A 1 kg rock is suspended by a massless string from one end of a meter stick at the 0 cm mark. What is the mass m suspended from the meter stick at the 75 cm mark if the system is balanced?

2 kg

Explanation:

Note: For the meter stick to be balanced,

Sum of clock wise moment must be equal to sum of anti clock wise moment

Wd = W'd' Equation 1

Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.

make W' the subject of the equation

W' = Wd/d' Equation 2

Taking our moment about the support,

Given: W = mg =  1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm

Substitute into equation 2

W' = 9.8(50)/25

W' = 19.6 N.

But,

m = W'/g

m = 19.6/9.8

m = 2 kg.

electrical power

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6