Answer:
Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his mdth of his model if he uses the same ratio?
Explanation:
Kent is making a scale model of his favorite train. The actual train is 12 feet long and 4 feet wide. Kent wants his model to be 6 inches in length. Which would be the width of his modelhe uses the same ratio?
Control is the comparison like 0 or the normal say you are testing salt on plants, no salt would be the comparison
Constant are what stay the same, like the things other then IV and DV
Answer:
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.
This is Millikan's oil drop experiment.
The downward force acting on the oil drop is its weight. The upward forces are air resistance, which is negligible due to the droplet's size, and the force due to the electric field present.
