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3 months ago

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

2 answers:

8 0

Yes, a small child can indeed play with a heavier child on a seesaw. This is because reducing the distance from the fulcrum for the heavier child increases the effort required, allowing the smaller child to play without much strain.

Ostrovityanka [3.2K]3 months ago

6 0

The weights of the small child and the heavier child differ. The smaller child has less mass in comparison to the heavier one. This indicates that the force exerted by the smaller child is lesser than that of the heavier child.

Further Explanation

Consequently, when both children sit equidistant from the pivot, their torques will not balance. As a result, the seesaw will tilt towards the child with greater weight (the heavier child).

Thus, for the two of them to use the seesaw together, they must position themselves at different distances from the pivot. This means the heavier child should sit closer to the pivot.

This topic pertains to torque. Torque is defined as a vector quantity, since it comprises both magnitude and direction. It measures the effectiveness of a force in causing an object to rotate around an axis.

Torque can be divided into two categories, which are

Static
Dynamic

Static Torque refers to a force that does not result in angular acceleration, meaning static torque is not related to acceleration, while dynamic torque results in angular acceleration.

LEARN MORE:

Can a small child play with a fat child on a see saw? explain how?
Can a small child play with a fat child on a see saw? explain how?

KEYWORDS:

fat child
small child
seesaw
torque
pivot

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If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

ValentinkaMS [3465]

Answer:

17.35 × 10^(-6) m

Explanation:

Mass; m = 50 kg

Weight; W = 554 N

From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

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3 months ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

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Answer:

(A) The tension's magnitude grows to four times the initial value, 4F.

Explanation:

When an object travels in a circular path, a centripetal force is exerted upon it. In this instance, the centripetal force acting on the stone can be represented by $\frac { m{ v }^{ 2 } }{ r }$ .

Here, m denotes the mass of the object

v is the velocity or speed of the object

r signifies the radius of the circular path

Importantly, the tension corresponds to the centripetal force.

Initially, the string completes one revolution each second, and subsequently, it accelerates to perform two revolutions in the same time frame. This signifies that the speed has increased twofold.

Applying our formula:F = $\frac { m{ v }^{ 2 } }{ r }$

where F indicates the tension in the string

assuming the starting speed is v, after doubling it becomes 2v

Maintaining the circle's radius, we arrive at:

F= $\frac { m{ (2v) }^{ 2 } }{ r } =\frac { 4m{ v }^{ 2 } }{ r }$

From this equation, it's clear that the initial tension has quadrupled.

Consequently, the magnitude of the tension increases to four times its original value, 4F.

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