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oksian1 [2.3K]
3 years ago
12

Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

A to the right and is located along the line y = 0.420 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
Physics
1 answer:
pishuonlain [190]3 years ago
3 0

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

\dfrac{F}{L}= \dfrac{KI_1I_2}{d}

I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}

I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}

    I₂ = 32.67 A

distance where the magnetic field is zero

\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}

y_1 = 0.248\ m

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

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Answer:

a. 1.80 %

b. 0.60 %

c. 1.20 %

d. 0.60 %

e. 0.60 %

Explanation:

In order to calculate percent uncertainties in each case, we forst need to calculate the average value:

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Average Diameter = 16.7 mm

Now, the formula for percent uncertainty is:

% Uncertainty = (Uncertainty/Average) * 100 %

where,

Uncertainty = |Value - Average|

For Each Case:

<u>a. 16.4 mm</u>:

Uncertainty = |16.4 mm - 16.7 mm| = 0.3 mm

Therefore,

% Uncertainty = (0.3 mm/16.7 mm) * 100%

<u>% Uncertainty = 1.80 %</u>

<u></u>

<u>b. 16.8 mm</u>:

Uncertainty = |16.8 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

<u>% Uncertainty = 0.60 %</u>

<u></u>

<u>c. 16.4 mm</u>:

Uncertainty = |16.9 mm - 16.7 mm| = 0.2 mm

Therefore,

% Uncertainty = (0.2 mm/16.7 mm) * 100%

<u>% Uncertainty = 1.20 %</u>

<u></u>

<u>d. 16.6 mm</u>:

Uncertainty = |16.6 mm - 16.7 mm| = 0.1 mm

Therefore,

% Uncertainty = (0.1 mm/16.7 mm) * 100%

<u>% Uncertainty = 0.60 %</u>

<u></u>

<u>e. 16.8 mm</u>:

Uncertainty = |16.8 mm - 16.7 mm| = 0.3 mm

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% Uncertainty = (0.1 mm/16.7 mm) * 100%

<u>% Uncertainty = 0.60 %</u>

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Answer attachment below
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Answer:

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Which describes two minerals that have different arrangements of their atoms?
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The best and most correct answer among the choices provided by your question is the third choice or letter C.
<span>
Two minerals that have different arrangements of their atoms may have different physical properties.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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4 years ago
A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I
melamori03 [73]
(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s

(b) The initial speed of the skater is
v_i = 4 m/s
while the final speed is
v_f = 5.3 m/s
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2

(c) The initial speed of the skater is 
v_i = 13.0 m/s
while the final speed is 
v_f=0
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
2aS=v_f^2 -v_i^2
from which we find
a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2
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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
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PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

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KE = \frac{1}{2}mv^2

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KE = 4.55 \times 10^{-27} kg

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So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
3 years ago
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