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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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Aleonysh [2.5K]

Answer:

d

Explanation:

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Pls I need help for the problem solving part.
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In a drill during basketball practice, a player runs the length of the 30-meter court and back. The player does this three times
Sergeeva-Olga [200]

Answer:

0 m/s

Explanation:

Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.

In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.

As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.

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Hence, the correct option is (1).

6 0
3 years ago
A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle
Aleks04 [339]

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Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = 60^oC

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0
3 years ago
What is the acceleration of a car initially traveling at -5m/s and<br> reaching -22m/s in 3s.
lozanna [386]

Answer:

a=-5.67\ m/s^2

Explanation:

Given that,

Initial velocity, u = -5 m/s

Final velocity, v = -22 m/s

Time, t = 3s

We need to find the acceleration of the car. The formula of it is given by :

Acceleration,

a=\dfrac{v-u}{t}\\\\a=\dfrac{(-22)-(-5)}{3}\\\\a=-5.67\ m/s^2

So, the acceleration of the car is -5.67\ m/s^2.

4 0
3 years ago
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