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Dmitry [639]
2 years ago
7

Which test element is used to apply test execution settings for the main business flow excluding Login and Log out 1) Run time c

ontroller 2) Loop controller 3) Transaction controller 4) Throughput controller
Engineering
1 answer:
Svet_ta [14]2 years ago
3 0

Test element is used to apply test execution settings for the main business flow excluding Login and Log out is "Transaction  controller"

<u>Explanation:</u>

Transaction controllers are a technoscientifically kind of controllers that produce an additional sample that measures the overall time is taken or response time to perform its nested samplers.

The Controller has two checkboxes,those are

1.Generate Parent Samples

2. Include term of the timer and pre-post processors in the produced sample - It is the time which involves all processing samples within the Transaction controller, not just the HTTP samples. In Case one sample is lost or crashed then the entire Transaction Controller will fail. In case anyone sample assertion is failed then the complete Transaction Controller will fail.

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A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting
sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s  = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

              F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

              F dx = m v dv

We replace and integrate

            -β ∫ x³ dx = m ∫ v dv

            β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

        -β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

         x_max⁴ = 2 m /β   v₀²

         

Let's look for the speed that the train can have for maximum compression

         x_max = 20 cm = 0.20 m

         

         v₀ =√(β/2m)   x_max²

Let's calculate

          v₀ = √(640 106/2 7.8 104)    0.20²

          v₀ = 64.05  0.04

          v₀ = 2,562 m / s

          v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

          v₀ = 9.2 km / h

5 0
2 years ago
Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th
Vilka [71]

Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

3 0
2 years ago
Develop a preliminary work breakdown structure (WBS) for a small one-story commercial building to be constructed on the site of
Zielflug [23.3K]

Answer:

The preliminary work breakdown structure will be divided into two steps, the first is to draw the first level and the second is to draw the second level.

Explanation:

Please look at attachment.

6 0
2 years ago
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True or False: Drag and tailwind are examples of a contact force.<br> tyy guyss
zloy xaker [14]

Answer:

False

Explanation:

8 0
2 years ago
Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20
Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

T_{\infty} = 10° C

h_{0} = 250 w/ m^{2} k

Pipe length = 20 m

Inner diameter d_{1} = 6 cm, r_{1} = 3 cm

Outer diameter d_{2} = 8 cm, r_{2} = 4 cm

The thickness of insulation is 4 cm.

r_{3} = r_{2} + 4

= 4+4

r_{3} = 8 cm

h_{0} is the heat transfer coefficient of  convection inside, h_{i} is the heat transfer coefficient of  convection outside.

The heat transfer rate between ambient and steam is

q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}} watt

=  \begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned} watt

= \frac{190}{0.0003317+0.0000458+0.0110+0.0004976} watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = \frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}

15999.86   =\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}

T_{3} - 10 = 7.96

T_{3} = 17.96 ° C.

4 0
2 years ago
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