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9966 [12]
3 years ago
14

Function 1: y = 4x + 5

Mathematics
1 answer:
pickupchik [31]3 years ago
5 0
The answer is b
hope this helps
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What fraction /12 the same as 32
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384/12 or 192/6 or 96/3

Step-by-step explanation:

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Question 5- is 9
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Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D is the trian
Alla [95]

Answer: mass (m) = 4 kg

              center of mass coordinate: (15.75,4.5)

Step-by-step explanation: As a surface, a lamina has 2 dimensions (x,y) and a density function.

The region D is shown in the attachment.

From the image of the triangle, lamina is limited at x-axis: 0≤x≤2

At y-axis, it is limited by the lines formed between (0,0) and (2,1) and (2,1) and (0.3):

<u>Points (0,0) and (2,1):</u>

y = \frac{1-0}{2-0}(x-0)

y = \frac{x}{2}

<u>Points (2,1) and (0,3):</u>

y = \frac{3-1}{0-2}(x-0) + 3

y = -x + 3

Now, find total mass, which is given by the formula:

m = \int\limits^a_b {\int\limits^a_b {\rho(x,y)} \, dA }

Calculating for the limits above:

m = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2(x+y)} \, dy \, dx  }

where a = -x+3

m = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {(xy+\frac{y^{2}}{2} )} \, dx  }

m = 2.\int\limits^2_0 {(-x^{2}-\frac{x^{2}}{2}+3x )} \, dx  }

m = 2.\int\limits^2_0 {(\frac{-3x^{2}}{2}+3x)} \, dx  }

m = 2.(\frac{-3.2^{2}}{2}+3.2-0)

m = 2(-4+6)

m = 4

<u>Mass of the lamina that occupies region D is 4.</u>

<u />

Center of mass is the point of gravity of an object if it is in an uniform gravitational field. For the lamina, or any other 2 dimensional object, center of mass is calculated by:

M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }

M_{y} = \int\limits^a_b {\int\limits^a_b {x.\rho(x,y)} \, dA }

M_{x} and M_{y} are moments of the lamina about x-axis and y-axis, respectively.

Calculating moments:

For moment about x-axis:

M_{x} = \int\limits^a_b {\int\limits^a_b {y.\rho(x,y)} \, dA }

M_{x} = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2.y.(x+y)} \, dy\, dx }

M_{x} = 2\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {y.x+y^{2}} \, dy\, dx }

M_{x} = 2\int\limits^2_0 { ({\frac{y^{2}x}{2}+\frac{y^{3}}{3})}\, dx }

M_{x} = 2\int\limits^2_0 { ({\frac{x(-x+3)^{2}}{2}+\frac{(-x+3)^{3}}{3} -\frac{x^{3}}{8}-\frac{x^{3}}{24}  )}\, dx }

M_{x} = 2.(\frac{-9.x^{2}}{4}+9x)

M_{x} = 2.(\frac{-9.2^{2}}{4}+9.2)

M_{x} = 18

Now to find the x-coordinate:

x = \frac{M_{y}}{m}

x = \frac{63}{4}

x = 15.75

For moment about the y-axis:

M_{y} = \int\limits^2_0 {\int\limits^a_\frac{x}{2}  {2x.(x+y))} \, dy\,dx }

M_{y} = 2.\int\limits^2_0 {\int\limits^a_\frac{x}{2}  {x^{2}+yx} \, dy\,dx }

M_{y} = 2.\int\limits^2_0 {y.x^{2}+x.{\frac{y^{2}}{2} } } \,dx }

M_{y} = 2.\int\limits^2_0 {x^{2}.(-x+3)+\frac{x.(-x+3)^{2}}{2} - {\frac{x^{3}}{2}-\frac{x^{3}}{8}  } } \,dx }

M_{y} = 2.\int\limits^2_0 {\frac{-9x^3}{8}+\frac{9x}{2}   } \,dx }

M_{y} = 2.({\frac{-9x^4}{32}+9x^{2})

M_{y} = 2.({\frac{-9.2^4}{32}+9.2^{2}-0)

M{y} = 63

To find y-coordinate:

y = \frac{M_{x}}{m}

y = \frac{18}{4}

y = 4.5

<u>Center mass coordinates for the lamina are (15.75,4.5)</u>

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3 years ago
The Canadian side of Niagara Falls has a flow rate of 600,000 gallons per second. How many gallons of water flow over the falls
Olenka [21]

Answer:

hope this helps 3,160 tons of water flows over Niagara Falls every second. This accounts for 75,750 gallons of water per second over the American and Bridal Veil Falls and 681,750 gallons per second over the Horseshoe Falls.

Step-by-step explanation:

brainlist would help thx

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It takes 7 hours for 12 men to paint a room.
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