Work out each map scale as a map ratio

1 answer:

6 0

1:500000
1cm:500000cm

2:250000
2cm:250000cm
1cm:125000cm

To get this you convert the km into cm as it is easier and then it becomes something like this 2cm : 250000cm
then divide or half to get to the simpler form like this 2cm divided by 2=1cm and 250000cm divided by 2= 125000cm
then put into ratio is 1:125000

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HELP URGENT PLEASE

ki77a [65]

Answer:

15 hours

108 hours

2074.29 miles

Step-by-step explanation:

Under the assumption the earth is a perfect circle, then in one complete rotation about its axis ( 24 hours) the Earth will cover 360 degrees or 2π radians.

In every 24 hours the earth rotates through 360 degrees ( a complete rotation). We are required to determine the length of time it will take the Earth to rotate through 225 degrees. Let x be the duration it takes the earth to rotate through 225 degrees, then the following proportions hold;

(24/360) = (x/225)

solving for x;

x = (24/360) * 225 = 15 hours

In 24 hours the earth rotates through an angle of 2π radians (a complete rotation) . We are required to determine the length of time it will take the Earth to rotate through 9π radians. Let x be the duration it takes the earth to rotate through 9π radians, then the following proportions hold;

(24/2π radians) = (x/9π radians)

Solving for x;

x = (24/2π radians)*9π radians = 108 hours

If the diameter of the earth is 7920 miles, then in a day or 24 hours a point on the equator will rotate through a distance equal to the circumference of the Earth. Using the formula for the circumference of a circle;

circumference = 2*π*R = π*D

= 7920*3.142

= 24891.43 miles

Therefore a point on the equator covers a distance of 24891.43 miles in 24 hours. This will imply that the speed of the earth is approximately;

(24891.43miles)/(24 hours) = 1037.14 miles/hr

The distance covered by the point in 2 hours will thus be;

1037.14 * 2 = 2074.29 miles

5 0

3 years ago

ZEARN MATH

joja [24]

Answer:

ok ill do it :-)

Step-by-step explanation:

3 0

3 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$