by its volume
Physics
2 answers:
faust18 [17]3 years ago
5 0

Which of these would best define sound intensity?

A) how fast the wave is travelling

<u>B) the frequency of the sound wave the power of the sound wave per unit area the mass of the wave divided by its volume</u>

STatiana [176]3 years ago
4 0

The correct answer is B. Hope this helps! :)

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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
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Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

       T  =  43.302 \ N

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3 years ago
Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta
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