Suppose you are given a 4 m long straight bar of uniform density balanced at its center and a weight of 4.6 kg hanging on the bar at a distance of 12.0 cm from the pivot. Where must we hang a 2.6 kg mass to have the system remain in balance?

21.23 cm

Explanation:

Being that the mass of the bar is equal on both sides, the moment produced by the masses cancel each other out, so you can just look at the hanging masses.

Sum of the moments about the fulcrum = 0... I'll say that moments produced on the 4.6 kg side are positive

0 = 4.6kg * g * 12.0 cm - 2.6kg * g * x

g's cancel. solve for x

x =( 4.6  * 12) / 2.6

x = 55.2 / 2.6

x = 21.23cm

so 2.6kg mass should be pivoted at a distance of 21.23 cm from pivot.

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day