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Dvinal [7]
3 years ago
12

An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most ra

pidly. But the stars are all v_________________.
Physics
1 answer:
sp2606 [1]3 years ago
4 0

An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most rapidly. But the stars are all very far and hence he can not identify the starts that spins more rapidly.

<u>Explanation:</u>

The things in the world rotates. the earth, sun star everything will be spinning. There is a star named Achenar. this is the start that spins at a faster rate. It is the biggest star that stays at the position 10. Its mass is 7 times greater than the sun's mass.The shining of the star can be measured by luminosity. It is the measure of the brightness of the star.

The energy that is given by a star in a second of time is the luminosity of the star. All stars are present in the different distances form the surface of the earth. but when we look from the earth or even when when go at certain altitude we can find all stars to be located at the same distance. hence if we need to find the star that spins more rapidly, then the star spectral lines can be observed. The stars that rotates will be having the wide lines in the spectra.

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3 years ago
A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

        B  =  \frac{\mu_o  * N  *  I}{2 r }

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

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The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, v, is \frac 12 m v^2

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So, the total energy of ball1,

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The total energy of ball1 hi higher than the total energy of ball2.

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