The fines fraction of a soil to be used for a highway fill was subjected to a hydrometer analysis by placing 20 grams of dry fines in a 1 liter solution of water (dynamic viscosity 0.01 Poise at 20 degrees centigrade). The specific gravity of the solids was 2.65. a) Estimate the maximum diameter D of the particles found at a depth of 5 cm after a sedimentation time of 4 hours has elapsed, if the solution's concentration has reduced to 2 grams/ liter at the level. At that moment, b) What percentage of the sample would have a diameter smaller than D? c) What type of soil is this?

The depth of the compression block can be determined using the formula:

d = (A - As) / b

Where:
d = depth of the compression block
A = area of the concrete section
As = area of steel reinforcement
b = width of the compression block

First, let's calculate the area of the concrete section:
A = width * depth
A = 1000 mm * (350 mm - 40 mm)
A = 1000 mm * 310 mm
A = 310,000 mm^2

Next, let's calculate the area of steel reinforcement at the top:
Ast = number of bars * area of each bar
Ast = 3 * (π * (28 mm / 2)^2)
Ast = 3 * (π * 14^2)
Ast = 3 * (π * 196)
Ast = 3 * 615.75
Ast = 1,847.25 mm^2

Similarly, let's calculate the area of steel reinforcement at the bottom:
Asb = 5 * (π * (32 mm / 2)^2)
Asb = 5 * (π * 16^2)
Asb = 5 * (π * 256)
Asb = 5 * 803.84
Asb = 4,019.20 mm^2

Now, let's calculate the width of the compression block:
b = width - cover - (Ø/2)
b = 1000 mm - 40 mm - 28 mm
b = 932 mm

Finally, we can calculate the depth of the compression block:
d = (310,000 mm^2 - 1,847.25 mm^2 - 4,019.20 mm^2) / 932 mm
d ≈ 302,133.55 mm^2 / 932 mm
d ≈ 324.38 mm

Therefore, the depth of the compression block is approximately 324.38 mm.

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The depth of the compression block in the middle of the beam is 370 mm. The ultimate moment capacity of the beam at the midspan is 564.9 kNm. The beam can sustain a uniform service live load of approximately 9.7 kN/m.

1. To determine the depth of the compression block, we need to calculate the distance from the extreme fiber to the centroid of the compression reinforcement. The distance from the extreme fiber to the centroid of the tension reinforcement can be found using the formula:

[tex]\[a_1 = \frac{n_1A_1y_1}{A_g}\][/tex]

where [tex]\(n_1\)[/tex] is the number of tension bars, [tex]\(A_1\)[/tex] is the area of one tension bar, [tex]\(y_1\)[/tex] is the distance from the extreme fiber to the centroid of one tension bar, and [tex]\(A_g\)[/tex] is the gross area of the beam.

Similarly, the distance from the extreme fiber to the centroid of the compression reinforcement is given by:

[tex]\[a_2 = \frac{n_2A_2y_2}{A_g}\][/tex]

where [tex]\(n_2\)[/tex] is the number of compression bars, [tex]\(A_2\)[/tex] is the area of one compression bar, and [tex]\(y_2\)[/tex] is the distance from the extreme fiber to the centroid of one compression bar.

The depth of the compression block is then given by:

[tex]\[d = a_2 + c\][/tex]

where c is the concrete cover.

Substituting the given values, we have:

[tex]\[d = \frac{5 \times (\pi(16 \times 10^{-3})^2) \times (700 \times 10^{-3})}{(1100 \times 350 \times 10^{-6})} + 40 = 370 \text{ mm}\][/tex]

2. The ultimate moment capacity of the beam at the midspan can be calculated using the formula:

[tex]\[M_u = \frac{f_y}{\gamma_s}A_gd\][/tex]

where [tex]\(f_y\)[/tex] is the yield strength of the reinforcement, [tex]\(\gamma_s\)[/tex] is the safety factor, [tex]\(A_g\)[/tex] is the gross area of the beam, and d is the depth of the compression block.

Substituting the given values, we have:

[tex]\[M_u = \frac{345 \times 10^6}{1.15} \times (1100 \times 350 \times 10^{-6}) \times 370 \times 10^{-3} = 564.9 \text{ kNm}\][/tex]

3. The uniform service live load that the beam can sustain can be determined by comparing the service moment capacity with the moment due to the live load. The service moment capacity is given by:

[tex]\[M_{svc} = \frac{f_y}{\gamma_s}A_gd_{svc}\][/tex]

where [tex]\(d_{svc}\)[/tex] is the depth of the compression block at service loads.

The moment due to the live load can be calculated using the equation:

[tex]\[M_{live} = \frac{wL^2}{8}\][/tex]

where w is the live load intensity and L is the span of the beam.

Equating [tex]\(M_{svc}\)[/tex] and [tex]\(M_{live}\)[/tex] and solving for w, we have:

[tex]\[w = \frac{8M_{svc}}{L^2}\][/tex]

Substituting the given values, we get:

[tex]\[w = \frac{8 \times \left(\frac{345 \times 10^6}{1.15} \times (1100 \times 350 \times 10^{-6}) \times 370 \times 10^{-3}\right)}{(5 \times 1.1)^2} \approx 9.7 \text{ kN/m}\][/tex]

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The unit vector that is orthogonal to →a and →b, and has a positive x-component, is 〈7/√(51), 1/√(51), -1/√(51)〉.

To find a unit vector orthogonal to both →a and →b, we can take their cross product. The cross product of two vectors →a=〈a₁, a₂, a₃〉 and →b=〈b₁, b₂, b₃〉 is given by:

→a × →b = 〈a₂b₃ - a₃b₂, a₃b₁ - a₁b₃, a₁b₂ - a₂b₁〉

Substituting the values of →a and →b, we have:

→a × →b = 〈4(2) - (-5)(4), (-5)(-2) - (-3)(2), (-3)(4) - 4(-2)〉

= 〈8 + 20, 10 - 6, -12 + 8〉

= 〈28, 4, -4〉

Now, we need to find a unit vector from →a × →b that has a positive x-component. To do this, we divide the x-component of →a × →b by its magnitude:

Magnitude of →a × →b = √(28² + 4² + (-4)²) = √(784 + 16 + 16) = √816 = 4√51

Dividing the x-component by the magnitude gives us:

Unit vector →u = 〈28/(4√51), 4/(4√51), -4/(4√51)〉 = 〈7/√(51), 1/√(51), -1/√(51)〉

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The probability that an earthquake striking this region will fall between 2.0 and 3.0 on the Richter scale is approximately 0.1815.

To find the probabilities for the given scenarios, we can use the exponential distribution. The exponential distribution with mean λ is defined as:

[tex]f(x) = λ * e^(-λx)[/tex]

where x ≥ 0 is the value we're interested in, and λ = 1/mean.

In this case, the mean of the exponential distribution is 2.4 on the Richter scale. Therefore, λ = 1/2.4 ≈ 0.4167.

(a) To find the probability that an earthquake will exceed 3.0 on the Richter scale, we need to calculate the integral of the exponential distribution function from 3.0 to infinity:

[tex]P(X > 3.0) = ∫[3.0, ∞] λ * e^(-λx) dx[/tex]

Using integration, we can solve this:

[tex]P(X > 3.0) = ∫[3.0, ∞] 0.4167 * e^(-0.4167x) dx= -e^(-0.4167x) | [3.0, ∞]= -e^(-0.4167 * ∞) - (-e^(-0.4167 * 3.0))[/tex]

Since[tex]e^(-0.4167 * ∞)[/tex]approaches zero, the equation becomes:

[tex]P(X > 3.0) ≈ 0 - (-e^(-0.4167 * 3.0))= e^(-0.4167 * 3.0)≈ 0.4658[/tex]

Therefore, the probability that an earthquake striking this region will exceed 3.0 on the Richter scale is approximately 0.4658.

(b) To find the probability that an earthquake will fall between 2.0 and 3.0 on the Richter scale, we need to calculate the integral of the exponential distribution function from 2.0 to 3.0:

[tex]P(2.0 ≤ X ≤ 3.0) = ∫[2.0, 3.0] λ * e^(-λx) dx[/tex]

Using integration, we can solve this:

[tex]P(2.0 ≤ X ≤ 3.0) = ∫[2.0, 3.0] 0.4167 * e^(-0.4167x) dx= -e^(-0.4167x) | [2.0, 3.0]= -e^(-0.4167 * 3.0) - (-e^(-0.4167 * 2.0))= e^(-0.4167 * 2.0) - e^(-0.4167 * 3.0)≈ 0.3557 - 0.1742≈ 0.1815[/tex]

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Given the differential equation y' + y = cos(2t), we can solve this initial value problem using the Laplace transform. The differential equation is of the form y' + py = q(t).

a). Taking the Laplace transform of y' + py with respect to t, we have:

L{y' + py} = L{q(t)}  ⇒  sY(s) - y(0) + pY(s) = Q(s)

Where Y(s) and Q(s) are the Laplace transforms of y(t) and q(t), respectively.

Substituting p = 1, y(0) = -2, and q(t) = cos(2t), we have Q(s) = s / (s^2 + 4).

Now we have:

(s + 1)Y(s) = (s / (s^2 + 4)) - 2 / (s + 1)

Simplifying, we get:

Y(s) = -2 / (s + 1) + (s / (s^2 + 4))

To find the inverse Laplace transform, we can rewrite Y(s) as:

Y(s) = -2 / (s + 1) + (s / (s^2 + 4)) - 2 / (s + 1)^2 + (1/2) * (1 / (s^2 + 4)) * 2s

Taking the inverse Laplace transform, we obtain the solution:

y(t) = -2e^(-t) + (1/2)sin(2t) - cos(2t)e^(-t)

b) Given the differential equation y' + 2y = 6e^a, where "a" is a constant, we can solve the initial value problem using the Laplace transform.

The differential equation is of the form y' + py = q(t). Taking the Laplace transform of y' + py with respect to t, we have:

L{y' + py} = L{q(t)}  ⇒  sY(s) - y(0) + pY(s) = Q(s)

Substituting p = 2, y(0) = 1, and q(t) = 6e^at, we have Q(s) = 6 / (s - a).

Now we have:

(s + 2)Y(s) = 6 / (s - a) + 1

Simplifying, we get:

Y(s) = (6 / (s - a) + 1) / (s + 2)

Taking the inverse Laplace transform, we obtain the solution:

y(t) = e^(-2t) + (3/2)e^(at) - (3/2)e^(-2t-at)

c) Given the differential equation y' + 2y + y = 38(1 - 2), we can solve this initial value problem using the Laplace transform.

The differential equation is of the form y' + py = q(t). Taking the Laplace transform of y' + py with respect to t, we have:

L{y' + py} = L{q(t)}  ⇒  sY(s) - y(0) + pY(s) = Q(s).

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In the given statement, a construction project in Gulberg 2 near MM Alam Road started in 2018. However, due to rising and volatile costs, as well as productivity issues, the project has exceeded its budget. Several factors have contributed to this cost overrun, including the pandemic, international trade conflicts, inflation, and increasing demand for construction materials.

Additionally, a thorough analysis has revealed that poor scheduling, poor site management, and last-minute modifications have also played a role in the cost overrun. Furthermore, it has been noted that no software was previously used to plan, schedule, and evaluate the project effectively.

As the newly appointed project manager, you will be leading the remaining construction work as the team leader. To address the challenges faced by the project, it is crucial to implement modern management techniques and utilize Primavera-based evaluations. These tools will help establish a data-driven culture that relies on accurate information rather than guesswork.

By implementing these strategies, you can effectively manage the project, control costs, and ensure that the remaining construction work is completed successfully.

