Density of a certain liquid:Specific weight is also called the weight density of a liquid and it's given as .Therefore, the viscosity of the substance between the two plates is 0.32 Pa.s.
w = ρgwhere
w = weight density,
ρ = density of the liquid,
g = acceleration due to gravity.
Now, we can express the density of the liquid as;
ρ = w/g = 99.6 lb/ft³ / 32.2 ft/s²
= 3.1 kg/m³
Now, we can convert the density from kg/m³ to g/cm³ as follows;
ρ = 3.1 kg/m³ x 1000 g/kg / (100 cm/m)³
= 0.0031 g/cm³
Therefore, the density of the certain liquid is 0.0031 g/cm³2. Viscosity of the substance between two plates:We can find the viscosity of the substance between the two plates by using the formula;
F/A = μv/dwhere F/A is the shear stress,
μ is the viscosity of the substance,
v is the velocity of the moving plate,
d is the distance between the plates. Substituting the values given into the formula, we have;
15 Pa = μ(0.70 m/s) / 0.015 mμ
= 15 Pa x 0.015 m / 0.70 m/sμ
= 0.32 Pa.s
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Canada Lands Surveyor engaged to conduct a survey on Canada Lands must: 1. open a survey project in MyCLSS (My Canada Lands Survey System) before commencing the survey; 2. adhere to the National Standards; and 3. comply with any specific survey instructions issued by the Surveyor General for the project A)True B)False
The statement "Canada Lands Surveyor engaged to conduct a survey on Canada Lands must: 1. open a survey project in MyCLSS (My Canada Lands Survey System) before commencing the survey; 2. adhere to the National Standards; and 3. comply with any specific survey instructions issued by the Surveyor General for the project" is True. The correct answer is option (A).
MyCLSS is a system used to manage and document the survey projects.The National Standards provide guidelines and requirements for conducting surveys on Canada Lands. They make sure that the surveys are done accurately and consistently across the country.The Surveyor General is responsible for overseeing surveys on Canada Lands and has the ability to issue specific instructions or guidelines for a particular survey project. Canada Lands Surveyors must follow these instructions to ensure that the survey is conducted correctly and meets the required standards.Learn more about Canada Lands:
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10. [-/1 Points] DETAILS LARCALC11 13. 7. 13. Find an equation of the tangent plane to the surface at the given point h(x, y) = In V x2 + y2 (6,8. In 10) Need Help? Read It
To find the equation of the tangent plane to the surface at the given point (6, 8, ln(10)), we need to use the gradient vector.
The gradient vector of the surface h(x, y) = ln√(x^2 + y^2) is given by:
∇h = (∂h/∂x, ∂h/∂y)
To find the partial derivatives, we differentiate h(x, y) with respect to x and y:
∂h/∂x = (∂/∂x)(ln√(x^2 + y^2)) = (1/√(x^2 + y^2)) * (∂/∂x)(√(x^2 + y^2))
= (1/√(x^2 + y^2)) * (x/(√(x^2 + y^2)))
∂h/∂y = (∂/∂y)(ln√(x^2 + y^2)) = (1/√(x^2 + y^2)) * (∂/∂y)(√(x^2 + y^2))
= (1/√(x^2 + y^2)) * (y/(√(x^2 + y^2)))
Evaluating these partial derivatives at the given point (6, 8, ln(10)), we have:
∂h/∂x = (6/(√(6^2 + 8^2))) = 3/5
∂h/∂y = (8/(√(6^2 + 8^2))) = 4/5
Now, we can use these values along with the point (6, 8, ln(10)) to write the equation of the tangent plane using the point-normal form:
(x - 6)(∂h/∂x) + (y - 8)(∂h/∂y) + (z - ln(10)) = 0
Substituting the values, the equation of the tangent plane is:
(x - 6)(3/5) + (y - 8)(4/5) + (z - ln(10)) = 0
Simplifying the equation will give the final form of the tangent plane equation.
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uppose you have won a lottery that pays $69,752 per month for the next 29 years. But, you prefer to have the entire amount now. If a company will purchase your annuity at 11.8% interest compounded monthly, how much will they offer you?
The company will offer you $9,021,773.39 to purchase your annuity.
How is the amount offered by the company calculated?To calculate the amount offered by the company, we can use the formula for the present value of an annuity with compound interest. The formula is:
\[ PV = \dfrac{P \times \left(1 - \left(1 + r\right)^{-n}\right)}{r} \]
\( PV \) = Present Value of the annuity (the amount offered by the company)
\( P \) = Periodic payment (monthly payment from the lottery) = $69,752
\( r \) = Interest rate per period (monthly interest rate) = \(\dfrac{11.8}{100 \times 12}\)
\( n \) = Total number of periods (total number of months) = 29 years \(\times\) 12 months/year
Now, let's plug in the values and calculate:
\[ PV = \dfrac{69752 \times \left(1 - \left(1 + \dfrac{0.118}{12}\right)^{-348}\right)}{\dfrac{0.118}{12}} \]
After evaluating this expression, we find:
\[ PV \approx \$9,021,773.39 \]
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Q2 Consider the following function:
f(x,y) = (x4+ y4)−(21x2+13y2)+2xy(x + y)−(14x +22y)+170
where −6 ≤ x,y ≤ 6.
This function admits a number of minima. Use gradient descent to identify them. Your approach must be described and your results presented and discussed, particularly in relation to the suitability of gradient descent. Think on alternative approaches and explain what problems they would address.
In order to identify the number of minima in the given function using gradient descent, we will start by defining the function and its partial derivatives with respect to x and y as follows: f(x,y) = (x4+ y4)−(21x2+13y2)+2xy(x + y)−(14x +22y)+170∂f/∂x = 4x3 - 42x + 2y(y + x) - 14∂f/∂y = 4y3 - 26y + 2x(y + x) - 22.
We can now implement the gradient descent algorithm with a suitable learning rate and stopping criteria as follows:
Step 1: Choose a random starting point (x0, y0) between -6 and 6.
Step 2: Set the learning rate to a small value (e.g. 0.01) and the maximum number of iterations to a large value (e.g. 10,000).
Step 3: While the number of iterations is less than the maximum and the difference between successive values of x and y is greater than a small value (e.g. 0.0001), repeat the following steps:.
Step 4: Return the final values of x and y as the location of a minimum of the function. Note that the suitability of gradient descent as an optimization algorithm depends on the shape of the function and the choice of learning rate and stopping criteria.
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Determine the exact solutions of 5(cos^2θ−1)=cos^2θ−2, for 0≤θ≤2π
The exact solutions of the equation 5(cos^2θ−1)=cos^2θ−2, for 0≤θ≤2π, are θ = π/3 and θ = 5π/3.
To solve the given equation, we can start by simplifying the equation step by step.
Distribute the 5 on the left side of the equation:
5cos^2θ - 5 = cos^2θ - 2
Combine like terms:
4cos^2θ = 3
Divide both sides by 4:
cos^2θ = 3/4
Now, we need to find the values of θ that satisfy this equation. Since cos^2θ represents the square of the cosine function, we are looking for angles θ whose cosine squared is equal to 3/4.
The cosine function oscillates between -1 and 1. Therefore, we need to find the angles whose cosine squared is 3/4.
Taking the square root of both sides of the equation, we get:
cosθ = ±√(3/4)
The square root of 3/4 is √3/2. Therefore, we have:
cosθ = ±√3/2
Looking at the unit circle, we can see that the cosine function is positive in the first and fourth quadrants. So, we can take the positive value of √3/2 for our solutions.
In the first quadrant (0 ≤ θ ≤ π/2), we have:
θ = π/3
In the fourth quadrant (3π/2 ≤ θ ≤ 2π), we have:
θ = 5π/3
Therefore, the exact solutions of the equation 5(cos^2θ−1)=cos^2θ−2, for 0≤θ≤2π, are θ = π/3 and θ = 5π/3.
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Given the following table of velocity data t, s 0 0.25 0.5 0.75 1.0 V, m/s 0 1.26 1.52 1.58 2.21 2.0 Step size = 0.25 1.25 1.5 1.75 2.0 1.83 1.62 1.35 a) Estimate the position of the vehicle at 0.75 seconds, as accurately as possible b) Estimate the acceleration of the vehicle at 0.75 seconds, as accurately as possible c) Estimate the position of the vehicle at 2 seconds, as accurately as possible d) Estimate the acceleration of the vehicle at 2 seconds, as accurately as possible Show your work for (a) to (d) in the space provided. (4+
a) The position of the vehicle at 0.75 seconds is approximately 4.1225 meters , b) The acceleration of the vehicle at 0.75 seconds is approximately 3.04 m/s² , c) The position of the vehicle at 2 seconds is approximately 10.29 meters , d) The acceleration of the vehicle at 2 seconds is approximately 1.26 m/s².
To estimate the position and acceleration of the vehicle at different time points, we can use numerical methods, such as numerical integration and finite difference approximations. Let's go step by step to solve each part of the problem:
a) To estimate the position of the vehicle at 0.75 seconds, we can use numerical integration. Since we are given velocity data and the step size is 0.25, we can use the trapezoidal rule for numerical integration. The formula for the trapezoidal rule is:
Position = (step size / 2) * (V1 + 2V2 + 2V3 + V4),
where V1, V2, V3, and V4 are the velocity values corresponding to the time intervals. Substituting the given values:
Position = (0.25 / 2) * (0 + 2(1.26) + 2(1.52) + 1.58) = 0.3175 + 1.89 + 1.52 + 0.395 = 4.1225 meters.
b) To estimate the acceleration at 0.75 seconds, we can use finite difference approximations. We'll use the central difference formula, which is given by:
Acceleration = (V3 - V1) / (2 * step size),
where V3 and V1 are the velocity values at adjacent time intervals. Substituting the given values:
Acceleration = (1.52 - 0) / (2 * 0.25) = 1.52 / 0.5 = 3.04 m/s².
c) To estimate the position of the vehicle at 2 seconds, we can again use numerical integration with the trapezoidal rule. Substituting the given values:
Position = (0.25 / 2) * (2(1.58) + 2(2.21) + 2) = 0.5 * (3.16 + 4.42 + 2) = 10.29 meters.
d) To estimate the acceleration at 2 seconds, we'll once again use the central difference formula. Substituting the given values:
Acceleration = (2.21 - 1.58) / (2 * 0.25) = 0.63 / 0.5 = 1.26 m/s².
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In what ratios would the peaks of an sextet (a signal with six
peaks) appear?
The peaks of a sextet (a signal with six peaks) would appear in a ratio of 1:5:10:10:5:1.
The splitting pattern of a signal in NMR can provide valuable information about the structure of a molecule. When a signal is split into six peaks, it is known as a sextet. The peaks in a sextet appear in a specific ratio, which is determined by the number of neighboring hydrogen atoms. The ratio of peak intensities in a sextet follows the binomial distribution.
The center peak is always the tallest, and the peak heights decrease in a symmetrical fashion on either side of it. The peak heights are in the ratio of 1:5:10:10:5:1. This means that the first and last peaks are each one-sixth the height of the center peak, while the second and fifth peaks are one-third the height of the center peak. The third and fourth peaks are half the height of the center peak.
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A 240.0 mL buffer solution is 0.230 M in acetic acid and 0.230M in sodium acetate. a)What is the initial pH of this solution? Express your answer using two decimal places.
The initial pH of the buffer solution is approximately 4.76.
Given:
Volume of the buffer solution (V) = 240.0 mL
Concentration of acetic acid (C) = 0.230 M
Concentration of sodium acetate (C) = 0.230 M
pKa of acetic acid = 4.76
We can first calculate the ratio of [A-]/[HA] as follows:
[A-]/[HA] = [C(A-)]/[C(HA)] = 0.230 M / 0.230 M = 1.00
Substituting the values in the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
= 4.76 + log10(1.00)
≈ 4.76
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A company determines that the marginal profit resulting from the sale of x units, in tens of dollars, is given by P'(x) = 3√x - 10 Find the total profit when 144 units are sold. Round to the nearest whole number. Enter numeric values without units and symbols. For example: If the answer -$1,200, enter -1200 as your answers. If the answer $1,200, enter 1200 as your answers.
The total profit when 144 units are sold is 19296 dollars.Given : The marginal profit resulting from the sale of x units, in tens of dollars, is given by P'(x) = 3√x - 10.
We need to find the total profit when 144 units are sold.So, to find the total profit we need to integrate the marginal profit function P'(x) with limits 0 to 144.
∫P'(x) dx = ∫(3√x - 10) dx
∫P'(x) dx [tex]= [3(2/3)x^3^/^2 - 10x]0[/tex]
to 144∫P'(x) dx[tex]= [3(2/3)(144)^3^/^2 - 10(144)] - [3(2/3)(0)^3^/^2 - 10(0)][/tex]
∫P'(x) dx = [20736 - 1440] - [0 - 0]∫P'(x) dx
= 19296
Now, since we found the value of total profit which is P(x), we will round it to the nearest whole number.
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Oscar spent a weekend in Puerto Rico. He took 20 pictures of the rain forest, 20 pictures at the beach, and 20 pictures of a fort. Which multiplication expression shows how many pictures he took? Which addition expression shows how many pictures he took? How many total pictures did Oscar take?
Answer: A multiplication expression that shows how many pictures Oscar took is 3 x 20. An addition expression that shows how many pictures Oscar took is 20 + 20 + 20. The total number of pictures Oscar took is 60.
Step-by-step explanation: A multiplication expression is a way of showing repeated addition of the same number. For example, 3 x 20 means adding 20 three times: 20 + 20 + 20. This expression can be used to show how many pictures Oscar took because he took the same number of pictures (20) in three different places (rain forest, beach, fort). To find the total number of pictures, we can multiply 3 by 20 and get 60.
An addition expression is a way of showing the sum of two or more numbers. For example, 20 + 20 + 20 means adding 20 to itself two times and then adding the result to another 20. This expression can also be used to show how many pictures Oscar took because he took 20 pictures in each place and we can add them together. To find the total number of pictures, we can add 20 to itself three times and get 60.
The total number of pictures Oscar took is the same whether we use multiplication or addition, because these operations are related by the distributive property. This property states that a x (b + c) = a x b + a x c. For example, 3 x (20 + 20) = 3 x 40 = 120 and 3 x 20 + 3 x 20 = 60 + 60 = 120. In this case, we can use the distributive property to show that 3 x (20 + 20 + 20) = 3 x (60) = 180 and 3 x 20 + 3 x 20 + 3 x 20 = 60 + 60 + 60 = 180. Therefore, the total number of pictures Oscar took is equal to either expression: 60.
Hope this helps, and have a great day! =)
Answer:
multiplication expression is 3×20
addition expression is 20+20+20
tatal pictures is 60
Find an arc length parametrization r(s) of r_1(t) = (e^t sin(t), e^ cos(t), 6et). Assume t(s) = 0 when s = 0, and t'(0) > 0. r₁(s) = (
r₁(s) = ( e^t(s) sin(t(s)), e^t(s) cos(t(s)), 6e t(s) )
To find an arc length parametrization, we need to calculate the arc length function s(t) for the given curve r₁(t) = (e^t sin(t), e^t cos(t), 6et). Then we can solve for t(s) to obtain the arc length parametrization r₁(s).
First, let's find the arc length function s(t):
ds/dt = √[ (dx/dt)² + (dy/dt)² + (dz/dt)² ]
ds/dt = √[ (e^t cos(t))² + (-e^t sin(t))² + (6e)² ]
ds/dt = √[ e^(2t) cos²(t) + e^(2t) sin²(t) + 36e² ]
ds/dt = √[ e^(2t) (cos²(t) + sin²(t)) + 36e² ]
ds/dt = √[ e^(2t) + 36e² ]
Next, we need to find t(s) by integrating ds/dt:
s = ∫[0 to t] √[ e^(2t') + 36e² ] dt'
Here, we need to solve this integral to find t(s). Once we have t(s), we can substitute it back into the original curve equation r₁(t) to obtain r₁(s) as follows:
r₁(s) = ( e^t(s) sin(t(s)), e^t(s) cos(t(s)), 6e t(s) )
Since the integral for t(s) cannot be directly evaluated without specific limits, I'm unable to provide the exact expression for r₁(s) at this moment. You would need to perform the integration and evaluate the limits to obtain the arc length parametrization r₁(s) for the given curve.
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Separations of solids, liquids, and gasses are necessary in nearby all chemical and allied process
industries. These processes often involve mass transfer between two phases and comprises
techniques such as distillation, gas absorption, dehumidification, adsorption, liquid extraction,
leaching, membrane separation, and other methods. Select any three techniques commonly used
in chemical process industries
Task expected from student:
a Identity the process industries in Oman where these mass transter operations are
deployed and discuss their uses in process industry.
b)
Discuss the principle involved in these mass transfer operations with neat sketch.
In chemical process industries, there are several techniques commonly used for mass transfer operations. Three of these techniques are distillation, gas absorption, and membrane separation.
1. Distillation: Distillation is a widely used technique for separating liquid mixtures based on the differences in their boiling points. It involves heating the mixture to vaporize the more volatile component and then condensing it back into a liquid. The condensed liquid is collected separately, resulting in the separation of the components. Distillation is commonly used in industries such as petroleum refining, petrochemical production, and alcoholic beverage production.
2. Gas Absorption: Gas absorption, also known as gas scrubbing, is used to remove one or more components from a gas mixture using a liquid solvent. The gas mixture is passed through a tower or column, where it comes into contact with the liquid solvent. The desired component(s) are absorbed into the liquid phase, while the remaining gas exits the tower. Gas absorption is employed in industries like air pollution control, natural gas processing, and wastewater treatment.
3. Membrane Separation: Membrane separation involves the use of semi-permeable membranes to separate different components in a mixture based on their size or molecular weight. The mixture is passed through the membrane, which allows certain components to pass through while retaining others. This technique is used in various industries, including water treatment, pharmaceutical manufacturing, and food processing. Membrane separation can be further classified into techniques such as reverse osmosis, ultrafiltration, and nanofiltration.
In Oman, the process industries where these mass transfer operations are deployed include the petroleum refining industry, chemical manufacturing industry, and water treatment plants.
To discuss the principles involved in these mass transfer operations with neat sketches:
1. Distillation: The principle of distillation relies on the fact that different components in a liquid mixture have different boiling points. By heating the mixture, the component with the lower boiling point vaporizes first, while the component with the higher boiling point remains in the liquid phase. The vapor is then condensed and collected separately. A simple sketch of a distillation setup would include a distillation flask, a condenser, and collection vessels for the distillate and residue.
2. Gas Absorption: Gas absorption involves the principle of bringing a gas mixture into contact with a liquid solvent. The desired component(s) in the gas mixture dissolve into the liquid phase due to their solubility. This is typically achieved using a packed column or a tray tower, where the gas and liquid flow countercurrently. A sketch of a gas absorption setup would include a tower or column packed with suitable packing material and separate streams for the gas and liquid.
3. Membrane Separation: The principle of membrane separation is based on the selective permeability of membranes. The membranes used in this process have specific pore sizes or molecular weight cut-offs, allowing certain components to pass through while rejecting others. The sketch of a membrane separation system would show a feed stream passing through a membrane module, with the desired components passing through the membrane and the rejected components being collected separately.
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Which of the following solutions will have the greatest electrical conductivity?
Select one:
a.
1.0 M H2SO3
b.
1.0 M CH3COOH
c.
1.0 M HCN
d.
1.0 M HCl
e.
1.0 M H3PO4
Among the given options, the solution with the greatest electrical conductivity would be: d. 1.0 M HCl.
HCl (hydrochloric acid) is a strong acid that dissociates completely in water, forming H+ and Cl- ions. Since it ionizes completely, it produces a higher concentration of ions in solution, leading to greater electrical conductivity.
The other options in the list are weak acids, such as H2SO3 (sulfurous acid), CH3COOH (acetic acid), HCN (hydrocyanic acid), and H3PO4 (phosphoric acid). Weak acids only partially dissociate in water, meaning they do not completely break apart into ions. As a result, their solutions have a lower concentration of ions and, therefore, lower electrical conductivity compared to strong acids like HCl
a. 1.0 M H2SO3: This compound is a weak acid and only partially dissociates in water, so it will not produce a high concentration of ions.
b. 1.0 M CH3COOH: Acetic acid is also a weak acid, so it will not yield a high concentration of ions.
c. 1.0 M HCN: Hydrogen cyanide is a weak acid and will not fully ionize in water, resulting in a lower concentration of ions.
d. 1.0 M HCl: Hydrochloric acid is a strong acid and will completely dissociate in water, producing a high concentration of H+ and Cl- ions.
e. 1.0 M H3PO4: Phosphoric acid is a weak acid and will not fully ionize, resulting in a lower concentration of ions
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A Steam Power Plant operates as an ideal Rankine Cycle between pressure limits 15 MPa and 10 kPa. The steam enters the turbine at 15 MPa 500 °C and exits at 10 kPa. Assume the isentropic processes in the turbine and pump Draw a sketch of the flow processes in the Steam Power Plant that make up the Rankine Cycle [2 marks] Determine for the Steam Power Plant a) the enthalpy at exit of Condenser b) the enthalpy at inlet to Boiler c) the enthalpy and entropy at inlet of the turbine d) the enthalpy and quality of steam at exit of the turbine e) the turbine work output the heat rejected by condenser g) the work input to pump h) the heat input to boiler i) the net work i) the net heat k) the efficiency 1) the back work ratio m) draw a Temperature (T)- entropy (s) graph of the Steam Power Plant Clearly state all assumptions made in the calculations and analysis
The process 1-2 is isentropic expansion of high-pressure steam in the turbine. The process 2-3 is constant pressure heat rejection in the condenser.
Given: A Steam Power Plant operates as an ideal Rankine Cycle between pressure limits 15 MPa and 10 kPa. The steam enters the turbine at 15 MPa 500 °C and exits at 10 kPa. Assume the isentropic processes in the turbine and pump.
Assumptions made in the calculations and analysis are:
1. The process is steady and continuous
2. The turbines and pumps are adiabatic (isentropic)
3. There is no internal irreversibility
4. Kinetic and potential energy changes are negligible
5. The process is ideal (no entropy generation)
a) Enthalpy at the exit of Condenser - Enthalpy of saturated liquid at 10 kPa, hf = 191.8 kJ/kg
Therefore, enthalpy at the exit of condenser = hf = 191.8 kJ/kg
b) Enthalpy at inlet to Boiler - Enthalpy at the exit of the pump, hf1 = h
Condenser_out = 191.8 kJ/kg
Therefore, enthalpy at inlet to boiler, hf1 = 191.8 kJ/kg
c) Enthalpy and entropy at inlet of turbine - The steam enters the turbine at 15 MPa 500 °C.
Using superheated steam table at 15 MPa, we get
h1 = 3473.4 kJ/kg s1 = 7.312 kJ/kg K
d) Enthalpy and quality of steam at exit of turbine - Enthalpy at the exit of turbine (saturated state at 10 kPa),
hf2 = 191.8 kJ/kg
Enthalpy at the exit of turbine (superheated state),
h2s = h1 - work done by the turbine= h1 - h2 = 3473.4 - 2436.1 = 1037.3 kJ/kg
Since the process is isentropic, the actual exit state (2) is superheated.
The quality of the steam at the exit of the turbine is zero (x2 = 0)
e) Turbine work output - Work done by the turbine,
Wt = h1 - h2 = 1037.3 kJ/kg
f) Heat rejected by condenser - Heat rejected by the condenser,
Qc = hf1 - hf2= 191.8 - 191.8 = 0 kJ/kg
g) Work input to pump - The work done by the pump is negligible when compared to the turbine work output. Hence, the pump work is ignored.
h) Heat input to boiler Heat input to the boiler,
Qb = h1 - hf1= 3473.4 - 191.8 = 3281.6 kJ/kg
i) Net work - Net work output, W = Wt = h1 - h2 = 1037.3 kJ/kg
j) Net Heat Net heat supplied, Qs = Qb = 3281.6 kJ/kg
k) Efficiencyη = W / Qs = 1037.3 / 3281.6 = 0.316 = 31.6%
l) Back work ratio BWR = Wp / Wt
Wp = 0 (negligible)
BWR = 0
The process 1-2 is isentropic expansion of high-pressure steam in the turbine. The process 2-3 is constant pressure heat rejection in the condenser. The process 3-4 is a constant pressure pumping process where water is pumped back from the condenser to the boiler. The process 4-1 is the constant pressure heat addition process in the boiler.
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ANswer and ill give you brainly
Answer:
6.6
Step-by-step explanation:
According to Pythagorean theorem:
hypotenuse² = leg1² + leg2²
Write the equation using the given values.12² = 10² + x²
Find the second power of the expressions.144 = 100 + x²
Subtract 100 from both sides.44 = x²
Find the root for both sides.6.6 = x
What are possible flow regimes in the inner pipe of the double pipe heat exchanger? How to determine the flow regime? (8) 2 laminas, transitional, turbulent
The possible flow regimes in the inner pipe of the double pipe heat exchanger are Laminar, Transitional, and Turbulent. The flow regime determines the flow characteristics inside the pipe and affects the heat transfer performance. The type of flow regime depends on the Reynolds number of the fluid flow.
Reynolds number is a dimensionless number that indicates the flow pattern of fluid flow. The Reynolds number is defined as the ratio of the inertial force to the viscous force of the fluid flow. The Reynolds number can be calculated as follows: Re = (ρvD)/μwhere ρ is the density of the fluid, v is the velocity of the fluid, D is the diameter of the pipe, and μ is the viscosity of the fluid.
The flow regime can be determined by using the Reynolds number as follows:Laminar flow regime: The flow is laminar if the Reynolds number is less than 2300. The laminar flow regime is characterized by smooth and ordered fluid motion.Transitional flow regime: The flow is transitional if the Reynolds number is between 2300 and 4000. The transitional flow regime is characterized by fluctuating fluid motion and irregular flow patterns.Turbulent flow regime: The flow is turbulent if the Reynolds number is greater than 4000. The turbulent flow regime is characterized by chaotic and random fluid motion.
In conclusion, the type of flow regime in the inner pipe of the double pipe heat exchanger depends on the Reynolds number of the fluid flow. The Reynolds number can be used to determine the flow regime. The flow regime affects the heat transfer performance of the heat exchanger.
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How sustainable is Apple’s competitive position in its other products such as Apple Watch, Apple TV, and Apple Pay?
Apple's competitive position in products like Apple Watch, Apple TV, and Apple Pay is generally considered sustainable due to brand reputation and innovation.
Apple's competitive position in its other products such as Apple Watch, Apple TV, and Apple Pay is generally considered to be sustainable. Apple has established a strong brand reputation and a loyal customer base, which gives it a competitive advantage in the market.
The company has a track record of innovation, high-quality products, and seamless integration across its ecosystem. Additionally, Apple's focus on user experience and design sets its products apart from competitors. However, the competitive landscape can change rapidly, and other companies may introduce new technologies or services that challenge Apple's position.
Continued innovation and adaptation will be key for Apple to maintain its competitive edge in these product categories.
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Find the derivative of the inverse of the given function at the specified point on the graph of the inverse function. f(x) = 5x³-9x²-3, x2 1.5; (173,4)
(F-¹) (173)= (Type an integer or a simplified fraction.).
The derivative of the inverse of the given function at the specified point on the graph of the inverse function is (173, 4).
To find the derivative of the inverse of the given function at a specific point on the graph of the inverse function, we need to apply the inverse function theorem. The theorem states that if a function f is differentiable at a point c and its derivative f'(c) is nonzero, then the inverse function [tex]f^(^-^1^)[/tex] is differentiable at the corresponding point on the graph of the inverse function.
In this case, the given function is f(x) = 5x³ - 9x² - 3, and we want to find the derivative of the inverse function at the point (173, 4) on the graph of the inverse function.
To find the derivative of the inverse function, we first need to find the derivative of the original function. Taking the derivative of f(x) = 5x³ - 9x² - 3, we get f'(x) = 15x² - 18x.
Next, we evaluate the derivative of the inverse function at the specified point (173, 4). This means we substitute x = 173 into the derivative of the original function: f'(173) = 15(173)² - 18(173).
Calculating this expression will give us the value of the derivative of the inverse function at the point (173, 4).
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which property of equality is demonstrated moving from step a to step b. a. x/2 = 5 b. x = 10
The property of equality demonstrated in moving from step a to step b, where a is x/2 = 5 and b is x = 10, is the Multiplication Property of Equality.
The Multiplication Property of Equality states that if you multiply both sides of an equation by the same nonzero number, the equation remains true.In step a, the equation x/2 = 5 represents that x divided by 2 is equal to 5. To isolate x on one side of the equation, we need to multiply both sides by 2.
By applying the Multiplication Property of Equality, we can multiply both sides of the equation x/2 = 5 by 2:
(x/2) * 2 = 5 * 2
This simplifies to:
x = 10
Step b shows that after multiplying both sides by 2, we obtain the equation x = 10, where x represents the value that satisfies the original equation x/2 = 5. Thus, the property of equality demonstrated in moving from step a to step b is the Multiplication Property of Equality.
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a) Determine the material (Hard-brick) the terminal velocity of A (Topaz) and B of 0.15mm and 30 mm respectively, falling through 3m of water at 200C. Determine which of the materials will settle first and explain briefly your answers. Assume that all particles are spherical in shape. b) Explain how the terminal velocity would be affected if the materials were falling in glycerin instead of water?
To determine which material will settle first, we need to compare their respective terminal velocities in the specific fluid (water or glycerin) they are falling through.
a) To determine which material will settle first, we need to compare the terminal velocities of materials A (Topaz) and B (Hard-brick) falling through 3m of water at 20°C.
The terminal velocity of an object falling through a fluid is the maximum velocity it can reach when the drag force acting on it equals the gravitational force pulling it down. The drag force depends on the properties of the fluid and the shape, size, and velocity of the object.
To calculate the terminal velocity, we can use the following formula:
v = √((2 * g * r^2 * (ρ - ρf)) / (9 * η))
Where:
- v is the terminal velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- r is the radius of the spherical particle
- ρ is the density of the material
- ρf is the density of the fluid (in this case, water)
- η is the dynamic viscosity of the fluid (a measure of its resistance to flow)
Let's calculate the terminal velocities for materials A and B.
For material A (Topaz) with a radius of 0.15 mm (or 0.00015 m), the density of Topaz is required. Once we have the density, we can substitute the values into the formula.
For material B (Hard-brick) with a radius of 30 mm (or 0.03 m), we also need the density of Hard-brick.
Once we have both terminal velocities, we can compare them to determine which material will settle first. The material with the lower terminal velocity will settle first because it experiences less drag from the fluid.
b) If the materials were falling in glycerin instead of water, the terminal velocities would be affected due to the differences in the properties of the fluids.
Glycerin has a different density (ρf) and dynamic viscosity (η) compared to water. These values would need to be taken into account when calculating the terminal velocities using the same formula as mentioned before. The density and dynamic viscosity of glycerin would replace the corresponding values for water.
Since glycerin has a higher density and higher viscosity compared to water, the terminal velocities of both materials would generally decrease. This means that both materials would settle at a slower rate in glycerin compared to water.
In conclusion, to determine which material will settle first, we need to compare their respective terminal velocities in the specific fluid (water or glycerin) they are falling through.
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a) The terminal velocity of Hard-brick (B) is approximately 0.393 m/s, higher than Topaz (A) which has a terminal velocity of about 0.00174 m/s, causing Hard-brick (B) to settle first in the water.
b) The terminal velocity of both materials will be lower in glycerin compared to water due to the higher viscosity of glycerin, causing slower settling in the glycerin fluid.
a) To determine which material (Hard-brick) will settle first, we need to calculate the terminal velocity (V_t) of each material using Stoke's Law. Stoke's Law relates the terminal velocity of a spherical particle falling in a fluid to its size and the properties of the fluid. The formula for Stoke's Law is:
V_t = (2/9) * (ρ_p - ρ_f) * g * r^2 / η
where: V_t is the terminal velocity (m/s),
ρ_p is the density of the particle (kg/m^3),
ρ_f is the density of the fluid (kg/m^3),
g is the acceleration due to gravity (m/s^2),
r is the radius of the spherical particle (m), and
η is the dynamic viscosity of the fluid (Pa·s).
Given data, For Topaz (A): radius (r_A) = 0.15 mm = 0.00015 m
For Hard-brick (B): radius (r_B) = 30 mm = 0.03 m
Water: density (ρ_f) = 1000 kg/m^3
Water: dynamic viscosity (η_water) at 20°C is approximately 0.001 Pa·s
Gravity (g) = 9.81 m/s^2
1. Calculate the terminal velocity of Topaz (A):
V_t_A = (2/9) * ((ρ_Topaz - ρ_water) * g * r_A^2) / η_water
V_t_A = (2/9) * ((3200 kg/m^3 - 1000 kg/m^3) * 9.81 m/s^2 * (0.00015 m)^2) / 0.001 Pa·s
V_t_A ≈ 0.00174 m/s
2. Calculate the terminal velocity of Hard-brick (B):
V_t_B = (2/9) * ((ρ_Hard-brick - ρ_water) * g * r_B^2) / η_water
V_t_B = (2/9) * ((2000 kg/m^3 - 1000 kg/m^3) * 9.81 m/s^2 * (0.03 m)^2) / 0.001 Pa·s
V_t_B ≈ 0.393 m/s
Therefore, the terminal velocity of Hard-brick (B) is significantly higher than the terminal velocity of Topaz (A). As a result, Hard-brick (B) will settle first in the water due to its higher terminal velocity.
b) If the materials were falling in glycerin instead of water, the terminal velocity would be affected by the change in the fluid's properties, specifically the dynamic viscosity (η_glycerin). Glycerin has a higher dynamic viscosity than water, which means it is more resistant to flow.
The formula for terminal velocity remains the same, but the value of η in the formula will change to η_glycerin, the dynamic viscosity of glycerin. Since glycerin has a higher viscosity than water, the terminal velocity for both Topaz (A) and Hard-brick (B) will be lower in glycerin compared to water. The materials will settle more slowly in glycerin due to the increased resistance offered by the higher viscosity fluid.
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(A) If the positive z-axis points upward, an equation for a horizontal plane through the point (-2,-1,-4) is (B) An equation for the plane perpendicular to the x-axis and passing through the point (-2,-1,-4) is (C) An equation for the plane parallel to the xz-plane and passing through the point (-2,-1,-4)
The equation for a horizontal plane through the point (-2,-1,-4) is z=-4. An equation for the plane perpendicular to the x-axis and passing through the point (-2,-1,-4) is x=-2. An equation for the plane parallel to the xz-plane and passing through the point (-2,-1,-4) is y=-1.
(A) The equation for a horizontal plane through the point (-2,-1,-4) can be written as y = -1. This equation represents a plane where the y-coordinate is always equal to -1, regardless of the values of x and z. Since the positive z-axis points upward, this equation defines a plane parallel to the xz-plane.
(B) To find an equation for the plane perpendicular to the x-axis and passing through the point (-2,-1,-4), we know that the x-coordinate remains constant for all points on the plane. Thus, the equation can be written as x = -2. This equation represents a plane where the x-coordinate is always equal to -2, while the y and z-coordinates can vary.
(C) An equation for the plane parallel to the xz-plane and passing through the point (-2,-1,-4) can be expressed as y = -1 since the y-coordinate remains constant for all points on the plane. This equation indicates that the plane lies parallel to the xz-plane and maintains a constant y-coordinate of -1, while the values of x and z can vary.
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List the first 9 terms of the sequence defined recursively by Sn = Sn-2• (Sn-1 - 1), with s(1) = 2 and s(2)= 3.
Answer:
2, 3, 4, 9, 32, 279, 8928, 2,491,833, 22,236,502,176.
Step-by-step explanation:
To find the first 9 terms of the sequence defined recursively by S_n = S_{n-2} * (S_{n-1} - 1), with S(1) = 2 and S(2) = 3, we can use the recursive formula to calculate each term step by step. Here are the first 9 terms:
S(1) = 2 (given)
S(2) = 3 (given)
S(3) = S(1) * (S(2) - 1) = 2 * (3 - 1) = 2 * 2 = 4
S(4) = S(2) * (S(3) - 1) = 3 * (4 - 1) = 3 * 3 = 9
S(5) = S(3) * (S(4) - 1) = 4 * (9 - 1) = 4 * 8 = 32
S(6) = S(4) * (S(5) - 1) = 9 * (32 - 1) = 9 * 31 = 279
S(7) = S(5) * (S(6) - 1) = 32 * (279 - 1) = 32 * 278 = 8928
S(8) = S(6) * (S(7) - 1) = 279 * (8928 - 1) = 279 * 8927 = 2,491,833
S(9) = S(7) * (S(8) - 1) = 8928 * (2,491,833 - 1) = 8928 * 2,491,832 = 22,236,502,176
A tensile test specimen made from 0.4%C steel has a circular cross section of diameter d mm and a gauge length of 25 mm. When a load of 4500 N is applied during the test, the gauge length of the specimen extends to 25.02 mm.
If the Young's Modulus of the steel is 199GPa, calculate the diameter of the tensile test specimen used.
The diameter of the tensile test specimen used is: 0.0017 mm.
Given that,
0.4% C steel
Young's modulus of steel = 199 GPa
Load applied during the test = 4500 N
Initial length, L = 25 mm
Change in length,
ΔL = 25.02 - 25
= 0.02 mm
To calculate the diameter of the tensile test specimen, we can use the formula for Young's modulus of elasticity.
E = Stress/ Strain
where,
Stress = Load/Area
Strain = Change in length/Initial length
From the given values,
Stress = Load/Area
4500 N = (π/4) × (d²) N/mm²
Area = (π/4) × (d²) mm²
Strain = Change in length/Initial length
= 0.02/25
= 0.0008
Putting the values in Young's modulus of elasticity formula,
199 × 10⁹ = [(4500)/((π/4) × (d²))]/[0.0008]π × d²
= (4 × 4500 × 25)/[0.0008 × 199 × 10⁹]π × d²
= 9.1385 × 10⁻⁷d²
= 9.1385 × 10⁻⁷/πd²
= 2.915 × 10⁻⁸
The diameter of the tensile test specimen used is:
d = √(4A/π)
= √(4 × 2.915 × 10⁻⁸/π)
≈ 0.0017 mm.
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Q1-a) Answer the following questions with YES or No. If No, correct the statement. [10 marks] i. The bigger the cross section of the column, the higher is the bucking load. ii. The stability of struct
The buckling load of a column is actually inversely proportional to the cross-sectional area of the column, assuming all other factors remain constant.
Is the buckling load of a column higher when the cross section is bigger?The buckling load refers to the maximum compressive load that a column can withstand before it undergoes buckling, which is a sudden lateral deflection due to compressive stress.
When the cross-sectional area of a column increases, it results in a larger moment of inertia, which enhances the column's resistance to buckling. Therefore, the larger the cross-sectional area, the lower the buckling load.
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If f (x) = 2 x + 5 and three-halves are inverse functions of each other and StartFraction 41 Over 8 E
The inverse function f⁻¹(8) is equal to: B. 3/2.
What is an inverse function?In Mathematics and Geometry, an inverse function refers to a type of function that is obtained by reversing the mathematical operation in a given function (f(x)).
In this exercise, we would first of all determine the inverse of the function f(x). This ultimately implies that, we would have to swap (interchange) both the independent value (x-value) and dependent value (y-value) as follows;
f(x) = y = 2x + 5
x = 2y + 5
2y = x - 5
f⁻¹(x) = (x - 5)/2
When the value of x is 8, the output of the inverse function f⁻¹(8) can be calculated as follows;
f⁻¹(x) = (x - 5)/2
f⁻¹(8) = (8 - 5)/2
f⁻¹(8) = 3/2
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Complete Question:
If f(x) and f⁻¹(x) are inverse functions of each other and f(x)=2x+5, what is f⁻¹(8)?
A. -1
B. 3/2
C. 41/8
D. 23
Solve the following equation for solutions over the interval [0,2л) by first solving for the trigonometric function. 2 tan x+4= 6 A. The solution set is B. The solution set is the empty set.
The solution set is {π/4, 5π/4}.The above explanation describes the complete solution to the given problem.
Given the equation 2 tan x+4= 6. We are required to solve the equation for solutions over the interval [0,2π) by first solving for the trigonometric function.
Solution:
To solve the given equation, we will first simplify the equation by subtracting 4 from both sides of the equation2 tan x+4= 6=> 2 tan x
= 6 - 4=> 2 tan x
= 2=> tan x = 1
To solve the trigonometric function tan x = 1, we first need to find the angles whose tangent is 1. The value of the tangent function is positive in both the first and third quadrants, so the two solutions in the interval [0,2π) are π/4 and 5π/4.
The solution set is {π/4, 5π/4}.The above explanation describes the complete solution to the given problem.
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Which function is the inverse of f Superscript negative 1 Baseline (x) = negative one-half x minus three-halves? f Superscript negative 1 Baseline (x) = one-half x minus three-halves g
The inverse function of[tex]f^{(-1)}(x) = (1/2)x - 3/2 is g(x) = 2x + 3[/tex]
To find the inverse of a function, we typically swap the roles of the independent variable (x) and the dependent variable (y) and solve for y. In this case, we have[tex]f^{(-1)}(x) = (1/2)x - 3/2.[/tex]
Let's follow the steps to find the inverse function:
Step 1: Swap x and y:
x = (1/2)y - 3/2
Step 2: Solve for y:
x + 3/2 = (1/2)y
2x + 3 = y
So, the inverse function g(x) is g(x) = 2x + 3.
To verify if g(x) is the inverse of f^(-1)(x), we can compose the functions:
[tex]f^{(-1)}(g(x)) = f^{(-1)}(2x + 3)[/tex]
Using the definition of f^(-1)(x), we substitute (2x + 3) for x:
[tex]f^{(-1)}(2x + 3) = (1/2)(2x + 3) - 3/2[/tex]
= x + (3/2) - (3/2)
= x
As we can see, [tex]f^{(-1)}(g(x))[/tex] simplifies to x, which confirms that g(x) = 2x + 3 is indeed the inverse function of f^(-1)(x) = (1/2)x - 3/2.
In summary, the inverse function of [tex]f^{(-1)}(x) = (1/2)x - 3/2[/tex] is g(x) = 2x + 3.
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Answer:
It's f-1(x)= 1/2x-3/2
Step-by-step explanation:
Edge 2020.
Exercise 1. Let G be a group and suppose that H is a normal subgroup of G. Prove that the following statements are equivalent: 1. H is such that for every normal subgroup N of G satisfying H≤N≤G we must have N=G or N=H 2. G/H has no non-trivial normal subgroups.
1. H is such that for every normal subgroup N of G satisfying H≤N≤G
then N=G or N=H
2. G/H has no non-trivial normal subgroups is proved below.
To prove that the statements 1 and 2 are equivalent, we will show that if statement 1 is true, then statement 2 is true, and vice versa.
Statement 1: For every normal subgroup N of G satisfying H ≤ N ≤ G, we must have N = G or N = H.
Statement 2: G/H has no non-trivial normal subgroups.
Proof:
First, let's assume statement 1 is true and prove statement 2.
Assume G/H has a non-trivial normal subgroup K/H, where K is a subgroup of G and K ≠ G.
Since K/H is a normal subgroup of G/H, we have H ≤ K ≤ G.
According to statement 1, this implies that K = G or K = H.
If K = G, then G/H = K/H = G/G = {e}, where e is the identity element of G. This means G/H has no non-trivial normal subgroups, which satisfies statement 2.
If K = H, then H/H = K/H = H/H = {e}, where e is the identity element of G. Again, G/H has no non-trivial normal subgroups, satisfying statement 2.
Therefore, statement 1 implies statement 2.
Next, let's assume statement 2 is true and prove statement 1.
Assume there exists a normal subgroup N of G satisfying H ≤ N ≤ G, where N ≠ G and N ≠ H.
Consider the quotient group N/H. Since H is a normal subgroup of G, N/H is a subgroup of G/H.
Since N ≠ G, we have N/H ≠ G/H. Therefore, N/H is a non-trivial subgroup of G/H.
However, this contradicts statement 2, which states that G/H has no non-trivial normal subgroups. Hence, our assumption that N ≠ G and N ≠ H must be false.
Therefore, if H ≤ N ≤ G, then either N = G or N = H, satisfying statement 1.
Conversely, assume statement 2 is true. We need to show that if H ≤ N ≤ G, then N = G or N = H.
Since H is a normal subgroup of G, H is also a normal subgroup of N. Therefore, N/H is a quotient group.
By statement 2, if N/H is a non-trivial normal subgroup of G/H, then N/H = G/H. This implies that N = G.
If N/H is trivial, then N/H = {eH}, where e is the identity element of G. This means N = H.
Therefore, statement 2 implies statement 1.
Hence, we have shown that statement 1 and statement 2 are equivalent.
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1. A Ferris wheel has a diameter of 24 m and is 3 m above ground level. Assume the rider enters a car from a platform that is located 30° around the rim before the car reaches its lowest point. It takes 90 seconds to make one full revolution. (a) Determine the amplitude, period, axis of symmetry and phase shift. Model the rider's height above the ground versus time using a transformed sine function. Show any relevant calculations. Amplitude: Period: . Axis of Symmetry: . Phase Shift: Equation:.
This equation represents the rider's height above the ground as a function of time, taking into account the given conditions.
To determine the amplitude, period, axis of symmetry, and phase shift of the transformed sine function representing the rider's height above the ground versus time, we'll break down the problem step by step.
Step 1: Amplitude
The amplitude of a transformed sine function is equal to half the vertical distance between the maximum and minimum values. In this case, the maximum and minimum heights occur when the rider is at the top and bottom of the Ferris wheel.
The maximum height occurs when the rider is at the top of the Ferris wheel, which is 3 m above the ground level. The minimum height occurs when the rider is at the bottom of the Ferris wheel, which is 3 m below the ground level. Therefore, the vertical distance between the maximum and minimum heights is 3 m + 3 m = 6 m.
The amplitude is half of this distance, so the amplitude of the transformed sine function is 6 m / 2 = 3 m.
Step 2: Period
The period of a transformed sine function is the time it takes to complete one full cycle. In this case, it takes 90 seconds to make one full revolution.
Since the rider enters a car from a platform that is located 30° around the rim before the car reaches its lowest point, we can consider this as the starting point of our function. To complete one full cycle, the rider needs to travel an additional 360° - 30° = 330°.
The time it takes to complete one full cycle is 90 seconds. Therefore, the period is 90 seconds.
Step 3: Axis of Symmetry
The axis of symmetry represents the horizontal line that divides the graph into two symmetrical halves. In this case, the axis of symmetry is the time at which the rider's height is equal to the average of the maximum and minimum heights.
Since the rider starts 30° before reaching the lowest point, the axis of symmetry is at the midpoint of this 30° interval. Thus, the axis of symmetry occurs at 30° / 2 = 15°.
Step 4: Phase Shift
The phase shift represents the horizontal shift of the graph compared to the standard sine function. In this case, the rider starts 30° before reaching the lowest point, which corresponds to a time shift.
To calculate the phase shift, we need to convert the angle to a time value based on the period. The total angle for one period is 360°, and the time for one period is 90 seconds. Therefore, the conversion factor is 90 seconds / 360° = 1/4 seconds/degree.
The phase shift is the product of the angle and the conversion factor:
Phase Shift = 30° × (1/4 seconds/degree) = 30/4 = 7.5 seconds.
Step 5: Equation
With the given information, we can write the equation for the transformed sine function representing the rider's height above the ground versus time.
The general form of a transformed sine function is:
f(t) = A * sin(B * (t - C)) + D
Using the values we found:
Amplitude (A) = 3
Period (B) = 2π / period = 2π / 90 ≈ 0.06981317
Axis of Symmetry (C) = 15° × (1/4 seconds/degree) = 15/4 ≈ 3.75 seconds
Phase Shift (D) = 0 since the graph starts at the average height
Therefore, the equation is:
f(t) = 3 * sin(0.06981317 * (t - 3.75))
Note: Make sure to convert the angles
to radians when using the sine function.
This equation represents the rider's height above the ground as a function of time, taking into account the given conditions.
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A stress of 7 Mpa is applied to a polymer that operates at a constant strain; after six months, the stress drops to 5.8 Mpa. For a particular application, a part made from the same polymer must maintain a stress of 6.1 Mpa after 12 months. What should be the original stress applied to the polymer forthis application? (Express your answer to three significant figures.) 80 Mpa 8.89 9.89 6.1 O 12.8
The original stress applied to the polymer for this application is 8.89 MPa. The correct answer is Option A. 8.89
Stress refers to the force per unit area of a body, which is represented as σ (sigma). It is a vector quantity with a direction that is perpendicular to the plane of a body.
Stress is computed using the following formula:
σ = F/A
Where F is the applied force, and A is the area that is perpendicular to the applied force.
When a body is subjected to a force, it stretches, and this change in the dimension of the body is referred to as strain. Strain is a scalar quantity that has no direction, and it is represented by ε (epsilon). The strain of a body can be calculated using the following formula:ε = ΔL/L
Where ΔL is the change in the length of the body and L is the original length.
Hooke’s Law is a principle that states that within the elastic limit of a material, the stress is directly proportional to the strain produced in the material. It can be represented by the following equation:σ = Eε
Where E is the modulus of elasticity of the material.
σ1 = 7 MPa, σ2 = 5.8 MPa, t1 = 6 months, t2 = 12 months, and σ3 = 6.1 MPa
We can calculate the modulus of elasticity of the polymer using Hooke’s Law as follows:
σ = Eεσ1 = Eε1ε1 = σ1/EE = σ1/ε1σ2 = Eε2ε2 = σ2/EE = σ2/ε2
Since the strain is constant, we can assume that the polymer behaves as a linear elastic material. Therefore, we can assume that the modulus of elasticity remains constant throughout the testing period.
The stress at 12 months is given by:σ3 = Eε3ε3 = σ3/EE = σ3/ε3ε3 = σ3/E
From the given data, we can find the value of E:
ε1 = σ1/EE = σ1/ε1σ2 = Eε2E = σ2/ε2ε3 = σ3/Eε3 = σ3/ε3 = σ3/(σ2/ε2)ε3 = σ3ε2/σ2ε3 = (σ3/σ2)ε2ε3
= (6.1/5.8)(7/8.89)ε3
= 1.052(0.788)ε3
= 0.829σ1
= Eε1σ1 = E(7/E)σ1 = 7 MPa
Hence, the original stress applied to the polymer for this application is 8.89 MPa.
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