a 0.50n weight is stuck to the 20cm mark of a uniform metre stick, which weighed 0.5n before the weight was added. you can balance the metre stick horizontally on your finger, if you put your finger in the right place.
how far from the 0.0cm end do you need to put your finger for the metre stick to be balanced?

The finger should place at the 35 cm in the stick to balance the forces.  

Given here,

There's a -0.5 N force at 20 cm.   and another -0.5 N force at 50 cm.  

Sum of the forces:  

∑F = ma

Where,

m - mass

a - acceleration due to gravity

F − 0.5 N − 0.5 N = 0  

F = 1.0 N  

Sum of the moments torques,  

∑τ = Iα

Where,

∑τ - sum of torques

I = moment of inertia (kg∙m2).

α = angular acceleration (radians/s2).

Put the values in the formula,  

(-0.50 N) (20 cm) + (-0.50 N) (50 cm) + (1.0 N) (x) = 0  

x = 35 cm

Therefore, the finger should place at the 35 cm in the stick to balance the forces.

To know more about torques,

link

35 cm

Explanation:

There's a -0.50N force at 20cm.

There's another -0.50N force at 50cm.

You apply a force of F at a position x.

Sum of the forces:

∑F = ma

F − 0.5 N − 0.5 N = 0

F = 1.0 N

Sum of the moments (or torques):

∑τ = Iα

(-0.50 N) (20 cm) + (-0.50 N) (50 cm) + (1.0 N) (x) = 0

x = 35 cm

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2