When discharging equivalent of 0% starting voltage will the capacitor reach after 5RC.
V(t) = V0e-t/RC = V0b initially.
<h3>What is capacitor?</h3>
- 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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Answer:
Honda won by 0.14 s
Explanation:
We are given that
Distance =S=200 m
Initial velocity of Honda=u=0m/s
Initial velocity of Porsche=u'=0m/s
Acceleration of Honda=
Acceleration of Porsche's=
Time taken by Honda to start=1 s
Substitute the values
Time taken by Honda=11.55 s
Now, time taken by Porsche
Total time taken by Porsche=10.69+1=11.69 s
Because it start 1 s late
Time taken by Honda is less than Porsche .Therefore, Honda won and
Time =11.69-11.55=0.14 s
Honda won by 0.14 s
Answer:
B, C, F
Explanation:
Conduction is heat transfer by means of molecular agitation within a material without any motion of the material as a whole. If one end of a metal rod is at a higher temperature, then energy will be transferred down the rod toward the colder end because the higher speed particles will collide with the slower ones with a net transfer of energy to the slower ones.
I would provide the source, but this is not allowed as it seems the website does not encourage learning of well-established materials.
Let me know if you require a further explanation
Answer:
SURE!!!...
But what to calculate!!!....
