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Anna35 [415]
3 years ago
6

A child is swinging back and forth on a swing. Changes in her kinetic and potential energy are happening each moment. At which p

oint, or multiple points, in her swinging would each of these two types of energy be at a maximum? In one to two sentences, describe these points and which type of energy is at a maximum at each point.
PLZ HELP I JUST NEED A SIMPE ANSWER

5O POINTS AND I WILL MARK Brainiest
Chemistry
2 answers:
dedylja [7]3 years ago
5 0

Answer:

The maximum potential energy of the child will be maximum at the two end points.  

The maximum kinetic energy of the <em>child </em>occurs at the lowest point of the swing.  

The potential energy of the child depends on the displacement of the child.  

P.E = mgh  

The maximum height attained occurs at the two end points of her swing motion.  

Thus, the maximum potential energy of the child will be maximum at the two end points.

The kinetic energy of the child depends on the velocity of the child  

K.E = ¹/₂mv²  

The maximum velocity of the swing occurs at the lowest point of the swing.  

Thus, the maximum kinetic energy of the child occurs at the lowest point of the swing.

Hope this helps!

Dovator [93]3 years ago
4 0
At the his best point because the higher you go, the more potential energy you have and when going down the swing you gain kinetic energy. The faster you go down, the more kinetic energy. Hope this helps!!!
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A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?
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<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125  mL = 125 x 10⁻⁶ m³

n = ?

<span> R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
                                          n = 4.0359 x 10</span>⁻³ mol

<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>

Moles = mass / molar mass

</span>Mass of the gas  = 0.529 g 

<span>Molar mass of the gas</span> = mass / number of moles<span>
                                    = </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span>                                    = </span>131.07 g mol</span>⁻¹<span>

Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹

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3 years ago
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