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faltersainse [42]

4 years ago

14

A circular loop of wire is positioned so the normal to its area points vertically upward. A bar magnet is dropped from above the

loop so that its north pole falls through the loop first. When looking at the loop from above, in which direction is the induced current in the loop just as the north pole of the magnet enters the loop?
a. Clockwise
b. There is no induced current Incorrect.

1 answer:

Kruka [31]4 years ago

8 0

B cause there is no induced current incorrect

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In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc

Harman [31]

Answer:

$A=\frac{A_0}{\sqrt 3}$

Explanation:

Initial amplitude= $A_0$

We are given that

Amplitude after 2 s=A= $\frac{1}{3}A_0$

We have to find the amplitude after 1 s.

We know that amplitude at any time t

$A=A_0e^{-\alpha t}$

Using the formula

$\frac{A_0}{3}=A_0e^{-2\alpha}$

$\frac{1}{3}=e^{-2\alpha}$

$3=e^{2\alpha}$

$ln 3=2\alpha$

$\alpha =\frac{ln 3}{2}=ln\sqrt 3$

$e^{\alpha}=\sqrt 3}$

When t=1 s

$A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}$

8 0

4 years ago

Describe an object that emits radiation in Astronomy

DochEvi [55]

Answer:

the sun

Explanation:

the sun emits all types of electromagnetic radiation but a vast majority of it comes in the for of in the for of visible light ultraviolet rays and infrarays

4 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

Read 2 more answers

A girl on a motorbike passes by at a speed of 15 m/sec. her mass is 55 kg what is her kinetic energy

MrRissso [65]

$K = m*v^{2} / 2

K = 50 * 225 / 2

K = 5625 J$

5 0

4 years ago

Give a quantitative definition of being in contact.

san4es73 [151]

Two things are said to be in contact if the smallest distance between a point in one of them and a point in the other one is zero.

7 0

3 years ago