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borishaifa [10]
3 years ago
13

Use the drop-down menus to rank the boiling points of the following hydrocarbons. Use a "1" to indicate

Chemistry
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

1. 4

2. 2

3. 3

4. 1

Explanation: just did it on edge 2021

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Wha do u do when u get in a fight wit yo best friend?
antiseptic1488 [7]
Talk to them and listen to each other. if they aren’t ready to talk, give them space. once both of you are ready, you can make up and forgive each other. don’t bother them by asking a lot of questions and forcing them to talk to you. and if they’re doing that, tell them you need time to think. just be sure to talk to them, listen, and understand. tell each other both sides of the stories. of course, different situations can require different solutions. so resolve it when it’s time :)
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In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead(II) oxide. Calculate the percent yi
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Let MM(x) be the molar mass of x.

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse
amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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