Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim working on now? A. design development B. schematic design C. mechanical D. structural

Answers

Answer 1

Answer:

Kim is working on the cost estimate. The stage of a construction plan is Kim working on now is schematic design. The correct option is B.

What is a schematic design?

A schematic design is an outline of a house or a building or another construction thing. The schematic design makes the outline map of the exterior or interior of the building. It is the foremost phase of designing something.

The design expert discusses the project three-dimensionally at this point in the process. To define the character of the finished project and an ideal fulfillment of the project program, a variety of potential design concepts are investigated.

The schematic design consists of a rough sketch with markings and measurements.

Therefore, the correct option is B. schematic design.

To learn more about schematic design, refer to the link:

brainly.com/question/14959467

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Answer 2

Answer:

B. schematic design

Explanation:

This correct for Plato/edmentum

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A wind turbine system has the following specifications: Diameter:45 m Rated power 700 kW at the wind speed of 12 m/s Turbine speed 1500 rpm Determine the swept area of the wind turbine. a)- 1640 m^2 B)- 1690 m^2 c)- 1590 m^2 d)- 1540 m^2

Answers

Answer:

1590 m^2

Explanation:

Given data in this question

Diameter = 45 m

power = 700 kW

wind speed = 12 m/s

turbine speed = 1500 rpm

To find out

swept area of the wind turbine

Solution

we know wind turbine is rotate circular form

and diameter is given so by the area of circular swept we will calculate it

we know area = $\pi /4$ × d²

put the value of d here

area = $\pi /4$ × 45²

swept area = 1590 m^2

What is the answer What two totally normal things become really weird if you do them back to back?

Answers

Answer: i like the way you scream

A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answers

Answer:

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Explanation:

From Open Channel Theory we know that critical depth of the rectangular channel ( $y_(c)$ ), measured in meters, is calculated by using this equation:

$y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }$ (Eq. 1)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$b$ - Channel width, measured in meters.

If we know that $\dot V = 15\,(m^(3))/(s)$ , $g = 9.807\,(m)/(s^(2))$ and $b = 2\,m$ , then the critical depth is:

$y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }$

$y_(c) \approx 1.790\,m$

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity ( $v$ ), measured in meters, is obtained from this formula:

$v = (\dot V)/(b\cdot y_(c))$ (Eq. 2)

If we know that $\dot V = 15\,(m^(3))/(s)$ , $b = 2\,m$ and $y_(c) = 1.790\,m$ , then the flow velocity in the rectangular channel is:

$v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))$

$v = 4.190\,(m)/(s)$

The flow velocity in the rectangular channel is 4.190 meters per second.

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 20.000 and 20.025 mm, respectively, and its final length is 74.96 mm, compute its original length if the deformation is totally elastic. The elastic and shear moduli for this alloy are 105 GPa and 39.7 GPa, respectively.

Answers

Answer:

L = 75.25 mm

Explanation:

First we need to find the lateral strain:

Lateral Strain = Change in Diameter/Original Diameter

Lateral Strain = (20.025 mm - 20 mm)/20 mm

Lateral Strain = 1.25 x 10⁻³

Now, we will find the Poisson's Ratio:

Poisson's Ratio = (E/2G) - 1

where,

E = Elastic Modulus = 105 GPa

G = Shear Modulus = 39.7 GPa

Therefore,

Poisson's Ratio = [(105 GPa)/(2)(39.7 GPa)] - 1

Poisson's Ratio = 0.322

Now, we find longitudinal strain by following formula:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

Longitudinal Strain = - Lateral Strain/Poisson's Ratio

Longitudinal Strain = - (1.25 x 10⁻³)/0.322

Longitudinal Strain = - 3.87 x 10⁻³

Now, we can fin the original length:

Longitudinal Strain = Change in Length/L

where,

L = Original Length = ?

Therefore,

- 3.87 x 10⁻³ = (74.96 mm - L)/L

(- 3.87 x 10⁻³)(L) + L = 74.96 mm

0.99612 L = 74.96 mm

L = 74.96 mm/0.99612

L = 75.25 mm

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage change in chip thickness when the coefficient of friction is doubled. Justify your answer. b. Determine the percentage change in chip thickness when the rake angle is increased to 25o . Justify your answer.

Answers

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α = 15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ + β - α = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

$r=(tan\phi)/(cos\alpha +sin\alpha\ tan\phi)$

$r=(tan48.08^(\circ))/(cos15^(\circ) +sin15^(\circ)\ tan48.08^(\circ))$

r=0.88

New coefficient of friction ,μ' = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' + β' - α = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

$r'=(tan\phi')/(cos\alpha +sin\alpha\ tan\phi')$

$r'=(tan44.15^(\circ))/(cos49.15^(\circ) +sin49.15^(\circ)\ tan44.15^(\circ))$

r'=0.70

Percentage change

$\Delta r=(r-r')/(r)* 100$

$\Delta r=(0.88-0.70)/(0.88)* 100$

Δr=20.45 %

Young students show a preference for which modality?

Answers

Answer. In the present study, it was found that 61% students had multimodal learning style preferences and that only 39% students had unimodal preferences. Amongst the multimodal learning styles, the most preferred mode was bimodal, followed by trimodal and quadrimodal respectively

Explanation:

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8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid $25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are $3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?

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