PHYSICS COLLEGE

A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Answers

Answer 1

Answer:

The energy stored is $E_s = 0.0064 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 8 \ \mu F = 8*10^(-6) \ F$

The resistance is R = 3.00-Ω

The emf is $E_t = 70.0 V$

The power is P = 300 W

Generally the total emf is mathematically represented as

$E_t = E_c + E_r$

Here $E_c$ is the emf across that capacitor which is mathematically represented as

$E_c = (q)/(C)$

and $E_r$ is the emf across the resistor which is mathematically represented as

$E_r = √(P R)$

$E_t = √(PR) + (q)/(C)$

=> $q = C[E_t - √(PR) ]$

Generally the energy stored in a capacitor is mathematically represented as

$E_s = (q^2)/(2C)$

=> $E_s = ([C [ E_t - √(PR) ]]^2)/(2C)$

=> $E_s = ([8.0*10^(-6) [ 70 - √(300 * 3))/(2 *(8.0*10^(-6)))$

=> $E_s = 0.0064 \ J$

Answer 2

Answer:

Final answer:

The energy stored in the capacitor is 0 J.

Explanation:

When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V

At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.

Learn more about Energy stored in a capacitor here:

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The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?

Answers

Answer:

$10.99* 10^6\ V/m$

Explanation:

Given:

Potential difference across the membrane (ΔV) = 0.078 V

Thickness of the membrane (Δx) = 7.1 × 10⁻⁹ m

Magnitude of electric field (|E|) = ?

We know that, the electric field due to a potential difference (ΔV) across a distance of Δx is given as:

$E=-(\Delta V)/(\Delta x)$

So, the magnitude of the electric field is calculated by ignoring the negative sign and thus is given as:

$|E|=(\Delta V)/(\Delta x)$

Plug in the given values and solve for '|E|'. This gives,

$|E|=(0.078\ V)/(7.1* 10^(-9)\ m)\n\n|E|=10.99* 10^6\ V/m$

Therefore, the magnitude of the electric field in the membrane is $10.99* 10^6\ V/m$ .

According to the U.S. Green Building Council, what percentage of the world’s energy use and greenhouse gas emissions can be attributed to buildings?

Answers

According to the US green building council, the US building account for 39% of world primary energy consumption . Electricity has approximately 78% of total building energy consumption and also contributes to GHG emissions

Answer:

40%

Explanation: United States Green Building Council is a body aimed at ensuring reduced green house gas emissions from activities taking place in building. they carry out surveys, carry out enlightenment activities and release the reports of and trending green house emission issues all these are to guarantee safe and healthy living for all. A total of 40% of Green house emissions are from buildings from the construction stage to it usage.

The radius of a typical human eardrum is about 4.15 mm. Calculate the energy per second received by an eardrum when it listens to sound that is at the threshold of hearing, assumed to be 1.20E-12 W/m2

Answers

The energy per second received by an eardrum is $6.4884 * 10^(-17) watt$

Calculation of the energy per second;

The area should be

$= \pi r^2\n\n= 3.14 * 0.00415m\n\n= 5.407 * 10^(-5)m^2$

Now

The power should be

$= 1.2 * 10^(-12) * 5.407 * 10^(-5)\n\n= 6.4884 * 10^(-17) watt$

Learn more about the energy here: brainly.com/question/14338287

Answer:

Power energy per second will be equal to $6.4884* 10^(-17)watt$

Explanation:

We have given radius of human eardrum r = 4.15 mm = 0.00415 m

Intensity at threshold of hearing $I=1.2* 10^(-12)w/m^2$

Area is given by $A=\pi r^2=3.14* 0.00415^2=5.407* 10^(-5)m^2$

We know that power is given by $P=I* A=1.2* 10^(-12)* 5.407* 10^(-5)=6.4884* 10^(-17)watt$

So power energy per second will be equal to $6.4884* 10^(-17)watt$

Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer:

$4U_0$

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P= $U_0$

We know that potential at point P= $V=(U)/(q)$

$U=Vq$

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially , $V=(U_0)/(q_0)$

New charge, q= $4q_0$

Then, work done,U= $(U_0)/(q_0)* (4q_0)=4U_0$

Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P= $4U_0$

Answer:

$4U_(0)$

Explanation:

V = u / q,

Work = P = V

1U / 1/4 = 4U

An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an instant, exceed 16 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit?A)8 sqrt 2 V

B) 16 sqrt 2 V

C) 256 V

D) 8

Answers

Answer:

A) $V_(rms)=8√(2) V$

Explanation:

Maximum voltage = $V_(max)=16 V$

Maximum voltage and rms voltage are related to each other by

$V_(max)=V_(rms) * √(2) \nV_(rms)=(V_(max))/( √(2))\nV_(rms)=(16)/(√(2)) \nV_(rms)=8√(2) V$

How much electrical energy is used by a 400 W toaster that is operating for 5minutes?
A. 2000 J
B. 75,000 J
C. 120,000 J
D. 300,000 J

Answers

The electrical energy used by a 400 W toaster that is operating for 5 minutes will be 120,000 J.Option C is correct.

What is the power output?

The rate of the work done is called the power output. It is denoted by P.Its unit of a watt. It is the ratio of the work done or the enrgy to the time period.

The given data in the problem is;

E is the electrical energy

P is the power output = 400 W

t is the time period = 5 minutes

The power output is given as;

$\rm P= (E)/(t) \n\n\ E= P * t \n\n\ E= 400 * 300 \n\n\ E=120,000 \ J$