+Math ConnectionsLaw of Universal Gravitation (continued)This law states that any two objects with mass exert a gravitational force on each other. The forcehas theto the product ofthe objects' masses. It is inversely proportional to the square of the distance between the objects.Where,mm₂)F = G(d²8Gravitational, Electrical, Magnetic, and NuclearF = gravitational force between two objectsG= gravitational constant = 6.673 x 10-11 Nx m²/kg²m, and m₂ = masses of the two objects in kgd= distance between the two objects in m5. Calculate the gravitational force between two 150 kg objects located 25 m apart.2.40228-10-2F. = 6.673x10 "((150)(150))N6. Calculate the change in the gravitational force between the two objects if the distance betweenthe two objects doubles.6.0057.10-10ons who out in or betsool elboidoF=6.673x10-|Arralerata I earninn Inc All Rinhte Racarvad(150)(150)50²(001) (0017. Use the inverse square law to describe the relationship between the gravitational force and thedistance between the two objects.What was the scale factor of the distances? 001)What is the inverse square of the scale factor?Multiply the gravitational force in question 6 by the inverse square of the scale factor. woh SHow does this relate to the gravitational force in question 7 when the distance is doubled?desnovni to lenoihogoro vipetib azam odi bre estol isnailsliveiped8. If the gravitational force between two objects is 16 N, and the distance between the objectsdoubles, what is the resulting gravitational force? Use the inference you made in question 7 tocalculate the resulting force.16Nelajdo el to ano to esem od 221 87 as owN_=($)612

Since you have posted a question with a multiple sub parts, we will solve first three sub - parts for you. To get the remaining sub parts solved please repost the complete question and mention the sub parts to be solved. 5) m1 = m2 = 150 kg distance (d) = 25 m using newton's law of gravitation : F = G×m1×m2d2F = 6.673×10-11×150×150(25)2F = 2.402×10-9 N 6) now distance between the two objects double new distance = 2×25 m = 50 m new force : F = 6.673×10-11×150×150(50)2F = 6.006×10-10 N Change in gravitational force = 6.006×10-10- 2.402×10-9= -1.8×10-9 N 7) scale factor of distance = 2 inverse square of the scale factor = 14 = 0.25F = 14×2.402×10-9 = 6.005×10-10 NThis is equal to the value for force obtained in question 6