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1 year ago

Q10 Please help me solve this….

Mathematics

1 answer:

kondor19780726 [428]1 year ago

5 0

Answer:

C. 0.23°

Step-by-step explanation:

We are given

$r=\frac{1}{32}v^{2}sin(2\theta)$

Re-arranging this equation (multiply by 32 on both sides and divide by $v^{2}$ on both sides gives us:

$\frac{32r}{v^{2}}=sin(2\theta)$ or

$sin(2\theta)=\frac{32r}{v^{2}}$

Substituting for, r and v we get

$sin(2\theta)=\frac{(32)5000}{4500^{2}}=0.00792\\\\\theta=sin^{-1}(0.0079)=0.4527\\\\\theta=\frac{{0.4527}}{2}=0.2263\approx0.23$

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Consider the following region R and the vector field F. a. Compute the two-dimensional curl of the vector field. b. Evaluate bo

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Looks like we're given

$\vec F(x,y)=\langle-x,-y\rangle$

which in three dimensions could be expressed as

$\vec F(x,y)=\langle-x,-y,0\rangle$

and this has curl

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which confirms the two-dimensional curl is 0.

It also looks like the region $R$ is the disk $x^2+y^2\le5$ . Green's theorem says the integral of $\vec F$ along the boundary of $R$ is equal to the integral of the two-dimensional curl of $\vec F$ over the interior of $R$ :

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA$

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of $R$ by

$\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle$

$\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt$

with $0\le t\le2\pi$ . Then

$\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt$