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KatRina [158]
3 years ago
15

T/F The outer planets have solid surfaces that are very rocky and dense

Physics
1 answer:
Luba_88 [7]3 years ago
6 0

Answer:

False

Explanation:

the inner planets have solid rocky dense surfaces

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Granite is uplifted by the movement of tectonic plates. It will most likely become _____. igneous rock sedimentary rock metamorp
pashok25 [27]

Answer:

The answer would be Igneous rock

Explanation:

3 0
3 years ago
Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+
babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• \mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j with 0\le x\le2 and 0\le y\le6-x;

• \mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2;

• \mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k with 0\le y\le 6 and 0\le z\le2;

• \mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2; and

• \mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k with 0\le u\le\frac\pi2 and 0\le y\le6-2\cos u.

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k

\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j

\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i

\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j

\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0

\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du

\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8

\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz

=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0

\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du

=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi

\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du

=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV

where <em>R</em> is the interior of <em>S</em>. We have

\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7

The integral is easily computed in cylindrical coordinates:

\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2

\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3

as expected.

4 0
3 years ago
Why was miasma theory replaced?
GaryK [48]
In the mid of the 19th century the miasma theory was replaced by the germ theory of diseases (Maia 2013) The Greek physician Hippocrates (c.460- 377 B.C.E) believed that bad air could be the cause of any pestilences, the fatal epidemic.


Hope that helps!
8 0
3 years ago
How much work is accplished when a force of 250 n pushes a box accross the floor for a distance of 50 meters
7nadin3 [17]
You just multiply these two numbers, it's 1250J
4 0
3 years ago
A jogger travels a route that has two parts. The first is a displacement of 3 km due south, and the second involves a displaceme
spayn [35]

Answer:

a) 2.41 km

b) 38.8°

Questions c and d are illegible.

Explanation:

We can express the displacements as vectors with origin on the point he started (0, 0).

When he traveled south he moved to (-3, 0).

When he moved east he moved to (-3, x)

The magnitude of the total displacement is found with Pythagoras theorem:

d^2 = dx^2 + dy^2

Rearranging:

dy^2 = d^2 - dx^2

dy = \sqrt{d^2 - dx^2}

dy = \sqrt{3.85^2 - 3^2}  = 2.41 km

The angle of the displacement vector is:

cos(a) = dx/d

a = arccos(dx/d)

a = arccos(3/3.85) = 38.8°

7 0
3 years ago
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