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Law Incorporation [45]
3 years ago
12

Select three possible applications of a capacitor. Select all that apply.

Physics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

I'm pretty sure it's all of them i'm not completely sure though hope it helps anyways! :)

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WHO IS GOOD AT PHYSICS, ELECTRIC CIRCUITS?​
Vanyuwa [196]

Answer: An electric circuit is a representation of how current moves from the source of the current( example a battery or a cell) through resistors and other devices before entering the source.

Explanation:

The two important types of electrical circuit includes:

--> open circuit and

--> closed circuit.

A circuit is said to be open when the electrical source, such as the battery or the cell, is not connected to any external conductor (or resistance). In this situation, any voltmeter connected across the terminal of the cell measures the total driving force of the cell.

In this type of circuit, current cannot flow from one end of power source to the other due to interruptions.

A circuit is said to be closed if the source of electricity is connected to an external conductor through which current is passed.

In a closed circuit, there is complete electrical connection which allows current to flow or circulate. Here, part of the total driving force of the source is used to drive current through the external resistance and the difference is used to overcome the internal resistance of the battery.

8 0
3 years ago
Describe how work done is related to a change in volume of a fluid. 100 points
Nookie1986 [14]

Answer: work is the pressure acting through the change in volume

Explanation:

In the same way that work is defined as force operating over a distance,  work is the pressure acting through the change in volume.   Pressure is comparable to force in pressure–volume work, while volume is analogous to distance in the classic definition of work.

4 0
2 years ago
Read 2 more answers
How much force must a locomotive exert on a 12840-kg boxcar to make it accelerate forward at 0.490 m/s2?
ddd [48]

Data given:

m=12840kg

a=0.490m/s²

Formula:

F=ma

Solution:

F= 12840kg×0.490m/s²

F=6291.6N

6 0
2 years ago
Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter
alukav5142 [94]

Answer: 0.258

Explanation:

The resistance R of a wire is calculated by the following formula:

R=\rho\frac{l}{s}    (1)

Where:

\rho is the resistivity of the material the wire is made of. For aluminium is \rho_{Al}=2.65(10)^{-8}m\Omega  and for copper is \rho_{Cu}=1.68(10)^{-8}m\Omega

l is the length of the wire, which in the case of aluminium is l_{Al}=12m, and in the case of copper is l_{Cu}=30m

s is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

s=\pi{(\frac{d}{2})}^{2}  (2) Where d  is the diameter of the circumference.

For aluminium wire the diameter is  d_{Al}=2.5mm=0.0025m  and for copper is d_{Cu}=1.6mm=0.0016m

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}

s_{Al}=0.000004908m^{2}   (3)

<u>For copper:</u>

s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}

s_{Cu}=0.00000201m^{2}    (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}     (5)

R_{Al}=0.0647\Omega     (6)  Resistance of aluminium wire

<u>Copper wire:</u>

R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}     (6)

R_{Cu}=0.250\Omega     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

Ratio=\frac{R_{Al}}{R_{Cu}}   (8)

\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}   (9)

Finally:

\frac{R_{Al}}{R_{Cu}}=0.258  This is the ratio

3 0
3 years ago
What is the net force on this object?​
Sergio039 [100]
200N

Explanation:
600N-400N = 200N
6 0
3 years ago
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