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olchik [2.2K]
2 years ago
13

Zero gauge wire has a diameter of 0.32in and can carry a sustained current of 150A safely. What is the magnetic field a distance

5cm from the center of a long straight segment of zero gauge wire carrying a current of 150A? This would be the field experienced by a bird sitting on the wire. Select One of the Following:
(a) 1.2 × 10−2T
(b) 1.9 × 10−3T
(c) 3.0 × 10−4T
(d) 6.0 × 10−4T
(e) 1.0 × 10−6T
Physics
1 answer:
DanielleElmas [232]2 years ago
5 0

Answer: d) 6*10^-4 T

Explanation: In order to explain this problem we have to use the Ampere law to calculate the magnetic field, which is given by:

∫B*dl=μo*I  then

B=μo*I/(2*π*r)  where r and I are the distance to the wire and the current in the wire. μo is equal 4*π*10^-7 Wb/A*m

Finally, we have:

B=4*π*10^-7*150/(2*π*0.05)=6*10^-4 T

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The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

  • the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

average \ speed = \frac{total \ distance }{total \ time }

The total distance from the graph is calculated as follows;

  • first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm
  • first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm
  • second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm
  • second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm
  • third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm
  • fourth downward distance from 12 cm to 9 cm = 3 cm
  • final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

  • This shortest distance is the straight line connecting the start and end position. Call this line P
  • From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm
  • Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of  line P.

                                                ↓end

                                                ↓

                                                ↓ 6cm

                                                ↓

  start -------------13 cm------------

Use Pythagoras theorem to solve for P.

P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm

The average velocity of the ant is calculated as;

average \ velocity= \frac{\Delta displacemnt  }{total\ time }= \frac{14.318 \ cm}{105 \  s} = 0.136 \ cm/s  \\\\

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
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The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
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  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
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The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

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Answer:

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