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3 years ago

9

An airplane flies 33 m/s due east while experiencing a tailwind of unknown velocity due northThe resultant velocity is determine

d to be 35.11 m/s northeastWhat is the velocity of the northern tailwind?

1 answer:

poizon [28]3 years ago

5 0

Answer22 MS:

Explanation:

22MS

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What kind of energy change occurs when a battery is operating a remote control toy?

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How does the law of inertia relate to how the planets in our solar system revolve around the sun?

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Point PP is a distance d1d1 = 4.0 mm above a large sheet of metal that carries a current of 35 AA in the positive xx direction a

zhenek [66]

Answer:

The width of the sheet is $w =0.8046m$

Explanation:

From the question we are told that

The distance of point P above a large sheet of metal is $D = 4.0mm =\frac{4}{1000} = 0.004m$

The current on the large metal sheet is $I =34A$

The distance of the the point P below a long wire $d = 3.0mm = \frac{3}{1000} = 0.003m$

The current on the long wire is $I_w = 0.41A$

The magnetic field at P is $B = 0T$

Generally magnetic field of P long wire is mathematically represented as

$B_w = \frac{\mu_o I_w}{2\pi r}$

Generally magnetic field of P large sheet of meta is mathematically represented as

$B_m = \frac{\mu_o K}{2}$

Where K is the current per unit width

The total magnetic field at P is

$\frac{\mu_o I_w}{2 \pi r} = \frac{\mu_o K}{2}$

Making K the subject of formula

$K = \frac{2 I_w }{2 \pi r }$

Substituting values

$K = \frac{2 * 0.41 }{2 * 3.142 * (0.0030) }$

$K = 43.4967 A/m$

Generally K is mathematically represented as

$K = \frac{I}{w}$

Where w is the width of the large sheet

Therefore the width of the metal sheet $w = \frac{I}{K}$

$= \frac{35}{43.4967}$

$w =0.8046m$

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3 years ago