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3 years ago

What does the balloon of the air capacitor represent in an electrical capacitor?

Physics

2 answers:

bulgar [2K]3 years ago

6 0

Answer:

The balloon prohibits the flow of air through the air capacitor.

Explanation:

Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.

Alex_Xolod [135]3 years ago

6 0

Answer: The insulation between plates

Explanation:i got it right on the test

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How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

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2 years ago

What player(s) can take a goal kick?

inysia [295]

Answer:

c) goalie

Explanation:

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3 years ago

Read 2 more answers

A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga

BartSMP [9]

Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

So total work done in expansion will be $3.60\times 10^5J$

5 0

2 years ago

a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric fiel

Lesechka [4]

Answer:

-1748*10^N/C

Explanation:

See attached file

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3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

2 years ago