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Nina [5.8K]
3 years ago
9

A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

This is the angle that the chains make with the vertical.) What is the childs maximum speed on the swing
Physics
1 answer:
kvv77 [185]3 years ago
6 0

Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          Em_{f} = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

           h = L - L cos 60

           h = L (1- cos 60)

energy is conserved

         Em₀ = Em_{f}

          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

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brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
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Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

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Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

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Combine equation (1) with (2)

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Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
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To learn more about Oscillation refer to:

brainly.com/question/28312746

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We have the equation of motion v^2=u^2+2as, where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

        Initial velocity, u = 0 m/s

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Substituting 40^2=0^2+2*a*2\\ \\ a=400m/s^2

So the baseball pitcher accelerates at 400m/s^2 to release a ball at 40 m/s.

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