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grigory [225]
1 year ago
14

Can a small child play with fat child on the seesaw?Explain how?

Physics
2 answers:
RoseWind [281]1 year ago
8 0
Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
Darya [45]1 year ago
6 0

The mass of the small child and that of the fat child is not the same. The small child would have a lesser mass compared to the fat child. This also implies that the weight force by the small child is smaller to the force of the fat child.

<h2>Further Explanation </h2>

Therefore, if both the small child and the fat child sit at equal distance from the pivot, the torques by them cannot be equal. The result is that the seesaw will rotate at the end of the child with a bigger weight (fat child).

Therefore, for both of them to play on the seesaw, they have to sit at an unequal distance from the pivot. This also implies that the child with a bigger weight (fat child) needs to sit near the pivot.

This question deals with torque. Torque is a vector quantity because it has both magnitude and direction. It can be described as the measure of a force that can make an object to rotate within an axis.

Torque can be classified into two and these include

  1. Static
  2. Dynamic

A static Torque is the type of force that does not produce an angular acceleration, that is, a static torque has nothing to with acceleration while dynamic torque produces angular acceleration.

LEARN MORE:

  • Can a small child play with a fat child on a see saw? explain how?  brainly.com/question/4267445
  • Can a small child play with a fat child on a see saw? explain how?  brainly.com/question/790091

KEYWORDS:

  • fat child
  • small child
  • seesaw
  • torque
  • pivot
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alexira [117]

Answer:

v_wind = 101.46 km / h   ,  θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

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           v = 750 / 3.14

           v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and v_{y}

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          v_N = v cos θ

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          V _N = 238.85 cos 22 = 221.46 km / h

           v_W = -238.85 sin 22 = -89.47

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in the direction to the East.

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Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

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This is the final speed of the plane, which can be written

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this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

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the address will then be found using trigonometry

             θ = Vy / vx

             θ = tan⁻¹ (vy / vx)

             θ = tan⁻¹1 (47.85 / 89.47)

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Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

                  90- θtea = 90- 28.2

                  θ = 61.8

East of South

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