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ale4655 [162]
2 years ago
9

(2.) Jerry first walks 5m to East, then 6m to N 60 E, then, she walks 16m to N30'W, 3 finally turns he face and moves 3V3m to so

uth. What is her displacement and direction?​
Physics
1 answer:
Vladimir79 [104]2 years ago
5 0

The displacement is determined to be 5.11 m after every direction is plotted on the page.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points.

The steps in which Jerry walks;

5m to East

6m to N 60' E,

16m to N30'W,

Turn

3√3 m to the south.

Each and every direction is plotted on the page and obtained that the displacement is found as 5.11 m.

To learn more about displacement refer to the link;

brainly.com/question/10919017

#SPJ1

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Explanation:

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This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

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3 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
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Answer:

a. B= 9.45 \times10^{-3} T

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Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

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μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

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The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

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B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

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c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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