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baherus [9]
3 years ago
11

An object having mass m is attached to a spring of force constant k oscillates with simple harmonic motion. The total mechanical

energy of the system is E and the maximum displacement from equilibrium is A. What is the system's potential energy when its kinetic energy is equal to 3/4E?
What is the system's potential energy when its kinetic energy is equal to ?
1. kA2
2. kA2/2
3. kA2/4
4. kA2/8
Physics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

The potential energy of the system is \dfrac{E}{4}.

Explanation:

Given :

Mass of object , m .

Spring constant , k .

Total energy , E .

Maximum displacement , A .

We have to find potential energy when its kinetic energy is equal to 3/4E .

Now, we know :

Total energy is constant in simple harmonic motion.

Therefore ,

Total energy = Kinetic energy + potential energy .

E=P.E+\dfrac{3E}{4}\\\\P.E=\dfrac{E}{4}

Hence , this is the required solution.

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Answer:

La intensidad del campo eléctrico es 70312.5 \frac{N}{C}.

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La perturbación que crea en torno a ella una carga eléctrica se representa mediante un vector denominado campo eléctrico.

Se dice que un campo eléctrico es uniforme en una región del espacio cuando la intensidad de dicho campo eléctrico es el mismo en todos los puntos de dicha región.

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E=k*\frac{q}{r^{2} }

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  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }
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Reemplazando:

E=9*10^{9}\frac{N*m^{2} }{C^{2} }  *\frac{5*10^{-6} C}{(0.8 m)^{2} }

Resolviendo:

E= 70312.5 \frac{N}{C}

<em><u>La intensidad del campo eléctrico es 70312.5 </u></em>\frac{N}{C}<em><u>.</u></em>

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