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Natalija [7]
2 years ago
8

Q.01 When charging a secondary cell, energy is stored within a dielectric material using an electric field. True or False

Physics
1 answer:
Nitella [24]2 years ago
4 0

True, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

<h3>Relationship between dielectric material and electric field</h3>

The electric field in a capacitor separates the negative and positive charges in the dielectric material, this causes an attractive force between each plate and the dielectric.

The dielectric material can store electric energy due to its polarization in the presence of external electric field, which causes the positive charge to store on one electrode and negative charge on the other.

Thus, when charging a secondary cell, energy can be stored within a dielectric material using an electric field.

Learn more about dielectric material here: brainly.com/question/17090590

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E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
\frac{1}{2}k(x_{max})^2 =  \frac{1}{2}m(v_{max})^2
v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s
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