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xxTIMURxx [149]
2 years ago
8

Which Of The Following Statements Are True?

Physics
1 answer:
aksik [14]2 years ago
5 0

Answer:

A) The north pole of a bar magnet will attract the south pole of another bar magnet.

B) Earth's geographic north pole is actually a magnetic south pole.

E) The south poles of two bar magnets will repel each other.

Explanation:

<u>According to </u><u>classical physics</u>, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This means that if we break a magnet in half, we will have two magnets, where each new magnet will have a new south pole, and a new north pole.

This is because <u>for classical physics, naturally, magnetic monopoles can not exist. </u>

In this context,  Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.  

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet.  

Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.  

And it is for this reason, moreover, that the magnetic field lines enter the Earth through its magnetic south pole (which is the geographic north pole).

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A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.
vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

           F = B i L

Rod is not necessarily vertical

F_x =i L B_d

F_y= i L B_w

the normal reaction N = mg-F y

static friction       f = μ_s (mg-F y )

horizontal acceleration is zero

F_x-f = 0

iLBd = \mu_s(mg-F_y )

 B_w = B sinθ

 B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}

\theta =tan{-1}{\mu_s}

\theta =tan{-1}{0.6}

\theta = 31^0

B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}

       B = 0.1 T

4 0
3 years ago
How does Mercury's close proximity to the sun and thin atmosphere affect its ability to maintain liquid water
alekssr [168]

Answer

A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.

5 0
3 years ago
A piece of aluminum foil has a known surface density of
bija089 [108]

The cube has 6 equal, square, foil faces. The mass of foil for each face is (380/6) milligrams.

The surface area of each piece is (380)/(6•11) cm^2.

The length of each side of the piece is √(380/6•11) cm

That's about 2.4 cm .

It's a cute little foil cube, just under 1-inch each way.


5 0
3 years ago
Help m-e-e people -_-!​
xxTIMURxx [149]

Answer:

the answer is

Explanation:For equilibrium

Weight = Tension

mg=T

∴T=4×3.1π=12.4πN (as can be inferred from the question)

Y=

△l/l

T/A

​

=

1000

0.031

​

/20

12.4π/π(

1000

2

​

)

2

​

=

4×0.031

12.4×20×1000×(1000)

2

​

=2×10

12

N/m

2

6 0
2 years ago
Use the dot product to find the magnitude of u if u = 6i - 3j
LuckyWell [14K]
Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | = \sqrt{6 ^{2}+(-3) ^{2}  } = \sqrt{36+9}= \\  \sqrt{45}= \sqrt{9*5}=3  \sqrt{5}
Answer:
The magnitude of vector u is 3√5. 
3 0
2 years ago
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