The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
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20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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On the floor because the force from the train moving and she wouldn't be able to catch it in time
Answer:
0.64 m
Explanation:
The first thing is calculate the center of mass of the system.
now multiplying every coordinate x by the mass of each object (romeo, juliet and the boat) and dividing all by the total mass taking by reference the position of juliet.
X_cm = 1.4589 m
When the forces involved are internals, the center of mass don't change
After the movement the center of mass remains in the same distance from the shore, but change relative to the rear of the boat.
X_cm= 2.10 m
this displacement is how the boat move toward the shore.
2.10-1.46= 0.64 m
Maybe there is people who used
Answer:
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Explanation:
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