Part one

objects with masses of 207 kg and 516 kg
are separated by 0.482 m. a 31.8 kg mass is
placed midway between them.
find the magnitude of the net gravitational
force exerted by the two larger masses on the
31.8 kg mass. the value of the universal gravitational constant is 6.672 × 10−11 n · m^2/kg^2.
answer in units of n.

part two

leaving the distance between the 207 kg and
the 516 kg masses fixed, at what distance from
the 516 kg mass (other than infinitely remote ones) does the 31.8 kg mass experience a net force of zero?
answer in units of m.

1.129×10⁻⁵ N

1.295 m

Explanation:

Take right to be positive.  Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

the answer is c. the lower legs are levers, and the knees are fulcrums. the ankles hold the loads.