ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach
ed at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
1 answer:
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
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d. Its magnitude and its direction both remained the same.
Explanation:
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Mathematically, momentum is given by the formula;
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This ultimately implies that, the law of conservation of momentum states that if objects exert forces only on each other, their total momentum is conserved.
In this scenario, a rubber ball moving at a speed of 5 m/s hit a flat wall and returned to the thrower at 5 m/s. Thus, the statement which correctly describes the momentum of the rubber ball is that its magnitude and its direction both remained the same because its velocity didn't change while returning to the thrower.