A uniform meter stick ( 100 cm long) that weighs 0.50 N has a 2 N weight located at the 30 cm mark (from the left) and a 5 N weight at the 85 cm mark (from the left). Where along the meter stick is the center of gravity of this system?

To find the center of gravity of the system, you need to consider the weights of the two objects and their distances from the left end of the meter stick. The center of gravity is the point at which the total weight of the system can be considered to act.

First, let's calculate the moments of the weights, which is the product of the weight and the distance from the left end.

For the 2 N weight located at the 30 cm mark:
Moment1 = 2 N × 30 cm = 60 N•cm

For the 5 N weight located at the 85 cm mark:
Moment2 = 5 N × 85 cm = 425 N•cm

Next, let's calculate the total weight and the total moment of the system.
Total Weight = Weight1 + Weight2 = 2 N + 5 N = 7 N
Total Moment = Moment1 + Moment2 = 60 N•cm + 425 N•cm = 485 N•cm

Now, we can find the center of gravity (CG) using the formula:
CG = Total Moment / Total Weight

CG = 485 N•cm / 7 N = 69.29 cm

Therefore, the center of gravity of this system is located at approximately 69.29 cm from the left end of the meter stick.