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Sagot :
Answer:
[tex]F=3.665 \times 10^{-7} N[/tex] acting along the line joining both the sphere and always attractive in nature.
Explanation:
The given radius of both the sphere, r= 65 mm = 0.065 m
So, the volume of the spheres, [tex]v= \frac 4 3 \pi r^3[/tex]
[tex]v= \frac 4 3 \pi (0.065)^3 = 1.150 \times 10^{-3} m^3[/tex]
The density of steel, [tex]\rho _s = 7850 kg/m^3[/tex]
and the density of copper, [tex]\rho_c= 8940 kg/m^3[/tex]
Let M be the mass of the copper ball and m is the mass of the steel ball.
So, [tex]M=\rho_c v= 8940\times 1.150 \times 10^{-3} = 10.281[/tex] kg
[tex]m=\rho_s v= 7850\times 1.150 \times 10^{-3} = 9.0275[/tex] kg
The gravitational force, F, between the two objects having masses M and m and separated by distance d is
[tex]F=\frac{GMm}{d^2}[/tex]
Where [tex]G= 6.674 30 x 10^{-11} m^3 kg^{-1} s^{-2}[/tex] is the universal gravitational constant.
When both the sphere touches each other, d = 2r= 2 x 0.065 = 0.13 m
Hence, the gravitational force between both the sphere,
[tex]F= \frac {6.674 30 x 10^{-11}\times 10.281 \times 9.0275}{0.13^2} \\\\F=3.665 \times 10^{-7} N[/tex]
The nature of gravitational force is always attractive and acting along the line joining the center of both the sphere.
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