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Inessa05 [86]
3 years ago
9

a 1.50*10^-5 C charge feels a 2.89*10^-3 N force when it moves 288m/s perpendicular (90) deg to a magnetic field. how strong is

the field?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
5 0
6.68, -1
Explanation: correct for acellus
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A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west
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The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}

(see the attached graphic)

We have

  • ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or

\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

  • wind (relative to Earth) = 5.0 m/s due East, or

\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

  • ducks (relative to earth) = some speed <em>v</em> due South, or

\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)

Then by setting components equal, we have

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

We only care about the direction for this question, which we get from the first equation:

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

\cos\theta=-\dfrac57

\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

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