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3 years ago

Generators use a magnetic field of which of the following to produce an electric current?

Physics

2 answers:

Scrat [10]3 years ago

8 0

The answer is going to be magnetic domain

Bas_tet [7]3 years ago

3 0

Answer:

<u><em>The answer is</em></u>: <u>Permanent magnet.</u>

<u />

Explanation:

<em>The rotor can be constituted by a</em><u> permanent magnet or more frequently, by an electromagnet</u>. <em>An electromagnet is a device formed by a coil wound around a ferromagnetic material through which a current is circulated, which produces a magnetic field</em>.

<u><em>The answer is</em></u>: <u>Permanent magnet.</u>

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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

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

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:

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

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

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:

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

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

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

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

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:

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

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

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

7 0

3 years ago

During low-intensity activity, your body obtains most of its energy from

Gemiola [76]

Carbohydrates, in cellular respiration.

3 0

2 years ago

It would really help if anyone could answers please and thanks

mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0

3 years ago

Read 2 more answers

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).

melamori03 [73]

Refer to the diagram shown below.

The given data is

mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer: (1.5, 0.5) m

6 0

3 years ago

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

7.5

×m

...........................(

)

in Above equation in x and t. Separating the variables and integrating,

∫

/7.5

×=

∫

−

4.7619

Here C =constant of integration.

0 at t

, we get: C

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

−

4.7619

(

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

−

4.7619

= −

3.7619

or x

7.15

(b) To find the jogger's acceleration in m

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

7.5×

Now differentiating above equation w.r.t time we get:

−

0.675

the joggers acceleration is :

−

0.675

−

4.34

6 in equation (2). We get:

−

4.7619

(

−

(

0.04

6 )

10=

−

4.7619

0.832

6 0

3 years ago