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In an electric field, a capacitor is a device that stores electrical energy. It has two terminals and is a passive electrical component.

Capacitance refers to a capacitor's effect.
The terminals in a capacitor are connected to two metal plates that are spaced apart by a dielectric, a non-conducting material.
Two pieces of aluminum foil, along with a piece of paper, can be used to create a capacitor (and some electrical clips).
It won't have a lot of storage capacity, so it won't be a great capacitor, but it will still function.

<h3>What are the different types of capacitors?</h3>

The numerous kinds of capacitors and their applications.

Frequently utilized in radio tuning circuits is air.

Mylar: The material most frequently used in timer circuits for clocks, alarms, and counting.

Good for high-voltage applications is glass.

Ceramic is used in X-ray, MRI, and other high-frequency devices.

a powerful capacitor fuels hybrid and electric vehicles.

Learn more about capacitors here:

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2 years ago

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marusya05 [52]

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3 years ago

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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

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2 years ago

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