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Natalija [7]
3 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]3 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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A light, inextensible cord passes over a light,
Musya8 [376]

Answer:          T = 93 N

Explanation:

Assuming the pulley is ideal meaning frictionless as mentioned and also negligible mass.

         ΣF = Σma

Mg - mg = Ma + ma

           a = g(M - m) / (M + m)

Now looking only at the larger mass as it falls

Mg - T = Ma

       T = Mg - Ma

       T = Mg - Mg(M - m) / (M + m)

       T = Mg(1  -(M - m) / (M + m))

       T = 16(9.8)(1 - (16 - 6.7) / (16 + 6.7))

       T = 93 N

or looking only at the smaller mass

T - mg = ma

T = m(g + a)

T = m(g +  g(M - m) / (M + m))

T = mg(1 +  (M - m) / (M + m))

T = 6.7(9.8)(1 +  (16 - 6.7) / (16 + 6.7))

T = 93 N

3 0
3 years ago
While on a train, you are holding a string with a ball attached to it. At first, the train’s velocity is constant and the string
mr Goodwill [35]

Answer:

a)  v = 45.37 m/s  and b)  T = 14.4 N

Explanation:

Let's use Newton's second law with centripetal acceleration

     F = ma

Let's set a reference system where the x-axis is horizontal and the y-axis is vertical, in that system the only force to decompose is tension, let's use trigonometry

     sin 35 = Tₓ / T

     cos 35 = T_{y} / T

     Tₓ = T sin35

     T_{y} = T cos 35

Y Axis

    T_{y} -W = 0

    T cos 35 = W

    T = mg / cos 35

X axis

   Tₓ = ma

   a = v² / r

   T sin 35 = m v² / r

   (mg / cos 35) sin 35 = m v² / r

    g tan 35 = v² / r

    v = √ g r tan 35

   

    v = √ (9.8 300 tan 35)

    v = 45.37 m / s

b) the tension of the rope is

    T = mg / cos 35

    T = 1.20 9.8 / cos 35

    T = 14.4 N

7 0
3 years ago
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Answer:

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Explanation:

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