Kate is sitting on a 1.5-m-wide porch swing. because of the rain the night before, the left side of the swing is wet, and kate sits close to the right side. the mass of the swing seat is 10 kg and kate's mass is 55 kg. let the center of mass of kate be 0.20 m from the right end.
estimate the magnitudes of the forces that the two supporting cables exert on the swing.

Explanation:

T(Tension in each cable from seat alone) = (10kg x g)/2

T= 49N.

Kate is 0.2m. from the right cable, so is 1.3m. from the left one. So Now to find extra tension in the left cable and right cable

T=(55kg x g)/((1.3/0.2) + 1)

T= 71.867N. extra tension in the left cable

And (55kg x g) - 71.867 = 467.133N. extra in the right cable.

a) Force of right cable = (467.133 + 49) = 516.13N.  

b) Force left cable = (71.867 + 49) = 120.87N.

9.8 used for g.  

Worked on ratio. (55*g) = 539N.,

and (1.3/0.2) = 6.5:1 ratio, so you add the sides of the ratio, = 7.5 'parts'.

(539/7.5) = minimum figure of 71.867N

and therefore the other tension is (539 - 71.867)= 467.133N.

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s