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PSYCHO15rus [73]
3 years ago
10

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on th

e left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?
 a. 4m
b. 5m
c. 2m
d. 3m​
Physics
1 answer:
Ostrovityanka [42]3 years ago
7 0

Answer:

We can say that when we multiply both the force given by the two body and along with the distinct lengths they are situated from centre of mass (fulcrum) will be the same

So the let the distance seperated from the centre of mass on the right side where the boy sits be x

so 60×3 = 45x

=> x = (180/45)

=> x = 4 metres

Hope it helps

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motikmotik

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
Evgesh-ka [11]

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For this problem, the circuit is charged to 39%, which means: q = 0.39 q_0

0.39q_0 = q_0 (1 - e^\frac{-t}{RC} )

0.39 = 1 - e^\frac{-t}{RC}

e^\frac{-t}{RC} = 0.61

\frac{-t}{RC} = ln(0.61)

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For the condition to be met it is needed 0.492 time constants must elapse.

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