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Soloha48 [4]
3 years ago
7

I really need help on this one

Physics
1 answer:
VLD [36.1K]3 years ago
4 0

the answer is A its correct because i looked it up

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A series LRC circuit consists of a 12.0-mH inductor, a 15.0-µF capacitor, a resistor, and a 110-V (rms) ac voltage source. If th
finlep [7]

Answer:

61.85 ohm

Explanation:

L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm

Let ω0 be the resonant frequency.

\omega _{0}=\frac{1}{\sqrt{LC}}

\omega _{0}=\frac{1}{\sqrt{12\times 10^{-3}\times 15\times 10^{-6}}}

ω0 = 2357 rad/s

ω = 2 x 2357 = 4714 rad/s

XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm

Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm

Impedance, Z = \sqrt{R^{2}+\left ( XL - Xc \right )^{2}}

Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm

Thus, the impedance at double the resonant frequency is 61.85 ohm.

7 0
3 years ago
A grating that has 3900 slits per cm produces a third- order fringe at a 28.0° angle. What wavelength of light is being used? Ex
Zolol [24]

Explanation:

Given that,

Number of slits per cm, N=3900\ lines/cm=390000\lines/m

The third fringe is obtained at an angle of, \theta=28

We need to find the wavelength of light used. The grating equation is given by :

d\ sin\theta=n\lambda

d=\dfrac{1}{N}=\dfrac{1}{390000}

d=2.56\times 10^{-6}\ m

\lambda=\dfrac{d\ sin\theta}{n}, n = 3

\lambda=\dfrac{2.56\times 10^{-6}\times \ sin(28)}{3}

\lambda=4.006\times 10^{-7}\ m

\lambda=400\ nm

So, the wavelength of the light is 400 nm. Hence, this is the required solution.

4 0
4 years ago
Mr.pointer has as mass of 50kg. his is at the top of a 10m hill. what is his PE
iragen [17]

Answer:

PE=4900J

Hope it helps please mark me as brainliest

Explanation:

P.E=m×g×h

Given m = 50 Kg , h = 10 m.

P.E=50×10×9.8

P.E=50×98=4900J

Hence increase in potential energy is 4900J.

6 0
3 years ago
DOES IT MATTER HOW WE TRAIN OUR MUSCLE
Sidana [21]

Answer:

it depends most of the time but i don't really have a clue either

3 0
3 years ago
A pendulum is 0.760 m long, and the bob has a mass of 1,00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. The tens
bixtya [17]

Answer:

  T = 13.17 N

the correct one is there is a digested centripetal acceleration towards upward

Explanation:

For this exercise we can use Newton's second law at the bottom of the pendulum's path

we will assume that the upward direction is positive

              T - W = m a

in this case it is describing a circle the acceleration is central

              a = v² / R

where the radius of the trajectories equals the length of the pendulum

             R = L

we substitute

            T = mg + m v² / L

             T = m (g + v² / L)

When reviewing the different statements, the correct one is there is a digested centripetal acceleration towards upward

let's calculate the tension of the rope

             T = 1 (9.8 + 1.6 2 / 0.760)

             T = 13.17 N

6 0
3 years ago
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