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2 years ago

5

The airport has a new plane that can go from 0 miles per hour to 100 miles per hour in 10 seconds. What is being described about

the new plane?

1 answer:

Nonamiya [84]2 years ago

6 0

<h3 />

<h2 /><h2 /><h2>Acceleration : </h2>

Acceleration is a vector quantity that is defined as ❝ rate of change in velocity ❞

And as per the question, the plane can go from 0 miles to 100 miles/hour in 10 seconds that shows rate of change in its velocity. that is :

from 0 miles/hour to 100 miles/hour. hence Acceleration is the quantity being described about the new plane.

$\qquad \sf \dashrightarrow \: a = \dfrac{v - u}{t}$

[ v = final velocity, u = initial velocity, t = time ]

$\qquad \sf \dashrightarrow \: a = \dfrac{100 - 0}{10}$

$\qquad \sf \dashrightarrow \: a = \dfrac{100}{10}$

$\qquad \sf \dashrightarrow \: a = 10 \:$

So, the Acceleration of the new plane is 10 miles/hour²

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A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

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Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

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Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

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In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

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<h3>a)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

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<h3>b)</h3>

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

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