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2 years ago

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

1 answer:

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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Is the following true or false?

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True, recycling allows products to be reused which saves alot the natural resources that are used.

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3 years ago

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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3 years ago

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Which is the best insulator? metal, glass, plastic, or Styrofoam?

Mashutka [201]

Answer:

Styrofoam would be the best insulator because it traps the air in small pockets, blocking the flow of heat energy.

Explanation:

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3 years ago

A football is place kicked with a velocity having a vertical component of 12 m/s and a horizontal component of 6 m/s. Find the r

SSSSS [86.1K]

The velocity is given by:

V = √(Vx²+Vy²)

V = velocity, Vx = horizontal velocity, Vy = vertical velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for V:

V = √(6²+12²)

V = 13.42m/s

Now find the direction:

θ = tan⁻¹(Vy/Vx)

θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for θ:

θ = tan⁻¹(12/6)

θ = 63.4°

The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.

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4 years ago

A 100-turn, 3.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60%u2218 away from vertical increase

Xelga [282]

Answer:

Induced emf in the coil, E = 0.157 volts

Explanation:

It is given that,

Number of turns, N = 100

Diameter of the coil, d = 3 cm = 0.03 m

Radius of the coil, r = 0.015 m

A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.

Due to this change in magnetic field, an emf is induced in the coil which is given by :

$E=-NA\dfrac{\Delta B}{\Delta t}$

$E=-100\times \pi (0.015)^2\times \dfrac{2.5-0.5}{0.9}$

E = -0.157 volts

Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.

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3 years ago