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charle [14.2K]
3 years ago
13

A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of

magnitude 4.0\times 10^{-5}~\text{T}4.0×10 ​−5 ​​ T that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 cm above the wire?
Physics
1 answer:
serg [7]3 years ago
8 0

Answer:

Explanation:

vertical magnetic field B_v = 4 x 10⁻⁵ T.

Magnetic field due to horizontal current at point 20 cm above

= (μ₀/4π ) x (2i / R)

= 10⁻⁷ x 2 x 20/ 20 x 10⁻²

= 2 x 10⁻⁵ T

It will act  coming out of paper. Hence it will be normal to magnetic field given .

So resultant magnetic field

= √ (4² + 2²) x 10⁻⁵ T

= 4.47 X 10⁻⁵ T .

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strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

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F_N =forceof the normal

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Generally the net force is given to be FC(force towards center)

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F_N=\frac{mv^2}{R} -mg

Mathematical we can now derive V

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v=10m/sec

Therefore the speed of the roller coaster is given ton be v=10m/sec

5 0
3 years ago
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IceJOKER [234]

Explanation:

Below is an attachment containing the solution

7 0
3 years ago
If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo
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Answer:

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Explanation:

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Force = mass * velocity / time.

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Going by the formula I stated, we can then say that

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8 0
2 years ago
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The mathematical relationship between force and extension for a spring is F = -kx , where F is the restoring force, k is the spring constant and x is the extension. If a box that weighs 40 N is hung from a spring of content 400 N/m, then the extension is equal to x = -F / k = -400 / 40 = -10 cm. The negative sign simply shows the extension and restoring force are in opposite directions. The answer is 10 cm. 
5 0
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