How thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.
<h3>Density of the Neutron star</h3>
Density
ρ = m / V
Where:
ρ= density
m = mass of the planet 5.98 10²⁴ km
V =volume of the spherical layer
Volume of a sphere
Volume = 4/3 π r³
Mass = 1.5 = 1.5 1,991 10³⁰
Mass= 2.99 10³⁰ kg
Density:
ρ = 2.99 10³⁰ / [4/3 π (10 10³)³]
ρ is = 7.13 10 17 kg / m³
V = 5.98 10²⁴ / 7.13 10¹⁷
V = 8,387 10⁶ m³
Thickness of the layer
V = 4π r² e
e = V / 4π r
e = 8,387 10⁶ / [4π (10 10³)²]
e = 6.67 10⁻³ m
Inconclusion how thick a layer would Earth form as it wraps around the neutron star’s surface is: 6.67 10⁻³ m.
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Answer:
The equilibrium in the system is a dynamic equilibrium. At equilibrium, the rate of the forward reaction (CH4 decomposing into C2H2 and H2) is equal to the rate of the reverse reaction (C2H2 and H2 reacting to form CH4). At equilibrium, the concentrations of all substances are not changing.
Explanation:
