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harina [27]
3 years ago
12

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

m, what must be its minimum mass so that material on its surface remains in place during the rapid rotation in kilograms?
Physics
1 answer:
Aleks [24]3 years ago
4 0

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

\frac{GM_{ns}}{R^2}= \omega^2 R

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

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What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?
sergey [27]

The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

<h3>How to calculate acceleration?</h3>

The acceleration of a freight train can be calculated using the following formula:

Force = mass × acceleration

According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:

100,000 = 600,000 × a

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Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².

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6 0
1 year ago
A 110-g bullet is fired from a rifle having a barrel 0.636 m long. Choose the origin to be at the location where the bullet begi
astraxan [27]
<span>Data:
mass =  110-g bullet
d = 0.636 m
Force = 13500 + 11000x - 25750x^2, newtons.

a) Work, W

W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =

W =  13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636  =

W = 8602.6 joule

b) x= 1.02 m

</span><span><span>W =  13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02

W = 10383.5

c) %

[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.


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3 years ago
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