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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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Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an

Anit [1.1K]

Answer:

b.only when the current in the first coil changes.

Explanation:

An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.

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3 years ago

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.680 Hz. The pendulum ha

sineoko [7]

Answer:

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

Explanation:

The period of an oscillation equation of a solid pendulum is given by:

$T=2\pi \sqrt{\frac{I}{Mgd}}$ (1)

Where:

I is the moment of inertia
M is the mass of the pendulum
d is the distance from the center of mass to the pivot
g is the gravity

Let's solve the equation (1) for I

$T=2\pi \sqrt{\frac{I}{Mgd}}$

$I=Mgd(\frac{T}{2\pi})^{2}$

Before find I, we need to remember that

$T = \frac{1}{f}=\frac{1}{0.680}=1.47\: s$

Now, the moment of inertia will be:

$I=2*9.81*0.340(\frac{1.47}{2\pi})^{2}$

Therefore, the moment of inertia is:

$I=0.37 \: kgm^{2}$

I hope it helps you!

7 0

2 years ago

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$

5 0

3 years ago

How many dots should be drawn in the lewis dot structure for an atom of sodium A)19 B)8 C)4 D)1

Stells [14]

Na is in the first column on the periodic table so therefore it would have 1 valence electron
D 1

6 0

2 years ago

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo

NISA [10]

Explanation:

The given data is as follows.

Mass of small bucket (m) = 4 kg

Mass of big bucket (M) = 12 kg

Initial velocity ( $v_{o}$ ) = 0 m/s

Final velocity ( $v_{f}$ ) = ?

Height $H_{o} = h_{f}$ = 2 m

and, $H_{f} = h_{o}$ = 0 m

Now, according to the law of conservation of energy

starting conditions = final conditions

$\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}$

$\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)$

235.44 = $8V^{2}_{f}$ + 78.48

$V_{f}$ = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0

2 years ago