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3 years ago

Which of the following are likely to be affected by a magnet? Check all that apply.

Physics

2 answers:

Gelneren [198K]3 years ago

6 0

<h3><u>Answer;</u></h3>

C.An access card you scan for entry into a building

<h3><u>Explanation;</u></h3>

<em><u>Magnets are materials that have the capability of producing magnetic field.</u></em> Magnets attract magnetic material such as iron and cobalt while they don't have an effect on non-magnetic materials such as wood and plastic.
<em><u>Access cards are plastic cards with a microchip or a magnetic strip containing an encoded data that is read for an individual to gain entry into a building.</u></em>
<em><u>Due to the presence of a magnetic strip that is swiped through the mechanism in the door, they may be affected by magnetic field from the magnets.</u></em>

stellarik [79]3 years ago

5 0

I think the correct answers from the choices listed above are options B and C. A credit card and an access card you scan for entry into a building are most likely to be affected by a magnet. Hope this answers the question. Have a nice day.

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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

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1 year ago

Which organisms play roles similar to organisms in water ecosystem

lina2011 [118]

Is there a multiple choice?

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3 years ago

If you pull horizontally on a desk with a force of 150 N and the desk doesn't move, the friction force must be 150 N. Now if you

s344n2d4d5 [400]

Answer:

The friction force is 250 N

Explanation:

The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):

$\sum F=ma$

In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:

$\sum F = 0$ (1)

We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:

- the pull, forward, F = 250 N

- the friction force, backward, $F_f$

Given (1), we have

$F-F_f = 0$

So the force of friction must be equal to the pull:

$F=F_f = 250 N$

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Virty [35]

Answer:

Explanation:

A combination of two controlled variables will make an experiment the most reliable.

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It is important to have them in pair so that two values can be kept on changing in terms of any constant condition. This will help to get better results in over all experiment data.

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