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anastassius [24]
3 years ago
9

Two bar magnets are placed side by side so that the north pole of one magnet is next to the south pole of the other magnet. If t

hese magnets are then pushed toward a coil of wire, would you expect an emf to be induced in the coil? Explain your answer.
(REAL ANSWERS PLEASE! I ALWAYS GET PEOPLE STEALING MY POINTS FOR THE HECK OF IT! IT'S A WASTE OF YOUR TIME! I will just report you if you do.)
Physics
1 answer:
Anika [276]3 years ago
4 0
The induced voltage is proportional to the time derivative of the net magnetic flux density. This is called faradays law of magnetic induction. Because you have a north and a south pole right next to each other (assuming they are magnets of equal strength) the total net magnetic flux will be zero. This is because the north end of one magnet adds a positive flux while the south end of the other magnet contributes negative flux.
As a result, the induced EMF will be zero. (I'm almost done with my PhD in electromagnetics engineering, so you can take this answer to the bank :) )
You might be interested in
Two loudspeakers on a concert stage are vibrating in phase. A listener 50.5 m from the left speaker and 26.0 m from the right on
mixer [17]

To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.

From the definition of constructive interference we know that

\Delta x = n \lambda \rightarrow \lambda = \frac{\Delta x}{n}

Where

\Delta x = Distance between speakers

n = Integer which represent he number of repetition of the spectrum

\lambda = Wavelength

At the same time the frequency is subject to the form,

f = \frac{v}{\lambda}

Where

v = 343m/s \rightarrow Velocity (of the sound at this case)

From our given values we have to \Delta x is

\Delta x = 50.5m-26m

\Delta x = 24.5m

The wavelength would be subject to the sound spectrum therefore for n = 1,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{1}

\lambda = 24.5m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{24.5}

f = 14Hz

For the value of n = 2,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{2}

\lambda = 12.25m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{12.25}

f = 28Hz

For n = 3,

\lambda = \frac{\Delta x}{n}

\lambda = \frac{24.5}{3}

\lambda = 8.167m

Then the frequency would be,

f = \frac{v}{\lambda}

f = \frac{343}{12.25}

f = 42Hz

From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz

5 0
3 years ago
A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat
Triss [41]

Answer:

2.1406 ×10^6 m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

qV=\frac{1}{2}mv^2

here V is the potential difference  

we know that mass of proton = 1.67×10^{-27} kg

we have given speed =50000m/sec

so potential difference V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045

now mass of electron =9.11×10^{-31}

so for electron

\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec

so the velocity of electron will be 2.1406×10^6 m/sec

4 0
4 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A space probe is traveling in outer space with a momentum that has a magnitude of 5.16 x 107 kg·m/s. A retrorocket is fired to s
kotegsom [21]

Answer:

p_{f}=99x10^6 N*s

Explanation:

When a force is applied to an object over a certain period of time, it is said to apply an impulse to the object. The magnitude of the impulse is given by:

J=F*Δt

When an impulse is provided to an object, the momentum of the object will change, as in:

J=Δp=p_{f}-p_{i}

So given:

Initial momentum

P_{i}=5.19x10^7 \frac{kg*m}{s}

Applied force

F=2.0x10^6 N

J=p_{f}-p_{i}

F*Δt=p_{f}-p_{i}

p_{f}=2.0x10^6N*12s+7.5x10^7 N*s

p_{f}=99x10^6 N*s

3 0
4 years ago
Which resource would be the best choice to learn more information about studying martial arts?
Mekhanik [1.2K]

I think videos help more than books when it comes to martial arts. You can get it on YouTu.be if your teacher doesn't teach you online, in this pandemic.

3 0
3 years ago
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