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Answers
Explanation:
What is the gravitation of two gold spheres of radius 20cm and 40 cm respectively, each separated by 170 m from their centre and density of both gold is 19.3G/cm^3?
On-premise and cloud data protection in times of uncertainty.
We know that in Newtonian mechanics, F = Gm1m2/r^2
Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2
IF, however, one mass is much greater than the other, we can write
g = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)
In this case, we have a gold sphere of R = 20cm, so Vol = 4/3 π R^3 = 33510.32 cm^3, and the mass of this sphere is m1 = 33510.32(19.3)/1000 = 646.75kg. Similarly, we have a gold sphere of R = 40cm, so vol = 4/3 π R^3 = 268082.57cm^3 and the mass of this sphere is m2 = 268082.57(19.3)/1000 = 5174.0kg.
Then the gravitational attractive force F = Gm1m2/r^2
F = (6.674x10^-11)(646.75)(5174.0)/170^2 = 772.77x10^-11
F = 7.73x10^-9 N
I hope it helps you mate.
On-premise and cloud data protection in times of uncertainty.
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can write
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can writeg = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can writeg = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)In this case, we have a gold sphere of R = 20cm, so Vol = 4/3 π R^3 = 33510.32 cm^3, and the mass of this sphere is m1 = 33510.32(19.3)/1000 = 646.75kg. Similarly, we have a gold sphere of R = 40cm, so vol = 4/3 π R^3 = 268082.57cm^3 and the mass of this sphere is m2 = 268082.57(19.3)/1000 = 5174.0kg.
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can writeg = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)In this case, we have a gold sphere of R = 20cm, so Vol = 4/3 π R^3 = 33510.32 cm^3, and the mass of this sphere is m1 = 33510.32(19.3)/1000 = 646.75kg. Similarly, we have a gold sphere of R = 40cm, so vol = 4/3 π R^3 = 268082.57cm^3 and the mass of this sphere is m2 = 268082.57(19.3)/1000 = 5174.0kg.Then the gravitational attractive force F = Gm1m2/r^2
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can writeg = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)In this case, we have a gold sphere of R = 20cm, so Vol = 4/3 π R^3 = 33510.32 cm^3, and the mass of this sphere is m1 = 33510.32(19.3)/1000 = 646.75kg. Similarly, we have a gold sphere of R = 40cm, so vol = 4/3 π R^3 = 268082.57cm^3 and the mass of this sphere is m2 = 268082.57(19.3)/1000 = 5174.0kg.Then the gravitational attractive force F = Gm1m2/r^2F = (6.674x10^-11)(646.75)(5174.0)/170^2 = 772.77x10^-11
and cloud data protection in times of uncertainty.We know that in Newtonian mechanics, F = Gm1m2/r^2Where F is the attractive force between 2 masses, m1 and m2, r is the distance between the centres of the masses and G is the universal gravitation constant, which is 6.674x10^-11 Nm^2/kg^2IF, however, one mass is much greater than the other, we can writeg = GM/r^2 where M is the larger mass (and we call g “gravity”, or properly “the acceleration due to gravity”; g has units of m/s^2)In this case, we have a gold sphere of R = 20cm, so Vol = 4/3 π R^3 = 33510.32 cm^3, and the mass of this sphere is m1 = 33510.32(19.3)/1000 = 646.75kg. Similarly, we have a gold sphere of R = 40cm, so vol = 4/3 π R^3 = 268082.57cm^3 and the mass of this sphere is m2 = 268082.57(19.3)/1000 = 5174.0kg.Then the gravitational attractive force F = Gm1m2/r^2F = (6.674x10^-11)(646.75)(5174.0)/170^2 = 772.77x10^-11F = 7.73x10^-9 N