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3 years ago

What part of an atom is responsible for producing magnetic fields

Physics

1 answer:

Nezavi [6.7K]3 years ago

6 0

Electrons is the part of the atom that creats a magnetic feild

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What happens when a chemical change takes place?

mart [117]

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3 years ago

Read 2 more answers

A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.

gulaghasi [49]

The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force $F_f$ acting against the motion. Since their resultant must be zero, we have:
$F-F_f = 0$
The frictional force is
$F_f = \mu mg$
where
$\mu=0.4$ is the coefficient of kinetic friction
$mg=400 N$ is the weight of the block.

Substituting these values, we find the magnitude of the force F:
$F=F_f = \mu mg=(0.4 )(400 N)=160 N$

4 0

3 years ago

6.) What is

Feliz [49]

Answer:

100 V

Explanation:

Hi there!

Ohm's law states that $V=IR$ where V is the voltage, I is the current and R is the resistance.

Plug the given information into Ohm's law (R=50, I=A)

$V=IR\\V=(2)(50)\\V=100$

Therefore, the voltage across this current is 100 V.

I hope this helps!

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3 years ago

What is the latin name for the hydra constellation

slava [35]

The latin name for hydra constellation is "Water snake"

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3 years ago

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago