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IRINA_888 [86]
3 years ago
14

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

the see saw in order to perfectly balance the child and rock?
Physics
1 answer:
ivann1987 [24]3 years ago
6 0

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock  

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

 (m)child(d)child)=(m)rock(d)rock)  

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

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Putting the given values for 71 dB

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A² = 1.1978 x 10⁻¹⁶

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To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be \frac{4.83598}{c^{\frac{1}{3} } }.

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Therefore, the ratio between the surface area and the volume for the sphere will be:

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To learn more about surface area and volume of sphere, refer to:

brainly.com/question/4387241

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