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ollegr [7]
2 years ago
10

According to the Law of Conservation of

Physics
2 answers:
timurjin [86]2 years ago
8 0

Answer:

option b is the answer which is 150

Maksim231197 [3]2 years ago
4 0
The answer would be B) 150 joules
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A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored i
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Explanation:

It is given that,

Inductance, L=40\ mH=40\times 10^{-3}\ H  

RMS value of voltage, v_{rms}=120\ V

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,  

The current flowing through the inductor is given by :

I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})

I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})

I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})

I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})    

I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})

I = 0.091 A

Energy stored in the inductor is, U=\dfrac{1}{2}LI^2

U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2

U = 0.000165 Joules

Hence, this is the required solution.

6 0
3 years ago
A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci
Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

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3 years ago
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Number 19 is frequency and not sure which question you asked!!!??
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Robotic technology is used for servicing spacecrafts. One group of scientists used robotic technology to design this model of a
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Many of the innovations developed by space programs had applicability beyond their original use. This is one such example, in which robotic technology used for servicing spaceships is helping to improve surgeries here on Earth. 
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3 years ago
PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m
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Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)

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3 years ago
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