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kondaur [170]
3 years ago
15

Including them, the mass of the balloon was 1890 kg and had a volume of 11,430 m3 . The balloon floats at a constant height of 6

.25m above the ground. What is the density of the hot air in the balloon
Physics
1 answer:
Monica [59]3 years ago
4 0

Given :

The mass of the balloon was 1890 kg and had a volume of 11,430 m3 .

The balloon floats at a constant height of 6.25m above the ground.

To Find :

The density of the hot air in the balloon.

Solution :

We know,

Volume × ( Density of surrounding air - Density of hot air ) = mass

Putting given values in above equation, we get :

11430\times ( 1.29 - \rho_{hot \ air } ) = 1890\\\\\rho_{hot \ air } = 1.29 - \dfrac{1890}{11430}\\\\\rho_{hot \ air } =  1.125\ kg\ m^3

Therefore, the density of hot air in the balloon is 1.125 kg m³.

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A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that incre
olga nikolaevna [1]

Answer:

The  induced current is I  =  6.25*10^{-4} \  A

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    The number of turns is  N  =  1

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    The  initial magnetic field is  B_i  =  0.500 \ T

     The  magnetic field at time =  1.02 s  is  B_t =  2.60 \ T

     The  resistance is  R  = 2.70\  \Omega

The  induced emf is mathematically represented as

       \epsilon  = - N  *   \frac{ d\phi }{dt}

The  negative sign tells us that the induced emf is moving opposite to the change in magnetic flux

      Here  d\phi is the change in magnetic flux which is mathematically represented as

        d \phi  =  dB  *  A

Where  dB  is the change in magnetic field which is mathematically represented as

        dB  =  B_t  - B_i

substituting values

        dB  =   2.60 -  0.500

        dB  =   2.1 \ T

Thus  

      d \phi  =  2.1 * 8.20 *10^{-4}

     d \phi  = 1.722*10^{-3} \ weber

So  

     |\epsilon|  = 1   *   \frac{ 1.722*10^{-3}}{1.02}

     |\epsilon|  = 1.69 *10^{-3} \  V

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      I  =  \frac{\epsilon}{ R }

  substituting values

       I  =  \frac{1.69*10^{-3}}{ 2.70 }

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