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Zina [86]
3 years ago
10

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

pacitor is zero. If a 12-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t
Physics
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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If it takes Jupiter 13 years to orbit the Sun. How long (in years) will it take Jupiter to go once around the sky as viewed from
irakobra [83]

We have that the time it will take Jupiter to around the sky  is mathematically given as

Psyn=295days

<h3>Time for Jupiter to around the sky</h3>

Question Parameters:

If it takes Jupiter 13 years to orbit the Sun.

Generally, The earth  moved a bit further, one Jupiter period ago, This is called synodic period

Psyn=\frac{psid}{Psid-1}\\\\Psyn=\frac{13}{13-1`}\\\\Psyn=1.08year\\\\

Psyn=295days

For more information on Arithmetic visit

brainly.com/question/22568180

3 0
2 years ago
For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot
Margarita [4]

Answer:

(a) -202 m/s²

(b) 198 m

Explanation:

Given data

  • Initial speed (v₀): 283 m/s

\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s

  • Final speed (vf): 0 (rest)
  • Time (t): 1.40 s

(a) The acceleration (a) is the change in the speed over the time elapsed.

a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²

(b) We can find the distance traveled (d) using the following kinematic expression.

y = v₀ × t + 1/2 × a × t²

y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²

y = 198 m

3 0
2 years ago
1. Balanced forces acting on an
SVETLANKA909090 [29]

Answer:

1. A

2. B

Explanation:

1. As long as The force is equally distributed the object will not move .

2. If the force in one side is greater than that of the other the object will move

3 0
3 years ago
Read 2 more answers
A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th
SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

   Volume = 470 cm³

   Density = 3.55 g/cm³

   Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

   Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

   Volume = 470 cm³

   Density of liquid = 1 g/cm³

   \texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N

c) Mass of the water displaced = Volume of body x Density of liquid

   Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

   Weight of rock under water = 16.37 - 4.61  =11.76 N

3 0
3 years ago
A future use of space stations may be to provide hospitals for severely burned persons. It is very painful for a badly burned pe
inessss [21]

Answer:

1.5min

Explanation:

To solve the problem it is necessary to take into account the concepts related to Period and Centripetal Acceleration.

By definition centripetal acceleration is given by

a_c = \frac{V^2}{r}

Where,

V = Tangencial velocity

r = radius

With our values we know that

a_c = \frac{V^2}{r}

\frac{V^2}{r} = \frac{1}{10}g

Therefore solving to find V, we have:

V = \sqrt{\frac{1}{10}g*r}

V = \sqrt{\frac{9.81*200}{10}}

V = 14m/s

For definition we know that the Time to complete are revolution is given by

t = \frac{Perimeter}{Speed}

t = \frac{2\pi R}{V}

t = \frac{2\pi * 200}{14}

t = 1.5min

6 0
3 years ago
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