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Vilka [71]
3 years ago
5

A damped LC circuit consists of a 0.17 μF capacitor and a 15 mH inductor with resistance 1.4Ω. How many cycles will the circuit

oscillate before the peak voltage on the capacitor drops to half its initial value? Express your answer using two significant figures.
Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

number of cycles = 4.68 × 10⁴ cycles

Explanation:

     In damped RLC oscillation

voltage (V(t)) = V_o\ e^{-\dfrac{tR}{2L}}............(1)

given,

C = 0.17μF = 0.17 × 10⁻⁶ F

R = 1.4 Ω

L = 15 m H = 15 × 10⁻³ H              V(t) = V₀/2

From the equation (1)

\dfrac{V_0}{2} = V_0\ e^{-\dfrac{tR}{2L}}

 2 = e^{\dfrac{tR}{2L}}

taking log both side

ln ( 2 ) = \dfrac{tR}{2L}

t = \dfrac{2 L ln(2)}{R}

t = \dfrac{2 \times 15 ln(2)}{1.4}

t = 14.85 sec

time period

T= 2\pi \sqrt{LC}

T= 2\pi \sqrt{0.015 \times 0.17 \times 10^{-6}}

T = 3.172 × 10⁻⁴

number of cycle =\dfrac{t}{T}

                           =  \dfrac{14.85}{3.172 \times 10^{-4}}

  number of cycles = 4.68 × 10⁴ cycles

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3 years ago
How would you describe the appearance of the substance after the phase change?
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Answer:

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3 years ago
A 0.00287 C charge is 6.52 m from
yKpoI14uk [10]

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7 0
3 years ago
In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0
3 years ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
3 years ago
Read 2 more answers
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