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2 years ago

10

A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11-

Calculate the magnitude of the acceleration of the car during this 5.0-second time interval.

1 answer:

Aleks04 [339]2 years ago

7 0

Answer:

$a=2.4\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2$

So, the acceleration of the car is $2.4\ m/s^2$ .

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2 years ago

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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Explanation:

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>

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

Take the natural logarithm of both sides:

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

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$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

<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:

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

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

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