Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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.
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
