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Viktor [21]
3 years ago
15

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

in them. Find the center of mass of the lamina if the density at any point is inversely proportional to its distance from the origin.
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

Required center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

Step-by-step explanation:

Given semcircles are,

y=\sqrt{1-x^2}, y=\sqrt{16-x^2} whose radious are 1 and 4 respectively.

To find center of mass, (\bar{x},\bar{y}), let density at any point is \rho and distance from the origin is r be such that,

\rho=\frac{k}{r} where k is a constant.

Mass of the lamina=m=\int\int_{D}\rho dA where A is the total region and D is curves.

then,

m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k

  • Now, x-coordinate of center of mass is \bar{y}=\frac{M_x}{m}. in polar coordinate y=r\sin\theta

\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta

=3k\int_{0}^{\pi}\sin\theta d\theta

=3k\big[-\cos\theta\big]_{0}^{\pi}

=3k\big[-\cos\pi+\cos 0\big]

=6k

Then, \bar{y}=\frac{M_x}{m}=\frac{2}{\pi}

  • y-coordinate of center of mass is \bar{x}=\frac{M_y}{m}. in polar coordinate x=r\cos\theta

\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA

=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta

=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta

=3k\int_{0}^{\pi}\cos\theta d\theta

=3k\big[\sin\theta\big]_{0}^{\pi}

=3k\big[\sin\pi-\sin 0\big]

=0

Then, \bar{x}=\frac{M_y}{m}=0

Hence center of mass (\bar{x},\bar{y})=(\frac{2}{\pi},0)

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Answer:

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Step-by-step explanation:

(-6+3)= -3

2--3=2+3=5

5+4c cant add them since they arent like terms

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If f(x) = 2x + 3 and g(x) = 3x - 9, find (f + g)(x).
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Answer:

5x-6

Step-by-step explanation:

f(x) = 2x + 3

g(x) = 3x - 9

f(x) + g(x) = 2x + 3  + 3x-9

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3 years ago
Select the correct answer.
Rashid [163]

Answer: D. (4,3)

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Repeat for the y coordinates. Add them up: 8+(-2) = 8-2 = 6. Then divide by two: 6/2 = 3. The midpoint has an y coordinate of 3.

Those two coordinates pair up to get (x,y) = (4,3) which is the midpoint of segment AB.

7 0
3 years ago
You have 108 seats, there are 12 times as many seats as rows, how many rows
nignag [31]
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6 0
3 years ago
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John has decided to prepare an old field for his son's new horse. the length of the field is 10 meters less than 4 times its wid
tatuchka [14]

Answer: The total money invested in the field will be $1,623.80

Step-by-step explanation:

step 1

Find the dimensions of the field

Let

x -----> the length of the field

y -----> the width of the field

p=2(x+y)

p(4.80)=1,584

p=330m

we know that

The perimeter of the field is equal to

so

330=2(x+y)----> equation A

x=4y-10 ------> equation B

substitute equation B in equation A and solve for y

330=2(4y-10+y)

165=(5y-10)

5y=165+10

y+35

Find the value of x

x=4(35)-10+130

therefore

The length of the field is 130 meters and the width of the field is 35 meters

step 2

Find the area of the field

The area of the field is

A=xy

substitute

A=(130)(35)=4,550m2

step 3

Find the number of bags of seed

Divide the area by 460

4,550/460=9.89 bags

Round up to the nearest whole number

9.89 bags=10 bags

step 4

Find the cost of the seed

(10 )(3.98)=$19.8

step 5

Find the total money invested in the field

1,584+$39.8=$1,623.80

7 0
2 years ago
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