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3 years ago

6

If a car is moving to the left with constant velocity, one can conclude that If a car is moving to the left with constant veloci

ty, one can conclude that there must be no forces applied to the car. the net force applied to the car is directed to the left. the net force applied to the car is zero. there is exactly one force applied to the car.

2 answers:

FinnZ [79.3K]3 years ago

6 0

Answer:

Explanation:

If a car is moving to the left with constant velocity, one can conclude that

there must be no forces applied to the car . The net force applied to the car is zero.

Since there is no acceleration ,no net force is applied , hence no force acts on the car.

egoroff_w [7]3 years ago

6 0

Answer:

the net force applied to the car is zero

Explanation:

Constant velocity means accleration = 0

F = ma

F = m(0)

F = 0

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Answer:

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

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

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

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>

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

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

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

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

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