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Vitek1552 [10]
3 years ago
7

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

d in tension. A true stress of 379 MPa causes the specimen to plastically elongate to a length of 51.7 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 49.7 mm to a length of 57.1 mm.
Chemistry
2 answers:
amm18123 years ago
6 0

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.

Citrus2011 [14]3 years ago
6 0

Answer:i ea

Explanation:

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Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance
ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0
3 years ago
Explain why 5.00 grams of salt does not contain the same number of particles as 5.0 grams of sugar​
Tom [10]

Answer:

5.00 grams of salt contain more particles than 5.0 grams of sugar​

Explanation:

Salt = NaCl

Molar mass = 58.45  g/mol

Sugar = C₁₂H₂₂O₁₁

Molar mass = 342.3 g/mol

Sugar's molar mass is higher than salt.

So 1 mol of sugar weighs more than 1 mol of salt

But 5 grams of salt occupies more mole than 5 grams of sugar

5 grams of salt = 5g / 58.45 g/m = 0.085 moles

5 grams of sugar = 5g/ 342.3 g/m = 0.014 moles

In conclusion, we have more moles of salt in 5 grams; therefore there are more particles than in 5 g of sugar.

3 0
3 years ago
What is the chemical formula of sodium azide?
Hunter-Best [27]
NaN₃ is the chemical formula for Sodium Azide
3 0
2 years ago
Which of the following properties would be the same among isotopes of the same element?
Viefleur [7K]

Answer : The properties which would be the same among isotopes of the same element are Flammability and Color

Explanation :

Isotopes are the elements that have same number of electrons and protons but they differ in number of neutrons.

The chemical properties of an element are determined by its valence electrons and since isotope have same number of electrons, they have similar chemical properties.

But isotopes have different number of neutrons. Neutrons are the subatomic particles which are present in the nucleus. They are charge less particles and they have mass. Therefore when the elements have different number of neutrons, their atomic masses differ.

As a result, isotopes have different mass related physical properties.

Let us discuss the given properties now

1) Flammability : Flammability in simple words means the ability of the substance to catch fire or the ability to burn. This is mainly dependent on the reactivity which is in turn dependent on the number of electrons.

Since the isotopes have same number of electrons, they show similar flammability.

2) Density : Density is defined as mass per unit volume. This property involves mass and we know that isotopes have different masses. Therefore they have different densities too.

3) Melting point : Melting point is a physical property which is mainly dependent on the molecular weight of the substance. Since isotopes have different masses, they tend to show different melting points.

4) Color : The color of the element is mainly dependent on the arrangement of electrons around the nucleus. Isotopes have same number of electrons which means the arrangement would also be the same. Therefore the isotopes of the same element do not show different colors.

The properties which would be same among isotopes of the same element are

1) Flammability

4) Color

5 0
3 years ago
Zn + I2 ---&gt; Znl2<br> Determine the theoretical yield of the product if 2g of Zn is used
egoroff_w [7]

Answer:

Mass = 9.58 g

Explanation:

Given data:

Mass of Zn = 2g

Theoretical yield of ZnI₂ = ?

Solution:

Chemical equation:

Zn + I₂       →     ZnI₂

Number of moles of Zn:

Number of moles = mass/molar mass

Number of moles = 2g / 65.38 g/mol

Number of moles = 0.03 mol

Now we will compare the moles of Zn and ZnI₂.

                   Zn           :          ZnI₂

                    1             :           1

                  0.03        :         0.03

Mass of ZnI₂:

Mass = number of moles × molar mass

Mass = 0.03 mol × 319.22 g/mol

Mass = 9.58 g

4 0
3 years ago
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