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Setler79 [48]
3 years ago
14

The young modulus of an elastic material is 2.5×10 raised to power 3 newton per meter square. What magnitude of load will cause

an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
the young modulus of an elastic material is  {2.5 \times 10}^{3}   nm ^{ - 2}  .  What magnitude of load will cause an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
​
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:.

200

Explanation:

.

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Answer:

  • The formula its f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816
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Explanation:

<h3>Obtaining the formula</h3>

We wish to find a formula that

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  • Reach 0 at 8 years. f( 8 \ years) \ = \ \$ \ 0
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We can cover all this requisites with a straight-line equation. (an straigh-line its the only curve that has a constant rate of change) :

f(t) \ = \ m\ t \ + \ b,

where m its the slope of the line and b give the place where the line intercepts the <em>y</em> axis.

So, we can use this formula with the data from our problem. For the first condition:

f ( 0 \ years ) = m \ (0 \ years) + b = \$ \ 2816

b = \$ \ 2816

So, b = $ 2816.

Now, for the second condition:

f ( 8 \ years ) = m \ (8 \ years) + \$ \ 2816 = \$ \ 0

m \ (8 \ years) = \ - \$ \ 2816

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \frac{\ - \$ \ 2816}{8 \ years}

m = \ - \ 352 \frac{\$ }{years}

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f(t) \ = \ - \ 352 \ \frac{\$ }{years} \ t \ + \ \$ \ 2816

<h3>After 5 years</h3>

Now, we just use <em>t = 5 years</em> in our formula

f(5 \ years) \ = \ - \ 352 \ \frac{\$ }{years} \ 5 \ years \ + \ \$ \ 2816

f(5 \ years) \ = \ - \$ \ 1760 + \ \$ \ 2816

f(5 \ years) \ = $ \ 1056

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