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Yuliya22 [10]
3 years ago
12

Calculate the second moment of area of a 4-in-diameter shaft about the x-x and yy axes, as shown.

Mathematics
1 answer:
Keith_Richards [23]3 years ago
6 0
Second moment of area about an axis along any diameter in the plane of the cross section (i.e. x-x, y-y) is each equal to (1/4)pi r^4.
The second moment of area about the zz-axis (along the axis of the cylinder) is the sum of the two, namely (1/2)pi r^4.

The derivation is by integration of the following:
int int y^2 dA
over the area of the cross section, and can be found in any book on mechanics of materials.

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Can you solve d-3/5=6 before multiplying by 5 reasoning yes or no and why?
Luda [366]

Answer:

yes you can

apart from getting rid of the denominator you can collect like terms

Step-by-step explanation:

d=6-3/5

6=30/5

30/5-3/5

27/5

d=27/5

5 0
3 years ago
How can you minimize the risk from investment?
adoni [48]
Don’t invest too much into it
3 0
3 years ago
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Can someone help me with this question?
arlik [135]

Try this solution:

1. m∠A=m∠L; m∠B=m∠M and m∠C=m∠N;

2. m∠B=m∠M=35° and m∠C=m∠N=95°;

3. m∠A=m∠L=180°-(m∠B+m∠C)=180-35-95=50°

answer: 50°

5 0
3 years ago
Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q. 3 + x - 3x2 = 9
Triss [41]
To <span>transform the quadratic equation into the equation form (x + p)2 = q we shall proceed as follows:
3+x-3x^2=9
putting like terms together we have:
-3x^2+x=6
dividing through by -3 we get:
x^2-x/3=-2
but
c=(b/2a)^2
c=(-1/6)^2=1/36
thus the expression will be:
x^2-x/3+1/36=-2+1/36
1/36(6x-1)</span>²=-71/36

the answer is:
1/36(6x-1)²=-71/36
7 0
3 years ago
Please help very urgent
Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by \sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}.

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

a_{n} = a_{1} • r^{n-1}

Where a_{n} is the nth term, a_{1} is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, a_{1} will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

\sqrt[5]{\frac{a_{6}}{a_{1}}} =

\sqrt[5]{\frac{1}{32}} =

\frac{1}{2}.

Therefore: r = \frac{1}{2}.

Using all the information we have, we can find the explicit rule:

a_{n} = a_{1} • r^{n-1}

  • a_{1} = 32
  • r = \frac{1}{2}

a_{n} = a_{1} • r^{n-1} →

\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

a_{1} = 32 • (\frac{1}{2})^{1-1}

a_{1} = 32 • (\frac{1}{2})^{0}

a_{1} = 32 • 1

a_{1} = 32

a_{6} = 32 • (\frac{1}{2})^{6-1}

a_{6} = 32 • (\frac{1}{2})^{5}

a_{6} = 32 • \frac{1}{32}

a_{6} = 1

3 0
2 years ago
Read 2 more answers
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