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Answer:</h2>
D. (1m, 0.5m)
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Explanation:</h2>
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
Answer:
500 N
Explanation:
Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)
Since W = 85.0 J and we need to find F,
F = W/x
= 85.0 J/0.170 m
= 500 N
So, the magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.
Answer:
all I know is C
are there more questions? anyone?
-KARL IS STOOPID
Explanation:
Similarity : inverse square law for strength of force compared with distance.