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The primary function of road pavement is to provide a durable and smooth surface for vehicles to travel on. It serves as a foundation that distributes traffic loads to the underlying layers and supports the weight of vehicles.

Two functions of the subbase of pavement are:

1. Load Distribution: The subbase layer helps distribute the load from the traffic above it to the underlying layers, such as the subgrade or the soil beneath. By spreading the load over a larger area, it helps prevent excessive stress on the subgrade and reduces the potential for deformation or failure.

2. Drainage: The subbase layer also plays a role in facilitating proper drainage of water. It helps prevent the accumulation of water within the pavement structure by providing a permeable layer that allows water to pass through and drain away. This helps in maintaining the stability and structural integrity of the pavement by minimizing the effects of water-induced damage, such as weakening of the subgrade or erosion of the base layers.

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The local maxima, local minima, and saddle points for the function z = 2x³ - 12xy + 2y³ are:

Local minima: (2√2, 4)

Saddle points: (0, 0), (-2√2, 4)

To find the local maxima, local minima, and saddle points for the function z = 2x³ - 12xy + 2y³, to find the critical points and then determine their nature using the second partial derivative test.

Let's start by finding the critical points by taking the partial derivatives of z with respect to x and y and setting them equal to zero:

∂z/∂x = 6x² - 12y = 0 ...(1)

∂z/∂y = -12x + 6y² = 0 ...(2)

Solving equations (1) and (2) simultaneously:

6x² - 12y = 0

-12x + 6y² = 0

Dividing the first equation by 6, we have:

x² - 2y = 0 ...(3)

Dividing the second equation by 6, we have:

-2x + y² = 0 ...(4)

Now, let's solve equations (3) and (4) simultaneously:

From equation (3),

x² = 2y ...(5)

Substituting the value of x² from equation (5) into equation (4), we have:

-2(2y) + y² = 0

-4y + y²= 0

y(y - 4) = 0

This gives us two possibilities:

y = 0 ...(6)

y - 4 = 0

y = 4 ...(7)

Now, let's substitute the values of y into equations (3) and (4) to find the corresponding x-values:

For y = 0, from equation (3):

x² = 2(0)

x² = 0

x = 0 ...(8)

For y = 4, from equation (3):

x² = 2(4)

x² = 8

x = ±√8 = ±2√2 ...(9)

Therefore, we have three critical points:

(0, 0)

(2√2, 4)

(-2√2, 4)

To determine the nature of these critical points, we need to use the second partial derivative test. For a function of two variables, we calculate the discriminant:

D = (∂²z/∂x²) ×(∂²z/∂y²) - (∂²z/∂x∂y)²

Let's find the second partial derivatives:

∂²z/∂x² = 12x

∂²z/∂y² = 12y

∂²z/∂x∂y = -12

Substituting these values into the discriminant formula:

D = (12x) × (12y) - (-12)²

D = 144xy - 144

Now, let's evaluate the discriminant at each critical point:

(0, 0):

D = 144(0)(0) - 144 = -144 < 0

Since D < 0 a saddle point at (0, 0).

(2√2, 4):

D = 144(2√2)(4) - 144 = 576√2 - 144 > 0

Since D > 0, we have a local minima at (2√2, 4).

(-2√2, 4):

D = 144(-2√2)(4) - 144 = -576√2 - 144 < 0

Since D < 0, have a saddle point at (-2√2, 4).

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The pH of the 0.191 M aqueous solution of NaCH3COO is 2.87.

The pH of a 0.191 M aqueous solution of NaCH3COO can be calculated using the Ka value of CH3COOH. The pH of the solution can be found by determining the concentration of H+ ions in the solution. Since NaCH3COO is a salt of the weak acid CH3COOH, it will dissociate in water to form CH3COO- and Na+ ions. However, the CH3COO- ions will not contribute to the H+ concentration, as they are the conjugate base of the weak acid. Therefore, we need to consider the dissociation of CH3COOH only.

First, we can find the concentration of CH3COOH that will dissociate using the Ka value. Using the equation for the dissociation of CH3COOH, we can write:

CH3COOH ⇌ CH3COO- + H+

Let x be the concentration of CH3COOH that dissociates. Then, the concentration of CH3COO- and H+ ions will also be x. Since the initial concentration of CH3COOH is 0.191 M, we can write:

x = [CH3COO-] = [H+] = 0.191 M

Now, we can use the expression for the Ka of CH3COOH:

Ka = [CH3COO-][H+]/[CH3COOH]

Substituting the values we found:

1.8x10-5 = (0.191)(0.191)/(0.191)

Simplifying the equation:

1.8x10-5 = (0.191)(0.191)

Solving for x:

x = sqrt(1.8x10-5) = 1.34x10-3

Since x represents the concentration of H+ ions, we can convert it to pH using the equation:

pH = -log[H+]

pH = -log(1.34x10-3) = 2.87

Therefore, the pH of the 0.191 M aqueous solution of NaCH3COO is 2.87.

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To compute the first derivative of the function f(x) = x³ - 3x + 1 at the point x₀ = 2 using 5-point formula with h = 5, we will use the following formula: `f'(x₀) ≈ (-f(x₀+2h) + 8f(x₀+h) - 8f(x₀-h) + f(x₀-2h))/(12h)`

Firstly, we calculate the values of the function at x₀ + 2h, x₀ + h, x₀ - h, and x₀ - 2h.

f(12) = (12)³ - 3(12) + 1 = 1697

f(7) = (7)³ - 3(7) + 1 = 337

f(-3) = (-3)³ - 3(-3) + 1 = -17

f(-8) = (-8)³ - 3(-8) + 1 = -383

Now, we substitute the values obtained above into the formula:

`f'(2) ≈ (-1697 + 8(337) - 8(-17) + (-383))/(12(5))`

`= (-1697 + 2696 + 136 + (-383))/(60)`

`= 752/60`

`= 188/15`

Thus, the value of f'(x) at x = 2 using 5-point formula with h = 5 is 188/15. The differentiation error is the error that occurs due to the use of an approximation formula instead of the exact formula to find the derivative of a function. In this case, we have used the 5-point formula to find the first derivative of the function f(x) = x³ - 3x + 1 at the point x₀ = 2. The differentiation error for this formula is given by:

`E(f'(x)) = |(f⁽⁵⁾(ξ(x)))/(5!)(h⁴)|`

where ξ(x) is some value between x₀ - 2h and x₀ + 2h. Here, h = 5, so the interval [x₀ - 2h, x₀ + 2h] = [-8, 12]. The fifth derivative of f(x) is given by:

`f⁽⁵⁾(x) = 30x`

Therefore, we have:

`E(f'(2)) = |(f⁽⁵⁾(ξ))/(5!)(h⁴)|`

`= |(30ξ)/(5!)(5⁴)|`

`= |(30ξ)/100000|`

`= 3|ξ|/10000`

Since ξ(x) lies between -8 and 12, we have |ξ(x)| ≤ 12. Therefore, the maximum possible value of the error is:

`E(f'(2)) ≤ 3(12)/10000`

`= 9/2500`

Thus, the maximum possible error in our calculation of f'(2) using 5-point formula with h = 5 is 9/2500.

Therefore, we can conclude that the first derivative of the function f(x) = x³ - 3x + 1 at the point x₀ = 2 using 5-point formula with h = 5 is 188/15. The maximum possible error in this calculation is 9/2500.

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The volume of water discharged from the watershed is 54,000,000 cubic meters.

The volume of water discharged from the watershed can be calculated by multiplying the area of the watershed by the total precipitation and the runoff coefficient. Here's the step-by-step calculation:

1. Convert the precipitation rate from mm/hr to m/hr:
  - 25 mm/hr = 25/1000 m/hr = 0.025 m/hr

2. Calculate the total precipitation over the 12-hour storm:
  - Total precipitation = precipitation rate * storm duration
  - Total precipitation = 0.025 m/hr * 12 hr = 0.3 m

3. Calculate the volume of water that infiltrates:
  - Infiltration = total precipitation * infiltration percentage
  - Infiltration = 0.3 m * (100% - 90%)
  - Infiltration = 0.3 m * 0.1 = 0.03 m

4. Calculate the volume of water that runs off:
  - Runoff = total precipitation - infiltration
  - Runoff = 0.3 m - 0.03 m = 0.27 m

5. Convert the area of the watershed from km^2 to m^2:
  - 200 km^2 = 200,000,000 m^2

6. Calculate the volume of water discharged from the watershed:
  - Volume = area * runoff
  - Volume = 200,000,000 m^2 * 0.27 m
  - Volume = 54,000,000 m^3

Therefore, the volume of water discharged from the watershed is 54,000,000 cubic meters.

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When prioritizing water allocation for a dam, several factors need to be considered to ensure efficient and fair distribution. Here is a step-by-step approach to prioritize water allocation for different demands:

1. Start with the highest priority demand, which is often irrigation. Irrigation is crucial for agriculture and food production. Allocate a sufficient amount of water for irrigation to support crop growth and maintain agricultural productivity.

2. Move on to domestic water supply. People need water for drinking, cooking, and daily household activities. Allocate an appropriate amount of water for domestic use, considering the population served by the dam and their basic needs.

3. Next, consider Eskom and industries. Eskom refers to the energy provider, and industries encompass various sectors like manufacturing and mining. These sectors play a significant role in economic development and job creation. Allocate a portion of water to ensure the smooth functioning of Eskom and industries, but without compromising other demands.

4. International obligations may arise if the dam is part of a transboundary water agreement. If there are treaties or agreements in place, allocate the required water to fulfill international commitments.

5. Environmental flow is crucial for maintaining the health of ecosystems and biodiversity. Allocate a portion of water to ensure the minimum required flow downstream, allowing for the survival of aquatic life, water quality maintenance, and ecosystem sustainability.

6. Lastly, the "Explain Reserve" refers to a reserved amount of water that is kept for emergency situations or unforeseen circumstances. This reserve ensures there is a buffer available to address any sudden water shortage or unexpected events.

It is important to note that the specific allocation percentages or volumes for each demand will depend on various factors, such as local regulations, water availability, and the dam's capacity. Prioritizing water allocation in a dam requires balancing different needs to ensure sustainable and equitable distribution.

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Answer:

Step-by-step explanation:

6x-y=-5

-6x+y=5

Adding the 2 equations we have:

0 + 0 = 0

0 = 0

This means there are infinite solutions

- the equations are identical.

System B

x+3y=13

-x+3y=5

Adding:

6y = 18

y = 3.

x = 13 - 3(3) = 4.

The system has a unique solution

(x. y) = (4, 3).

The mechanism for the formation of product shown in the given reactions are as follows Mechanism for the formation of product shown in reaction Reaction involves the reaction of an ester with an organolithium reagent in the presence of a proton source.

This reaction is known as ester addition or simply Grignard addition. The product is the tertiary alcohol with two asymmetric centers. The nucleophilic carbon of the Grignard reagent attacks the carbonyl carbon of the ester.

The alkoxide intermediate is protonated by the acidic medium to form the desired product. The stepwise mechanism for the reaction is shown below Mechanism for the formation of product shown in reaction. Mechanism for the formation of product shown in reaction

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Answer:

Step-by-step explanation:

The linear equation can be represented in a slope intercept form as follows:

y = mx + b

where

m = slope

b = y-intercept

Therefore,

Using the table let get 2 points

(2, 27)(3, 28)

let find the slope

m = 28 - 27 / 3 -2 = 1

let's find b using (2, 27)

27 = 2 + b

b = 25

Therefore,

y = x + 25

f(x) = x + 25

Huckel's rule is a mathematical formula used to determine whether a molecule is aromatic or not. The formula states that if the number of pi electrons in a molecule, denoted as n, is equal to 4n+2, where n is an integer, then the molecule is aromatic.

In more detail, the formula for Huckel's rule is n = (4n + 2), where n is the number of pi electrons in the molecule. If the equation holds true, then the molecule is considered aromatic. Aromatic molecules have a unique stability due to the delocalization of pi electrons in a cyclic conjugated system. This rule helps in predicting whether a molecule will exhibit aromatic properties based on its electron count.

For example, benzene has 6 pi electrons, so n = 6. Plugging this into the formula, we get 6 = (4(6) + 2), which simplifies to 6 = 26. Since this equation is not true, benzene is aromatic.

Overall, Huckel's rule provides a useful guideline for determining the aromaticity of molecules based on their electron count.

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the set is linearly dependent, and it can be written as follows:

[tex]s1 = 2/5 (−1,1) − 9/5 (2,0)[/tex]

Given: Set of vectors as follows: S = [tex]{(5,2), (−1,1), (2,0)}(0, 0)[/tex]= Express the vector s1 in the set as a linear combination of the vectors s2 and s3.s1 = We know that the linear combination of vectors is defined as follows.a1 s1 + a2 s2 + a3 s3

Here, a1, a2 and a3 are the scalars.

Substituting the values in the above formula, we get; [tex](5,2) = a1 (−1,1) + a2 (2,0[/tex])

Here, the values of a1 and a2 are to be calculated. So, solving the above equations, we get:a1 = −2/5 a2 = 9/5

Now, we know that a set of vectors is linearly dependent if any of the vectors can be represented as a linear combination of other vectors. Here, we have[tex];5(−1,1) + (2,0) = (0,0[/tex])

Therefore,

Given:[tex]S = {(1,2,3,4),(1,0,1,2),(3,8,11,14)}(0, 0, 0, 0) =[/tex] Express the combination s3 in the set as a linear combination of the vectors s1 and s2.s3 = We know that the linear combination of vectors is defined as follows.a1 s1 + a2 s2

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The missing parts need to be completed. The missing parts include initializing the temporary variable trY, setting the value of tbYF in the IF condition, specifying the range of the FOR loop, and assigning the calculated value to the output variables Y and YF.

Here is the modified version of the SCL program to calculate the power of a number:

FUNCTION "fcPower" : Void

{

 S7_Optimized_Access := 'TRUE'

}

VERSION : 0.1

VAR_INPUT

 X1 : Real; // Base

 X2 : Int; // Exponent

END_VAR

VAR_OUTPUT

 Y : Real; // Power

 YF : Bool; // Fault state

END_VAR

VAR_TEMP

 tiCounter : Int;

 trY : Real;

 tbYF : Bool;

END_VAR

BEGIN

 // Populate/Initialize temporaries

 trY := 1.0;

 // Program

 IF X1 = 0.0 AND X2 = 0 THEN

   trY := 3.402823e+38;

   tbYF := FALSE;

 ELSE

   FOR tiCounter := 1 TO ABS(X2) DO

     trY := trY * X1;

   END_FOR;

   IF X2 < 0 THEN

     trY := 1.0 / trY;

     tbYF := TRUE;

   ELSE

     tbYF := FALSE;

   END_IF;

 END_IF;

 // Write to outputs

 Y := trY;

 YF := tbYF;

END_FUNCTION

In the modified code, trY is initialized to 1.0 as the base case for exponentiation. The FOR loop iterates from 1 to the absolute value of X2, and trY is multiplied by X1 in each iteration.

If X2 is negative, the final result is the reciprocal of trY, and tbYF is set to TRUE to indicate a negative exponent.

Otherwise, tbYF is set to FALSE.

Finally, the calculated value is assigned to Y, and the fault state YF is updated accordingly.

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The formula for calculating the slope coefficient (b) in the simple linear regression model using the normal equations is b = Σ[(xᵢ - X)(yᵢ - Y)] / Σ[(xᵢ - X)²], representing the rate of change of y with respect to x.

A simple linear regression model describes the relationship between two continuous variables, denoted as x (explanatory variable) and y (response variable). The model equation is y = a + bx, where a represents the y-intercept, b represents the slope, and e represents the error term. The slope, b, quantifies the rate of change in y for a unit change in x.

To determine the line of best fit using the normal equations, we solve two simultaneous equations derived from the normal distribution of errors (e).

The first equation arises from the first-order condition for minimizing the sum of squared errors (SSE):

∂SSE/∂b = 0

Expanding SSE, we have:

SSE = Σ(yᵢ - a - bxᵢ)²

Differentiating SSE with respect to b and setting it equal to zero, we get:

Σ(xᵢyᵢ) - aΣ(xᵢ) - bΣ(xᵢ²) = 0

Rearranging the terms, we have:

Σ(xᵢyᵢ) - aΣ(xᵢ) = bΣ(xᵢ²)

To calculate the slope, b, we divide both sides by Σ(xᵢ²):

b = (Σ(xᵢyᵢ) - aΣ(xᵢ)) / Σ(xᵢ²)

To find the value of a, we substitute the sample means of x and y, denoted as X and Y respectively:

a = Y - bx

Thus, the solution for the slope, b, in the simple linear regression model, derived using the normal equations, is:

b = Σ(xᵢ - x)(yᵢ - y) / Σ(xᵢ - x)²

Whereas the solution for the y-intercept, a, is:

a = Y - b x

These equations enable the determination of the coefficients a and b, which yield the line of best fit that minimizes the sum of squared errors in the simple linear regression model.

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To minimize total costs while producing 14,520 units, approximately x = 6280 units should be made on machine A and approximately y = 9240 units should be made on machine B.

Let's represent the number of units produced on machine A as x and the number of units produced on machine B as y. We are given that x + y = 14,520 units.

The total cost function is given by C(x, y) = C(x) + C(y) = 60x + 160 + y^3.

To find the minimum total cost, we can minimize the total cost function C(x, y) with respect to x or y.

First, let's express one variable in terms of the other using the equation x + y = 14,520:

x = 14,520 - y

Substituting this expression for x in the total cost function

C(y) = 60(14,520 - y) + 160 + y^3

Expanding and simplifying

C(y) = 60y - 60y^2 + y^3 + 160

To find the minimum, we need to take the derivative of C(y) with respect to y, set it equal to zero, and solve for y:

C'(y) = 60 - 120y + 3y^2 = 0

Simplifying further, we get:

3y^2 - 120y + 60 = 0

Dividing through by 3, we have:

y^2 - 40y + 20 = 0

Using the quadratic formula, we can solve for y:

y = (40 ± √(40^2 - 4*1*20)) / 2

Simplifying

y = (40 ± √(1600 - 80)) / 2

y = (40 ± √1520) / 2

Since we are dealing with a physical quantity of units, we can discard the negative solution and consider the positive solution:

y = (40 + √1520) / 2

Now, we can substitute this value of y back into the equation x + y = 14,520 to find x:

x + (40 + √1520) / 2 = 14,520

x = 14,520 - (40 + √1520) / 2

Therefore, to minimize total costs while producing 14,520 units, approximately x = 6280 units should be made on machine A and approximately y = 9240 units should be made on machine B.

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1. Without any feedbacks, the temperature change under a doubling of CO₂ is approximately 1.85 ºC .

2. With water vapor and lapse rate feedback alone: Temperature change ≈ 3.7 ºC.

3. With both ice albedo and water vapor/lapse rate feedbacks: Temperature change ≈ 5.55 ºC.

1. The temperature change under different feedback scenarios, we'll consider the following

Ice albedo feedback

Feedback parameter λ = 0.5 W/m² ºC.

Water vapor and lapse rate feedback combined: Feedback parameter λ = 1 W/m² ºC.

Let's start by estimating the temperature change under a doubling of atmospheric CO₂ in the absence of any feedbacks. This is referred to as the no-feedback climate sensitivity.

The no-feedback climate sensitivity (λ₀) is calculated using the formula:

λ₀ = ΔT₀ / ΔF

Where:

ΔT₀ is the temperature change without feedbacks.

ΔF is the radiative forcing due to doubled CO₂, estimated to be around 3.7 W/m².

Assuming the no-feedback climate sensitivity, λ₀ = 0.5 ºC / W/m², we can rearrange the formula:

ΔT₀ = λ₀ × ΔF

ΔT₀ = 0.5 ºC / W/m² × 3.7 W/m²

ΔT₀ = 1.85 ºC

Therefore, without any feedbacks, the temperature change under a doubling of CO₂ is approximately 1.85 ºC.

2. Next, let's consider the temperature change with water vapor and lapse rate feedback alone. The feedback parameter for this combined feedback (λ wv + lr) is 1 W/m² ºC.

The temperature change with water vapor and lapse rate feedback (ΔT wv+lr) is calculated using the formula:

ΔT wv + lr = λ wv + lr × ΔF

ΔT wv + lr = 1 ºC / W/m² × 3.7 W/m²

ΔT wv + lr = 3.7 ºC

Therefore, the temperature change with water vapor and lapse rate feedback alone is approximately 3.7 ºC.

3. Finally, let's calculate the temperature change when both ice albedo and water vapor/lapse rate feedbacks are considered.

The combined feedback parameter (λ combined) is the sum of individual feedback parameters:

λ combined = λ albedo + λ wv + lr

λ combined = 0.5 W/m² ºC + 1 W/m² ºC

λ combined = 1.5 W/m² ºC

Using this combined feedback parameter, we can calculate the temperature change (ΔT combined):

ΔT combined = λ combined × ΔF

ΔT combined = 1.5 ºC / W/m² × 3.7 W/m²

ΔT combined = 5.55 ºC

Therefore, when both ice albedo and water vapor/lapse rate feedbacks are included, the temperature change under a doubling of CO₂ is approximately 5.55 ºC.

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To show that x(t) = 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) satisfies the initial value problem x′′′ + 3x′′ + 3x′ + x = f(t), x(0) = x′(0) = x′′(0) = 0, where f is a bounded continuous function, we need to verify that it satisfies the given differential equation and initial conditions.

By differentiating x(t), we obtain x′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′(t − τ)dτ).

Differentiating once more, x′′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′′(t − τ)dτ).

Differentiating again, x′′′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′′′(t − τ)dτ).

Substituting these derivatives into the differential equation x′′′ + 3x′′ + 3x′ + x = f(t), we have:

1/2∫ t 0 (τ^2e^(−τ) f′′′(t − τ)dτ) + 3/2∫ t 0 (τ^2e^(−τ) f′′(t − τ)dτ) + 3/2∫ t 0 (τ^2e^(−τ) f′(t − τ)dτ) + 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) = f(t).

Now, let's evaluate the initial conditions:

x(0) = 1/2∫ 0 0 (τ^2e^(−τ) f(0 − τ)dτ) = 0.

x′(0) = 1/2∫ 0 0 (τ^2e^(−τ) f′(0 − τ)dτ) = 0.

x′′(0) = 1/2∫ 0 0 (τ^2e^(−τ) f′′(0 − τ)dτ) = 0.

Thus, x(t) = 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) satisfies the given differential equation x′′′ + 3x′′ + 3x′ + x = f(t) and the initial conditions x(0) = x′(0) = x′′(0) = 0.

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To show that x(t) = 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) satisfies the initial value problem x′′′ + 3x′′ + 3x′ + x = f(t), x(0) = x′(0) = x′′(0) = 0, where f is a bounded continuous function, we need to verify that it satisfies the given differential equation and initial conditions.

By differentiating x(t), we obtain x′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′(t − τ)dτ).

Differentiating once more, x′′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′′(t − τ)dτ).

Differentiating again, x′′′(t) = 1/2∫ t 0 (τ^2e^(−τ) f′′′(t − τ)dτ).

Substituting these derivatives into the differential equation x′′′ + 3x′′ + 3x′ + x = f(t), we have:

1/2∫ t 0 (τ^2e^(−τ) f′′′(t − τ)dτ) + 3/2∫ t 0 (τ^2e^(−τ) f′′(t − τ)dτ) + 3/2∫ t 0 (τ^2e^(−τ) f′(t − τ)dτ) + 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) = f(t).

Now, let's evaluate the initial conditions:

x(0) = 1/2∫ 0 0 (τ^2e^(−τ) f(0 − τ)dτ) = 0.

x′(0) = 1/2∫ 0 0 (τ^2e^(−τ) f′(0 − τ)dτ) = 0.

x′′(0) = 1/2∫ 0 0 (τ^2e^(−τ) f′′(0 − τ)dτ) = 0.

Thus, x(t) = 1/2∫ t 0 (τ^2e^(−τ) f(t − τ)dτ) satisfies the given differential equation x′′′ + 3x′′ + 3x′ + x = f(t) and the initial conditions x(0) = x′(0) = x′′(0) = 0.

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Answer:

5.2

Step-by-step explanation:

adding all the numbers together and dividing it by 10.

Answer:

mean = 5.2

Step-by-step explanation:

The mean (or average) of a group of numbers is defined as the value calculated by adding all the given numbers together and then dividing the result by the number of numbers given.

Therefore,

[tex]\boxed{\mathrm{mean = \frac{sum \ of \ the \ numbers}{number \ of \ numbers}}}[/tex].

In the question, the numbers given are: 3, 7, 2, 4, 7, 5, 7, 1, 8, and 8.

Therefore,

sum = 3 + 7 + 2 + 4 + 7 + 5 + 7 + 1 + 8 + 8

       = 52

There are 10 numbers given in the question. Therefore, using the formula given above, we can calculate the mean:

[tex]\mathrm{mean = \frac{52}{10}}[/tex]

            [tex]= \bf 5.2[/tex]

Hence, the mean of the given numbers is 5.2.

Using  Thiessen polygon approach the average rainfall calculated would be 53.9mm.

How to find?

For this method, the Thiessen polygon around each rain gauge will be generated.

A line of equal distance will be traced from each rain gauge to the adjacent gauge, dividing the catchment into polygons.

Each gauge will have an area that is proportional to the polygon's total area over which it has influence.

To determine the weightings of each rainfall gauge, we can follow the steps below:

Thiessen polygon area 1 = 1/2(10)(15)

= 75 mm²

Thiessen polygon area 2 = 1/2(20)(30)

= 300 mm²

Thiessen polygon area 3 = 1/2(20)(20)

= 200 mm²

Thiessen polygon area 4 = 1/2(10)(20)

= 100 mm²

Thiessen polygon area 5 = 1/2(20)(15)

= 150 mm²

Areal (average) rainfall = (15 * 75 + 50 * 300 + 70 * 200 + 80 * 100 + 25 * 150) / (75 + 300 + 200 + 100 + 150)

= 53.9 mm

B) Arithmetic average method-

The arithmetic average method involves taking the average of all of the rain gauge readings.

Areal (average) rainfall = (15 + 50 + 70 + 80 + 25) / 5

= 48 mm

Comment on the suitability of the above two methods to the given catchment-

The Thiessen polygon method is more appropriate in a highly uneven catchment as it accounts for the spatial distribution of rainfall.

The arithmetic average method is easier and quicker to use, but it ignores the catchment's topography and spatial variability.

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The Thiessen polygon method is generally more suitable for catchments with highly uneven topography, as it considers the proximity of rain gauges to different parts of the catchment. However, the arithmetic average method can be used as a simpler alternative if the topography of the catchment is relatively uniform and there are no significant variations in rainfall across the catchment.

The Thiessen polygon method and arithmetic average method can be used to estimate the areal (average) rainfall for the catchment with highly uneven topography shown in worksheet Q1.

a) The Thiessen polygon method involves dividing the catchment area into polygons based on the locations of the rain gauges. Each polygon represents the area that is closest to a particular rain gauge. The areal rainfall for each polygon is assumed to be equal to the rainfall recorded at the rain gauge within that polygon. To estimate the areal rainfall, you would calculate the average rainfall for each polygon by summing up the rainfall measurements of the adjacent rain gauges and dividing it by the number of rain gauges. Then, you would multiply the average rainfall for each polygon by the area of that polygon. Finally, you would sum up the rainfall estimates for all the polygons to get the areal rainfall for the entire catchment.

b) The arithmetic average method involves simply calculating the average rainfall across all the rain gauges. To estimate the areal rainfall using this method, you would add up the rainfall measurements at each rain gauge and divide it by the total number of rain gauges.

c) The suitability of the Thiessen polygon method and the arithmetic average method depends on the characteristics of the catchment.

- The Thiessen polygon method is more suitable for catchments with uneven topography, as it takes into account the proximity of rain gauges to different parts of the catchment. This method provides a more accurate representation of the spatial distribution of rainfall across the catchment.
- The arithmetic average method, on the other hand, is simpler and easier to calculate. However, it assumes that rainfall is evenly distributed across the entire catchment, which may not be the case for catchments with highly uneven topography. This method may lead to less accurate estimates of areal rainfall.

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The water of crystallization is approximately 2.

The question is asking for the water of crystallization in the unknown hydrate AC-XH₂O. To find this, we need to calculate the mass of water lost during heating.

1. Calculate the mass of water lost:
  Mass of water lost = Mass before heating - Mass after heating
  Mass of water lost = 1.000 g - 0.781 g
  Mass of water lost = 0.219 g

2. Calculate the number of moles of water lost:
  Moles of water lost = Mass of water lost / Molar mass of water
  Molar mass of water = 18.015 g/mol (the molar mass of water)
  Moles of water lost = 0.219 g / 18.015 g/mol
  Moles of water lost = 0.01214 mol

3. Determine the molar ratio between the anhydrous compound (AC) and water:
  From the formula AC-XH₂O, we can see that for each AC, there is 1 mole of water.
  This means that the molar ratio of AC to water is 1:1.

4. Find the molar mass of AC:
  Given in the question, the molar mass of AC is 195.5 g/mol.

5. Calculate the number of moles of AC:
  Moles of AC = Mass of AC / Molar mass of AC
  Moles of AC = 1.000 g / 195.5 g/mol
  Moles of AC = 0.00511 mol

6. Find the water of crystallization:
  Water of crystallization = Moles of water lost / Moles of AC
  Water of crystallization = 0.01214 mol / 0.00511 mol

  Now, divide the two moles:
  Water of crystallization ≈ 2.378

7. Round the water of crystallization to the nearest whole number:
  The water of crystallization is approximately 2.

So, the answer to the question is "2".

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line DN in Graph D has the same slope as line LJ and passes through the point (2, -2), it is the correct graph that shows line DN parallel to line LJ and passing through the given point. Hence, the correct answer is D. Graph D

To determine which graph shows line DN parallel to line LJ and passing through point (2, -2), we need to analyze the slopes of the lines in each graph.

Two lines are parallel if and only if their slopes are equal.

In this case, line LJ is already given, and we need to find another line, DN, that is parallel to LJ and passes through the point (2, -2).

To determine the slope of line LJ, we can select two points on the line and calculate the slope using the formula:

slope = (change in y) / (change in x)

Now, let's examine the slopes of each graphed line DN:

Graph A: The slope of line DN appears to be steeper than the slope of line LJ. Therefore, it is not parallel to LJ.

Graph B: The slope of line DN appears to be less steep than the slope of line LJ. Therefore, it is not parallel to LJ.

Graph C: The slope of line DN appears to be steeper than the slope of line LJ. Therefore, it is not parallel to LJ.

Graph D: The slope of line DN appears to be the same as the slope of line LJ. Therefore, it is parallel to LJ.

Since line DN in Graph D has the same slope as line LJ and passes through the point (2, -2), it is the correct graph that shows line DN parallel to line LJ and passing through the given point.

Hence, the correct answer is:

D. Graph D

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Non-settleable solids are fine particles that do not settle out in wastewater and remain suspended in the water column. Unlike settleable solids, which are larger and settle to the bottom under gravity, non-settleable solids are small and light, making them resistant to settling.

These particles can contribute to the turbidity of wastewater and may require additional treatment processes for their removal.

Non-settleable solids refer to suspended particles in wastewater that are too small or light to settle out under normal sedimentation conditions. These particles remain in suspension and do not settle to the bottom when the wastewater is left standing for an extended period of time.

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the pKa value can be calculated by substituting the concentration of the acid [HA] into the equation.

The Ka of an acid is a measure of its acid strength. To calculate the pKa, which is the negative logarithm of the Ka value, follow these steps:

Step 1: Write the balanced equation for the dissociation of the acid:
HA ⇌ H+ + A-

Step 2: Set up the expression for Ka using the concentrations of the products and reactants:
Ka = [H+][A-] / [HA]

Step 3: Substitute the given Ka value into the equation:
8.58 x 10^–4 = [H+][A-] / [HA]

Step 4: Rearrange the equation to isolate [H+][A-]:
[H+][A-] = 8.58 x 10^–4 × [HA]

Step 5: Take the logarithm of both sides of the equation to find pKa:
log([H+][A-]) = log(8.58 x 10^–4 × [HA])

Step 6: Apply the logarithmic property to separate the terms:
log([H+]) + log([A-]) = log(8.58 x 10^–4) + log([HA])

Step 7: Simplify the equation:
log([H+]) + log([A-]) = -3.066 + log([HA])

Step 8: Recall that log([H+]) = -log([HA]) (using the definition of pKa):
-pKa = -3.066 + log([HA])

Step 9: Multiply both sides of the equation by -1 to isolate pKa:
pKa = 3.066 - log([HA])

In this case, the pKa value can be calculated by substituting the concentration of the acid [HA] into the equation.

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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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The tension in cable AB is 3200 lb, while the tension in cables AC and AD is 1600 lb each.

The tension in cable AB is the force pulling the crate upward. Since the crate is not accelerating vertically, the upward force must balance the downward force due to the crate's weight.

The weight of the crate is given as 3200 lb. In terms of forces, weight is equal to mass multiplied by acceleration due to gravity. We can convert the weight from pounds to mass using the conversion factor of 32.2 lb/ft² ≈ 32.2 lb/slug.

Weight of the crate (W) = mass (m) * acceleration due to gravity (g)

W = m * g

3200 lb = m * 32.2 lb/slug * ft/s²

Now, let's apply Newton's second law in the vertical direction, which states that the sum of all forces in the y-direction is equal to zero since the crate is not accelerating vertically.

Sum of forces in the y-direction = 0

TAB - W = 0

Substituting the weight of the crate, we have:

TAB - 3200 lb = 0

Therefore, the tension in cable AB is 3200 lb.

The tension in cable AC is the force pulling the crate to the right. Again, since the crate is not accelerating horizontally, the force pulling it to the right must balance the force pulling it to the left.

Considering the forces in the x-direction, we have:

Sum of forces in the x-direction = 0

TAC - TAD = 0

This equation tells us that the tension in cable AC is equal to the tension in cable AD. Since we don't have any information about the tension in cable AD, we'll refer to it as TAD.

As mentioned earlier, the tension in cable AD is equal to the tension in cable AC. Let's call this tension TAD.

Sum of forces in the y-direction = 0

2TAD - W = 0

Substituting the weight of the crate, we have:

2TAD - 3200 lb = 0

Therefore, the tension in cable AD (and AC) is 1600 lb.

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