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lora16 [44]
3 years ago
11

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.

The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0x10^5 km (comparable to our sun); its final radius is 16 km .If the original star rotated once in 35 days, find the angular speed of the neutron star.
Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

The angular speed of the neutron star is 3976.978\ rads^{-1}

Solution:

As per the question:

Initial radius of the star, r = 7.0\times 10^{5}\ km = 7\times 10^{8}\ m

Final Radius, r' = 16 km

Period of rotation of the star, T = 35 days = 35\times 24\times 3600 = 3024000\ s

Now,

To calculate the angular speed of the neutron star:

We use the principle of conservation of angular momentum:

I\omega = I'\omega'

where

I = Moment of inertia

\omega = \frac{2\pi}{T}

(7\times 10^{8})^{2}\times \frac{2\pi}{T} = (16000)^{2}\omega'

\omega = 3976.978\ rads^{-1}

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Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

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Final velocity v = 20 m/s

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Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

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2 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

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          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

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having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

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we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

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8 0
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What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?
PtichkaEL [24]

Answer:

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This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

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V(t)=ve^{-\frac{t}{RC} }

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