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UkoKoshka [18]
3 years ago
7

If a material has a high index of refraction, it has a 1 low optical density, and light travels more slowly through it. . high o

ptical density, and light travels more slowly through it. Olow optical density, and light travels more quickly through it. high optical density, and light travels more quickly through it.​
Mathematics
1 answer:
Archy [21]3 years ago
8 0

Answer:

B. High optical density, and light travels more slowly through it.

Step-by-step explanation:

Refractive index is the property of a material that describes the change in the initial path of light and the rate at which light travels through it. And optical density of a material describes its absorption ability to rays or beam of light passing through it.

As the index of refraction of a material increases, therefore its optical density would increase. This would cause light to travel through it slowly. Unless the light has a high frequency.

A material with a high index of refraction has a high optical density, and light travels more slowly through it.

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John plays basketball at the local gym. The gym charges a monthly membership fee of $15.00, plus each player pays $2.50 per game
Katyanochek1 [597]
Answer:
Step-by-step explanation:
Assuming that for each option, you play the same number of games,x
Let y represent the cost of playing x games using option A
Option A is to buy a membership card and pay $2 every time you go to the gym. The membership card costs $20. It means that
y = 20 + 2x
Let z represent the cost of playing x games using option B
Option B is to pay $4 each time you go. It means that
z = 4x
To determine how many games will be played before cost of option A equal to the cost of option B, we would equate y to z. It becomes
20 + 2x = 4x
4x - 2x = 20
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It will take 10 games for both to be the same
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Please help and explain!
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K is the slope of the line formed by connecting the points.
Use slope formula:
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Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation
Gemiola [76]

To find:

(a) Equation for the sphere of radius 5 centered at the origin in cylindrical coordinates

(b) Equation for a cylinder of radius 1 centered at the origin and running parallel to the z-axis in spherical coordinates

Solution:

(a) The equation of a sphere with center at (a, b, c) & having a radius 'p' is given in cartesian coordinates as:

(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=p^{2}

Here, it is given that the center of the sphere is at origin, i.e., at (0,0,0) & radius of the sphere is 5. That is, here we have,

a=b=c=0,p=5

That is, the equation of the sphere in cartesian coordinates is,

(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=5^{2}

\Rightarrow x^{2}+y^{2}+z^{2}=25

Now, the cylindrical coordinate system is represented by (r, \theta,z)

The relation between cartesian and cylindrical coordinates is given by,

x=rcos\theta,y=rsin\theta,z=z

r^{2}=x^{2}+y^{2},tan\theta=\frac{y}{x},z=z

Thus, the obtained equation of the sphere in cartesian coordinates can be rewritten in cylindrical coordinates as,

r^{2}+z^{2}=25

This is the required equation of the given sphere in cylindrical coordinates.

(b) A cylinder is defined by the circle that gives the top and bottom faces or alternatively, the cross section, & it's axis. A cylinder running parallel to the z-axis has an axis that is parallel to the z-axis. The equation of such a cylinder is given by the equation of the circle of cross-section with the assumption that a point in 3 dimension lying on the cylinder has 'x' & 'y' values satisfying the equation of the circle & that 'z' can be any value.

That is, in cartesian coordinates, the equation of a cylinder running parallel to the z-axis having radius 'p' with center at (a, b) is given by,

(x-a)^{2}+(y-b)^{2}=p^{2}

Here, it is given that the center is at origin & radius is 1. That is, here, we have, a=b=0,p=1. Then the equation of the cylinder in cartesian coordinates is,

x^{2}+y^{2}=1

Now, the spherical coordinate system is represented by (\rho,\theta,\phi)

The relation between cartesian and spherical coordinates is given by,

x=\rho sin\phi cos\theta,y=\rho sin\phi sin\theta, z= \rho cos\phi

Thus, the equation of the cylinder can be rewritten in spherical coordinates as,

(\rho sin\phi cos\theta)^{2}+(\rho sin\phi sin\theta)^{2}=1

\Rightarrow \rho^{2} sin^{2}\phi cos^{2}\theta+\rho^{2} sin^{2}\phi sin^{2}\theta=1

\Rightarrow \rho^{2} sin^{2}\phi (cos^{2}\theta+sin^{2}\theta)=1

\Rightarrow \rho^{2} sin^{2}\phi=1 (As sin^{2}\theta+cos^{2}\theta=1)

Note that \rho represents the distance of a point from the origin, which is always positive. \phi represents the angle made by the line segment joining the point with z-axis. The range of \phi is given as 0\leq \phi\leq \pi. We know that in this range the sine function is positive. Thus, we can say that sin\phi is always positive.

Thus, we can square root both sides and only consider the positive root as,

\Rightarrow \rho sin\phi=1

This is the required equation of the cylinder in spherical coordinates.

Final answer:

(a) The equation of the given sphere in cylindrical coordinates is r^{2}+z^{2}=25

(b) The equation of the given cylinder in spherical coordinates is \rho sin\phi=1

7 0
2 years ago
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