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grigory [225]
3 years ago
14

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

Physics
2 answers:
RoseWind [281]3 years 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]3 years 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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nadezda [96]

Answer:

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Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

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