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NikAS [45]
3 years ago
7

Multiplicative inverse a^m is Please answer this fast

Mathematics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

a^-m

Step-by-step explanation:

a^m ×a^-m = a^m-m = a^0 = 1

a^-m is the multiplicative inverse of a^m

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A particle moves so that its position vector with respect to the origin of a reference frame Oxyz is r(t)=bcos w t+bsin w t+vt k
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(i) Velocity is the rate of change of position, so if

<em>r</em><em>(t)</em> = <em>b</em> cos(<em>ω t </em>) <em>i</em> + <em>b</em> sin(<em>ω t </em>) <em>j</em> + <em>v</em> <em>t</em> <em>k</em>

then

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em>

<em>v</em><em>(t)</em> = -<em>b</em> <em>ω </em>sin(<em>ω t</em> ) <em>i</em> + <em>b</em> <em>ω</em> cos(<em>ω</em> <em>t</em> ) <em>j</em> + <em>v</em> <em>k</em>

The speed of the particle is the magnitude of the velocity, given by

|| <em>v</em><em>(t)</em> || = √[(-<em>b</em> <em>ω </em>sin(<em>ω t</em> ))² + (<em>b</em> <em>ω</em> cos(<em>ω</em> <em>t</em> ))² + <em>v</em> ²]

… = √[<em>b </em>²<em>ω </em>² + <em>v</em> ²]

(ii) The path is a helix. Suppose you zero out the <em>k</em> component. Then the path is a circle of radius <em>b</em>, and the value of <em>ω</em> determines how quickly a particle on the path traverses the circle. Now if you reintroduce the <em>k</em> component, the value of <em>v</em> will determine how far from the plane <em>z</em> = 0 the particle moves in a helical path as <em>t</em> varies.

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<em>a</em><em>(t)</em> = d<em>v</em>/d<em>t</em>

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3 years ago
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