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Sloan [31]
3 years ago
13

Which of the following best describes a capacitor?

Physics
2 answers:
noname [10]3 years ago
7 0

Answer:

APEX two conductors seperated by an insulator

Explanation:

galben [10]3 years ago
4 0

Answer:

B

Explanation:

The capacitor is a component which has the ability to store energy in the form of an electrical charge  making a potential difference on those two metal plates

A capacitor consists of two or more parallel conductive (metal) plates. They are electrically seperated by an insulating material (ex: air, mica,ceramic etc.) which is called as Dielectric Layer

Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.

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How do objects move under the influence of gravity?
Nady [450]

Under the influence of gravity, objects just move down to the earth.

PLEASE RATE AS THE BRAINLIEST ANSWER! THANK YOU! :)

4 0
3 years ago
a current of 5 ampere is passed for 2 hours in an electric iron having a resistance of 100 ohms calculate the heat produced
Alina [70]

Answer:

\boxed{\sf Heat \ produced \ (H) = 5 \ kWh}

Given:

Resistance (R) = 100 Ω

Current (I) = 5 A

Time (t) = 2 hours

To Find:

Heat developed (H) in the electric iron

Explanation:

Formula:

\boxed{ \bold{ \sf H = I^2Rt}}

Substituting values of I, R & t in the equation:

\sf \implies H =  {5}^{2}  \times 100 \times 2 \\  \\ \sf \implies H = 25 \times 100 \times 2 \\  \\  \sf \implies H = 5000  \\  \\ \sf \implies H = 5 \: kWh

\therefore

Heat developed (H) in the electric iron = 15 kWh

3 0
3 years ago
For the following types
Ad libitum [116K]

Answer: In order of increasing frequency and decreasing wavelength these are: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

4 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
4 years ago
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi
maria [59]

Answer:

a) a = 5.03x10¹³ m/s²

b) V_{f} = 4.4 \cdot 10^{5} m/s

Explanation:    

a) The acceleration of the positron can be found as follows:

F = q*E    (1)

Also,

F = ma    (2)

By entering equation (1) into (2), we have:

a = \frac{F}{m} = \frac{qE}{m}

<u>Where:</u>

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C    

E: is the electric field = 286 N/C

a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2}

b) The positron's speed can be calculated using the following equation:

V_{f} = V_{0} + at

<u>Where</u>:

V_{f}: is the final speed =?

V_{0}: is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s

I hope it helps you!

4 0
3 years ago
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