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3 years ago

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

2 answers:

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

Static
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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The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b

ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})$

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) = \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V$

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

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2 years ago

Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm

Leya [2.2K]

If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).

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2 years ago

a 65 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 7 m/s. What is the velocity of t

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3 years ago

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Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

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The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

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$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

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3 years ago

If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river,

expeople1 [14]

Answer:

t=40s,

Explanation:

If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river, if the water flows downstream at a rate of 1.5m/s, is most nearly:

from the question the swimmer will have a velocity which is equal to the sum of the speed of the water and the velocity to swi across the bank

Vt=v1+v2

the time is takes to swim across the bank will be

DY=Dv*t

DY=distance across the bank

Dv=ther velocity of the swimmer across the bank

t=20/ 0.5m/s,

t=40s, time it takes to swim across the bank

velocity is the rate of displacement

displacement is distance covered in a specific direction

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2 years ago

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