Two students stting in adjacent seats in a lecture room have weights of 569 N and 673 N. Assume that Newton's law of universal gravitation can be applied to these two students and find the gravitational force (in Newtons) that one student exerts on the other when they are separated by 0.628 m.

Answer: F = 6.746 * 10^-7 N

Step-by-step explanation: assuming g = 9.8m/s²

The weight of the first student is 569 N, hence his mass is calculated using the same formula.

W = mg, where W = weight = 569 N, m = mass = ? and g = acceleration due gravity = 9.8m/s²

569 = m*9.8

m = 569/9.8

m = 58.06 kg.

When the second weight is 673 N, the mass is gotten by using the formulae W=mg

m = 673/9.8

m = 68.67kg.

The gravitational force between 2 objects is given as

F = G*m1*m2/r²

Where G = 6.674*10^-11 and r = 0.628m

F = 6.674*10^-11 * 68.67 * 58.06 / (0.628) ²

F = 26,609.10 * 10^-11/0.394384

F = 67469. 77*10^-11

F = 6.746 * 10^-7 N

6.73 x 10^-7 N.

Step-by-step explanation:

F1 = M1 * a

M1 = 569/9.81

= 58 kg

F2 = M2 * a

M2 = 673/9?81

= 68.6 kg

r = 0.628 m

Using force of attraction, F = (GM1M2) / r^2

Where,

G = 6.67408 * 10-11 m3/kg. s

F = (6.67408 * 10-11 * 68.6 * 58) / (0.628) ^2

= 6.73 x 10^-7 N.