To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.
Mathematically it can be expressed as

Where
m = Mass (Neutron at this case)
V = Volume
The mass of the neutron star is 1.4times to that of the mass of the sun
The volume of a sphere is determined by the equation
Replacing at the equation we have that



Therefore the density of a neutron star is 
(B) Away from the observer.
The closer- the higher
The farther- the lower
Answer:
formula used K=F/∆l
∆l is the elongation of the spring
- F=10N
- ∆l=20mm===> 0.02m
- K=10N divided 0.02m= 500N/m
The frequency will not change
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg