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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.

6 0

2 years ago

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An animal cell has a 0.9% saline environment is placed into a solution that is 9% saline. What will happen to the cell

Softa [21]

The concentration of cell is less than that of the solution .

Hence the cell will be called as hypotonic and the solution will be called as hypertonic.

in order to balance the concentration on the two sides of cell (inside and outside in the solution) there will be movement of solvent particles (through semipermeable membrane ) from cell (lower concentration of solute) to solution (higher concentration of solute).

Thus cell will shrink.

7 0

3 years ago

9. How many grams of potassium sulfate are needed to make 250 mL of a 0.150 M

antiseptic1488 [7]

Answer:

6.53g of K₂SO₄

Explanation:

Formula of the compound is K₂SO₄

Given parameters:

Volume of K₂SO₄ = 250mL = 250 x 10⁻³L

= 0.25L

Concentration of K₂SO₄ = 0.15M or 0. 15mol/L

Unknown:

Mass of K₂SO₄ =?

Methods:

We use the mole concept to solve this kind of problem.

>>First, we find the number of moles using the expression below:

Number of moles= concentration x volume

Solving for number of moles:

Number of moles = 0.25 x 01.5

= 0.0375mole

>>Secondly, we use the number of moles to find the mass of K₂SO₄ needed. This can be obtained using the expression below:

Mass(g) = number of moles x molar mass

Solving:

To find the molar mass of K₂SO₄, we must know the atomic mass of each element in the compound. This can be obtained using the periodic table.

For:

K = 39g

S = 32g

O = 16g

Molar mass of K₂SO₄ = (39x2) + 32 + (16x4)

= 78 +32 + 64

= 174g/mol

Using the expression:

Mass(g) = number of moles x molar mass

Mass of K₂SO₄ = 0.0375 x 174 = 6.53g

5 0

3 years ago

Balance the following chemical

poizon [28]

Answer:

Explanation:

A chemical formula can be defined as a notation that is used to show which element and how many is contained in a chemical compound.

Also, in chemistry, the sum of charges of the anion and the cation of any ionic compound is always equal to zero.

A chemical equation is considered to be balanced when the amount of reactants on the left is equal to the amount of products on the right.

Therefore;

[2]FeBr3 + [3]Na2S → [1]Fe2S3 + [6]NaBr

In the above chemical equation, we will balance the reactants in the chemical equation with the smallest coefficients possible;

$2FeBr_{3} + 3Na_{3}S ----> Fe_{2}S{3} + 6NaBr$

Two (2) moles of Iron (III) Bromide reacts with two (2) moles of Sodium Sulfide to form Iron (III) Sulfide and Sodium Bromide.

4 0

2 years ago

El hipoclorito de sodio (NaClO), es vendido en una solución clara de ligero color verde-amarillento y un olor característico que

vivado [14]

Respuesta:

16,7 mL

Explicación:

Paso 1: Información provista

Concentración inicial (C₁): 6%

Volumen inicial (V₁): ?

Concentración final (C₂): 0,5%

Volumen final (V₂): 200 mL

Paso 2: Calcular el volumen de la solución concentrada

Queremos preparar una solución diluida de hipoclorito de sodio a partir de una concentrada. Podemos calcular el volumen inicial que debemos tomar usando la regla de dilución.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 0,5% × 200 mL / 6% = 16,7 mL

6 0

2 years ago