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3 years ago

When a wire with a current is placed in a magnetic field,

2 answers:

8 0

When a current carrying wire is placed in a magnetic field, it experiences a mechanical force. This was discovered by Michael Faraday in 1821. The wire will not experience any mechanical force if it is placed parallel to the magnetic field. And the force will be maximum if the current carrying wire is placed in a direction perpendicular to that of the magnetic field. The force exerted on the wire is known as the Lorentz force.

cupoosta [38]3 years ago

7 0

Answer:

It makes it fast and has a bigger field

Explanation:

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A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little l

Y_Kistochka [10]

The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E = P.E
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3 years ago

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atroni [7]

3.8 is the smallest number because if you multiple the others would be large

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3 years ago

You move a 25 N object 5 meters. If it takes 8 s how much power did you do?

klasskru [66]

Answer:

15.625 watts

Explanation:

Recall that power is defined as the worked performed per unit of time:

Power = Work / time

The work done is Force * distance, so in our case the work is:

Work = 25 M * 5 m = 125 J

Then the power will be:

Power = 125 J / 8 sec = 15.625 watts

6 0

2 years ago

The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$