Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa
1 answer:
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Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
Answer:
The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.
Explanation:
Given that,
Mass of object = 5 kg
Speed = 3 m/s
Mass of stationary object = 3 kg
Moving object deflected = 30°
Stationary object deflected = 31°
We need to calculate the velocity of each ball after collision
Using conservation of momentum
Along x-axis
Put the value into the fomrula
....(I)
Along y -axis
Put the value into the formula
...(II)
From equation (I) and (II)
Put the value of v₁ in equation (I)
Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.