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3 years ago

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A homogeneous 1300 kg bar AB is supported at either end by a cable. calculate the stress acting in bronze bar in N/mm² if stress

should not exceed to 120MPa for steel and the area of the bronze bar and steel bar is 40.05 mm² and 36.33 mm² respectively.

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer:

4553332 is the answer

Step-by-step explanation:

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Step-by-step explanation: 2= 41/7 3= 37/9 4= 62/7

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Find the unit rate in each situation below in miles per hour.

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Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

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Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

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2nd subtraction property.

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According to the United Nations, in the year 2002, the population of the world was 6.1 billion people and was growing at an annu

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Answer:

(a) 6.2843 billion

(b) 6.8716 billion

Step-by-step explanation:

Since there is a constant growth rate of 1.5% per year, and the population in 2002 was 6.1 billion people. The general equation for the total world population, in billions, after 2002 is:

$P(t) = 6.1*(1+0.015)^t$

With t being the time, in years, after 2002.

a) What was the population in 2004?

$t=2004-2002=2\\P(2) = 6.1*(1.015)^2\\P(2) = 6.2843 \ billion$

b) What was the population in 2010?

$t=2010-2002=8\\P(8) = 6.1*(1.015)^8\\P(8) = 6.8716 \ billion$

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