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Vinvika [58]
4 years ago
6

Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive δg at rt be made f

avorable by increasing the reaction temperature?
Physics
1 answer:
Over [174]4 years ago
7 0
I believe that the answer to the question provided above is yes, an unfavorable reaction that has a positive δg at rt be made favorable by increasing the reaction temperature.
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
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Explanation:

A magnet has a magnetic field around it which originates at the north pole and enters through the south pole.

In a magnet, like poles will repel each other and unlike poles will attract.

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A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it
White raven [17]

Answer:

The positions are  0.0194 m  and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

v = \omega \sqrt{A^2-x^2}

the maximum speed of the simple harmonic motion is given as;

v_{max} = \omega A

when the speed equal one fourth of its maximum speed

v =\frac{v_{max}}{4}

\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m  \ \ or\  - 0.0194  \ m

Thus, the positions are  0.0194 m and - 0.0194 m.

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