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3 years ago

A 63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s.

Physics

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

10.1 m/s

Explanation:

Every moving body or an object has a momentum. The simplest way to calculate it is to multiply the mass of an object with its velocity (p = m•v). So, just after this ice skater finished her performance, her momentum was 63 kg•10.8 m/s which equals to 680.4 kg•m/s. Then, she was given a huge bouquet which changed her total mass and speed. The thing we need to know is that the momentum before receiving the bouqet and after that are the same. With the bouqet, the mass increased and now is 63 kg + 4.4 kg = 67.4 kg. If p=m•v, then v=p/m, so her new speed is 680.4/67.4 = 10.0949, or approximately 10.1 m/s.

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3 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

-BARSIC- [3]

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 $\displaystyle \frac{7}{9}\, q_0$ .

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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:

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

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

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:

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

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

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

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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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3 years ago

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Softa [21]

Answer:

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$B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}$

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r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

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3 years ago