A uniform steel rod has mass 0.300 kg and length 40.0 cmand is horizontal. A uniform sphere with radius 8.00 cm and mass 0.700 kg is welded to one end of the bar, and a uniform sphere with radius 6.00 cm and mass 0.580 kg is welded to the other end of the bar. The centers of the rod and of each sphere all lie along a horizontal line. How far is the center of gravity of the combined object from the center of the rod?
Express your answer with the appropriate units. Enter negative value if the center of gravity is toward the 0.700 kg sphere and positive value if the center of gravity is toward the 0.580 kg sphere.

1.86 cm

Explanation:

The center of gravity of the combined object x = m₁x₁ + m₂x₂ + m₃x₃/m₁ + m₂ + m₃ where m₁ = mass of sphere on left end of rod = 0.700 kg, x₁ = distance of center of mass of sphere from left end of combined object = radius of sphere = 8.00 cm, m₂ = mass of rod = 0.300 kg, x₂ = distance of center of mass of rod from left end of combined object = diameter of sphere on left end + length of rod/2 = 16.00 cm + 40 cm/2 = 16.00 cm + 20 cm = 36 cm, m₃ = mass of sphere on right end of rod = 0.580 kg, x₃ = distance of center of mass of sphere from left end of combined object = length of rod + diameter of first sphere + radius of sphere = 40 cm + 16 cm + 6.00 cm = 62 cm

x = (m₁x₁ + m₂x₂ + m₃x₃)/m₁ + m₂ + m₃

Substituting the values of the variables into the equation, we have

x = 0.700 kg × 8.00 cm + 0.300 kg × 36 cm + 0.580 kg × 62 cm/(0.700 kg + 0.300 kg + 0.580 kg)

x = 5.6 kg cm + 10.8 kg cm + 35.96 kg cm/1.580 kg

x = 52.36 kgcm/1.580 kg

x = 33.14 cm

Since the center of the rod is at x' = (40 cm + 16.00 cm + 12.00 cm)/2 = 70.00 cm/2 = 35 cm

The distance between the center of the rod and the center of gravity is x' - x = 35 cm - 33.14 cm = 1.86 cm

So, the center of gravity is 1.86 cm away from the center of the rod and closer to the 0.700 kg sphere.

f=ma newton

f=100x2.5=250n

where’s the list ? the question is incomplete