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n200080 [17]
3 years ago
13

I need Help with this physics question please !

Physics
1 answer:
Elena-2011 [213]3 years ago
7 0

I'll denote vectors in boldface. So the given vector field is

<em>B</em> = <em>x</em>² <em>x</em> - <em>y</em>² <em>y</em> + 3<em>z</em> <em>z</em>

<em></em>

Compute the divergence:

∇ • <em>B</em> = (∂/∂<em>x</em> <em>x</em> + ∂/∂<em>y</em> <em>y</em> + ∂/∂<em>z</em> <em>z</em>) • (<em>x</em>² <em>x</em> - <em>y</em>² <em>y</em> + 3<em>z</em> <em>z</em>)

∇ • <em>B</em> = ∂/∂<em>x </em>[<em>x</em>²] + ∂/∂<em>y</em> [-<em>y</em>²] + ∂/∂<em>z</em> [3<em>z</em>]

∇ • <em>B</em> = 2<em>x</em> - 2<em>y</em> + 3

Compute the curl:

∇ × <em>B</em> = (∂/∂<em>x</em> <em>x</em> + ∂/∂<em>y</em> <em>y</em> + ∂/∂<em>z</em> <em>z</em>) × (<em>x</em>² <em>x</em> - <em>y</em>² <em>y</em> + 3<em>z</em> <em>z</em>)

∇ × <em>B</em> = ∂/∂<em>x</em> [<em>x</em>²] (<em>x</em> × <em>x</em>) + ∂/∂<em>x</em> [-<em>y</em>²] (<em>x</em> × <em>y</em>) + ∂/∂<em>x</em> [3<em>z</em>] (<em>x</em> × <em>z</em>)

… … … + ∂/∂<em>y</em> [<em>x</em>²] (<em>y</em> × <em>x</em>) + ∂/∂<em>y</em> [-<em>y</em>²] (<em>y</em> × <em>y</em>) + ∂/∂<em>y</em> [3<em>z</em>] (<em>y</em> × <em>z</em>)

… … … + ∂/∂<em>z</em> [<em>x</em>²] (<em>z</em> × <em>x</em>) + ∂/∂<em>z</em> [-<em>y</em>²] (<em>z</em> × <em>y</em>) + ∂/∂<em>z</em> [3<em>z</em>] (<em>z</em> × <em>z</em>)

∇ × <em>B</em> = 0

(since each partial derivative not along the main diagonal vanishes, and for any vector <em>a</em> we have <em>a</em> × <em>a</em> = 0)

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Resultant displacement is 29.2 km at 83.1^{\circ} north of west

Explanation:

To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.

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R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km

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