PHYSICS COLLEGE

A 13.0-Ω resistor, 13.5-mH inductor, and 50.0-µF capacitor are connected in series to a 55.0-V (rms) source having variable frequency. If the operating frequency is twice the resonance frequency, find the energy delivered to the circuit during one period.

Answers

Answer 1

Answer:

Answer:

E = 0.13 J

Explanation:

At resonance condition we have

$\omega = \sqrt{(1)/(LC)}$

$\omega = \sqrt{(1)/((13.5 * 10^(-3)){50* 10^(-6))}}$

$\omega = 1217.2 rad/s$

now if the frequency is double that of resonance condition then we have

$x_L = 2\omega L$

$x_L = 2(1217.2)(13.5* 10^(-3))$

$x_L = 32.86 ohm$

now we have

$x_c = (1)/(2(1217)(50* 10^(-6)))$

$x_c = 8.22 ohm$

now average power is given as

$P = i_(rms)V_(rms)* (R)/(z)$

$P = (55)/(√((32.86 - 8.22)^2 + 13^2)))(55)* (13)/(√((32.86 - 8.22)^2 + 13^2))$

$P_(avg) = 50.67 Watt$

Now time period is given as

$T = (2\pi)/(\omega)$

so total energy consumed is given as

$E_(avg) = 50.67((2\pi)/(2(1217.2)))$

$E = 0.13 J$

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Answers

Answer: ∆p2 = 2* ∆p1

Explanation:

Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L

Therefore,

∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

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Eliminating L1 we have,

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Assuming the same initial conditions as described in FNT 2.2.1-1, use the energy-interaction model in two different ways (parts (a) and (b) below) to determine the speed of the ball when it is 4 meters above the floor headed down: a) Construct a particular model of the entire physical process, with the initial time when the ball leaves Christine’s hand, and the final time when the ball is 4 meters above the floor headed down.
b) Divide the overall process into two physical processes by constructing two energy-system diagrams and applying energy conservation for each, one diagram for the interval corresponding to the ball traveling from Christine’s hand to the maximum height, and then one diagram corresponding to the interval for the ball traveling from the maximum height to 4 meters above the floor headed down.
c) Did you get different answers (in parts (a) and (b)) for the speed of the ball when it is 4 meters above the floor headed down?

Answers

Answer:

(a). Vf = 7.14 m/s

(b). Vf = 7.14 m/s

(c). same answer

Explanation:

for question (a), we would be applying conservation of energy principle.

but the initial height is h = 1.5 m

and the initial upward velocity of the ball is Vi = 10 m/s

Therefore

(a). using conservation law

Ef = Ei

where Ef = 1/2mVf² + mghf ........................(1)

also Ei = 1/2mVi² + mghi ........................(2)

equating both we have

1/2mVf² + mghf = 1/2mVi² + mghi

eliminating same terms gives,

Vf = √(Vi² + 2g (hi -hf))

Vf = √(10² + -2*9.8*2.5) = 7.14 m/s

Vf = 7.14 m/s

(b). Same process as done in previous;

Ef = Ei

but here the Ef = mghf ...........(3)

and Ei = 1/2mVi² + mghi ...........(4)

solving for the final height (hf) we relate both equation 3 and 4 to give

mghf = 1/2mVi² + mghi ..............(5)

canceling out same terms

hf = hi + Vi²/2g

hf = 1.5 + 10²/2*9.8 = 6.60204m ............(6)

recalling conservation energy,

Ef = Ei

1/2mVf² + mghf = mghi

inputting values of hf and hi we have

Vf = √(2g(hi -hf)) = 7.14 m/s

Vf = 7.14 m/s

(c). From answer in option a and c, we can see there were no changes in the answers.

(PLEASE HELP ITS DUE SOON ILL MARK BRAINLIEST AND 5 STARS & PLEASE SHOW WORK!!)(And the answer is not 44 I already tried that and it doesn’t start with 4 either)

Answers

Lol I would help you but I have no clue

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Answers

Answer:

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