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dusya [7]
2 years ago
9

A current runs from the left to the right in a long, straight wire. How does the magnetic field at point X compare with the magn

etic field at point Y?
A. The magnetic field at point X points into the page, and the magnetic field at point Y points into the page.

B. The magnetic field at point X points out of the page, and the magnetic field at point Y points out of the page.

C. The magnetic field at point X points out of the page, and the magnetic field at point Y points into the page.

D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.
Physics
1 answer:
vitfil [10]2 years ago
7 0
I think the answer is <span>D. The magnetic field at point X points into the page, and the magnetic field at point Y points out of the page.</span>
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vfiekz [6]

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Diameter of the rod d=2.54\ cm

length of rod is l=20\ cm

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cross-section of the rod A

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Resistance of rod is  R

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3 0
2 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

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Thus;

ma = qVB

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m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
2 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

7 0
2 years ago
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