A boy on board a cruise ship drops a 30. 0 gm marble into the ocean. If the resistive force proportionality constant is 0. 500 k
2 answers:
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Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ =
let's calculate
x₂ =
x₂ = 1.33 m
Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.