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2 years ago

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A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

me constant for the process is about:

1 answer:

vichka [17]2 years ago

5 0

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

$\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}$

$Ln(\frac{1}{2}) = -\frac{0.01}{RC}$

$- RC = \frac{0.01}{Ln(1/2)}$

$RC = 0.014s$

We know that RC is equal to the time constant, then

$T = RC = 0.014s = 14ms$

Therefore the time constant for the process is about 14ms

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If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will attract each other.

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1 year ago

Convert 4 µL to mL<br> Please help meeeee

stiv31 [10]

Answer: the answer would be four thousand

Explanation: hope this helps

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2 years ago

Read 2 more answers

a crane lifts four pallets of bricks each of which weigh 5000 N. the crane lifts each pallet a height of 30m. the crane takes 4

love history [14]

Answer:

625 W

Explanation:

Applying

P = W/t.................... Equation 1

Where p = power, W = Work, t = time

But,

W = Force (F) × distance (d)

W = Fd........................ Equation 2

Substitute equation 2 into equation 1

P = Fd/t.................... Equation 3

From the question,

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

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I Hope this helps, greetings ... Dexteright02! =)

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3 years ago

An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a

Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile $a=3m/sec^2$

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So $v=5.13+3\times 4.9=19.83m/sec$

So the final velocity is 19.83 m/sec

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2 years ago