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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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2. A hanging chair is suspended on a spring from the ceiling. The 5880 N/m spring stretches
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Answer:

a)= 29.4J

b)F = 588 N

c)= 60 Kg

Explanation:

Force constant of the spring (k) = 5880 N/m

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This work done on the spring as it is stretched (or compressed) can be recovered. This is stored work that can be used to do work on something else by this spring. That means the stretched (or compressed) spring has energy -- potential energy. This is spring potential energy or elastic potential energy.

a) Work done in pulling the body W = 1/2kx²

= 1/2 (5880)(0.1)2

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b)From Hook's Law,

F = ke

Where F = applied force, k = spring constant, e = extension.

Given: k = 5880 N/m, e = 25-15 = 10 cm = 0.1 m.

Substitute into the formula above

F = 5880(0.1)

F = 588 N

c)By using the formula, F = -kx

Hence mg = kx

Thus m x 9.8 =5880 x0.1

Hence mass of the body

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4 years ago
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<u>Answer:</u>

It took \bf 4.07 \space\ \mathrm{s} to travel.

<u>Explanation:</u>

To calculate the time taken to travel a distance of 122 m at a speed of 30 m/s, we can use the formula for speed:

\boxed{speed = \frac{distance}{time}}.

We can substitute the known values, and then solve for time:

30 = \frac{122}{\mathrm {time}}

⇒ \mathrm{time } = \frac{122}{30}

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1 year ago
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