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vaieri [72.5K]
3 years ago
9

Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If

it were to undergo gravitational collapse to a neutron star of radius 15 km, losing 3/4 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times
Physics
1 answer:
Tanya [424]3 years ago
4 0
Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end 
Conservation of angular momentum: 
I1 w1 = I2 w2 
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get 
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2) 
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
= 3.48E10 revs/(17days)
= 2.04705882 x 10^9 revs/sec
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A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

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