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Maurinko [17]
3 years ago
11

28 points

Physics
1 answer:
dem82 [27]3 years ago
7 0
Answer: C. 1.64 x 10-3 m/s2
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Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each
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Answer with Explanation:

We are given that

I_1=29 A

I_2=78 A

d=38 cm=\frac{38}{100}=0.38 m

1 m=100 cm

a.Length of segment,l=20 m

Magnetic force ,F=\frac{2\mu_0I_1I_2 l}{4\pi d}

\frac{\mu_0}{4\pi}=10^{-7}

Substitute the values

F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

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Use this table of a school bus during morning pickups to answer the question.
AlexFokin [52]

Answer:

3.5

Explanation:

The answer should be 3.5

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PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

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