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3 years ago

5

Certain neutron stars (extremely dense stars) are believed to be rotating at about 1000 rev/s. If such a star has a radius of 14

km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation

1 answer:

Sever21 [200]3 years ago

8 0

Answer:

minimum mass of the neutron star = 1.624 × 10^30 kg

Explanation:

For a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.

This can be represented by the explanations in the attached document.

minimum mass of the neutron star = 1.624 × 10^30 kg

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a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !

The formula is D = 1/2 a T²

   Distance = (1/2 acceleration) x (time in seconds)²

   Acceleration = (change in speed) / (time for the change)
    = (18 m/s) / (12 sec)
    = 1.5 m/s² .

Distance = (1/2 x 1.5 m/s²) x (12 sec)²
      =     (0.75 m/s²) x (144 sec²) = 108 meters .

5 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

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3 years ago

On planet Q the standard unit of volume is called guppi. Space travelers from Earth have determined that one liter = 38.2 guppie

ankoles [38]

Answer:

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Explanation:

1 liter= 38.2 guppies

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3 years ago

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Rasek [7]

Answer:

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3 years ago

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A line of charge starts at x = +x0 and extends to positive infinity. The linear charge density is λ = λ0x0/x. Determine the elec

kari74 [83]

Explanation:

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Firstly, we can define the electric field for a small element and then integrate for the whole. The very small electric field is given by :

$dE=\dfrac{k\ dq}{x^2}$ ..........(1)

The linear charge density is given by :

$\lambda=\dfrac{dq}{dx}$

$dq=\lambda.dx=\dfrac{\lambda_ox_o}{x}dx$

Integrating equation (1) from x = x₀ to x = infinity

$E=\int\limits^\infty_{x_o} {\dfrac{k\lambda_ox_o}{x^3}}.dx$

$E=-\dfrac{k\lambda_ox_o}{2}\dfrac{1}{x^2}|_{x_o}^\infty}$

$E=\dfrac{k\lambda_o}{2x_o}$

Hence, this is the required solution.

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3 years ago

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