Amass m is split into two parts, m and m-m, which are then separated by a certain distance. what ratio m/m maximizes the magnitude of the gravitational force between the parts

Ratio to maximize force = 1/1 The equation for gravitational attraction is F = G*m1*m2/r^2 where G = gravitational constant m1,m2 = mass of each mass r = distance between mass centers. Since the values of G and r are remaining constant as well as the sum of the masses. Let's simplify the problem as follows. m = mass 1 (1-m) = mass 2 So we're looking for a value of m that maximizes m(1-m) where m is somewhere between 0 and 1. Let's do it now: F = m(1-m) F = m - m^2 Now when you're looking to maximize a quantity, that screams "First derivative". So let's calculate that now. F = m - m^2 F' = 1 - 2m And set it to 0. F' = 1 - 2m 0 = 1 - 2m 2m = 1 m = 1/2 So the optimal value of m is 1/2. So m1 = 1/2 and m2 = 1 - 1/2 = 1/2. So the optimal ratio between the two masses to maximize the gravitational force is 1/1.

Sorry i don’t know what to do

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.