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Tatiana [17]
3 years ago
15

A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 18 k

m in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm^3?
Physics
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Density=6.51*10^{14}g/cm^{3

Explanation:

Sun mass:

Ms=1.989 × 10^30 kg

Neutron star has the same mass.

Radius Neutron:

R=9Km (because diameter is 18Km)

R=9*10^3m

Density neutron star:

D=Mn/Vol=Ms/(4/3*\pi*R^3)=1.989*10^{30} /(4/3*\pi*(9*10^{3})^{3})=6.51*10^{17}kg/m^{3}

D=6.51*10^{17}kg/m^{3}*(1000g/1Kg)*(1m^{3}/1000000cm^{3})=6.51*10^{14}g/cm^{3

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The simplest form of a substance
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Answer :

C) Atom

Actual answer should be element but depends on the way you interpret the question and the options given to answer it

Explanation :

Atom is the most basic unit of any substance and is what molecules are made of.

Hope it helps if it does let me know by thanking
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3 years ago
Does it take more force to slow something down than to speed it up.
ANTONII [103]
To bring something to a stop the same force that was applied to speed it up can be used to stop it. If a greater force is used it will stop quicker.
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3 years ago
Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the
castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}.

Now, making the substitutions

d \ = \ 2.00 * 10 ^{-15} \ m,

q_1 = q_2 = 1.60 * 10 ^ {-19} \ C,

\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2},

we can find:

F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}.

F \ = 57.536 N.

Not so big for everyday life, but enormous for subatomic particles.

4 0
3 years ago
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The author writes that each part of the ecosystem is as important as another. Based
Masja [62]
I agreeeeeee do you?
3 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
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