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Viefleur [7K]
3 years ago
12

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

meter. If you weigh 675 N on the earth, what would you weigh at the surface of a neutron star that has the same mass as our sun and a diameter of 20 km
Physics
1 answer:
Crank3 years ago
3 0

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

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Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

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Then:

$d u=-\sin (\omega t) d t

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Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

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The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>
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B=(μ₀ I/4π)× (0.9/0.006)

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Answer:

<h2>1.17 m/s²</h2>

Explanation:

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We have the final answer as

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