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LUCKY_DIMON [66]
3 years ago
11

A straight 2.20 m wire carries a typical household current of 1.50 A (in one direction) at a location where the earth's magnetic

field is 0.550 gauss from south to north.. . Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east.
Physics
1 answer:
Snowcat [4.5K]3 years ago
7 0
Use F = I*B*L*sinθ 

<span>I is the current, B is the magnitude of the magnetic field (Note: 1 gauss is equal to 1E-4 Tesla, the SI unit), and L is the length of wire. Adjust θ for each separate orientation.

All you need to do is just plug the given values to the equation. You should receive the same answer for parts a and b.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

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