All categories

2 years ago

How do you produce a well-constructed survey questionnaire?

Engineering

1 answer:

Anni [7]2 years ago

5 0

Answer:

Explanation:

Define the purpose of the survey.

Make every question count.

Keep it short and simple.

Ask direct questions.

Ask one question at a time.

Avoid leading and biased questions.

You might be interested in

1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used

bazaltina [42]

Answer:

F_x = 100,200 N

x' = 21.321 cm ... Length

y' =0.7825 cm

z' = 1.565 cm

A' = ( 0.783 x 1.565 ) cm

Explanation:

Given:

- The Modulus of Elasticity E = 645 MPa

- The poisson ratio v = 0.28

- The Yield Strength Y = 501 MPa

- The Length along x-direction x = 12 cm

- The length along y-direction y = 1 cm

- The length along z--direction z = 2 cm

Find:

The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

Solution:

- The Tensile forces within the limit of proportionality is given as:

F_i = б_i*A_jk

- A maximum tensile Force F_x along x direction can be given as:

F_x = Y*A_yz

F_x = 501*( 0.01*0.02)*10^6

F_x = 100,200 N

- The corresponding strains in x, y and z direction due to F_x are:

ξ_x = Y / E

ξ_x = 501 / 645 = 0.7767

ξ_y = ξ_z = -v*Y / E

ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175

- The corresponding change in lengths at tensile elastic stress are:

Δx = x*ξ_x = 12*0.7767 = 9.321 cm

Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm

Δz = z*ξ_z = - 2*0.2175 = -0.435 cm

- The final lengths are:

x' = x + Δx = 12 + 9.321 = 21.321 cm

y' = y + Δy = 1 - 0.2175 = 0.7825 cm

z' = z + Δz = 2 - 0.435 = 1.565 cm

3 0

4 years ago

Savabuck University has installed standard pressure-operated flush valves on their water closets. When flushing, these valves de

Dvinal [7]

Answer:

Cost = $2527.2 per month.

Explanation:

Given that

Discharge ,Q = 130 L/min

$Q=0.13\ m^3/min$

Cost = $0.45 per cubic meter

1 month = 30 days

1 days = 24 hr = 24 x 60 min

1 month = 30 x 24 x 60 min

1 month = 43,200 min

Lets x $m^3\ water\ waste\ in\ a\ month$

x = 0.13 x 43,200

$x=5616\ m^3$

So the total cost = 5616 x 045 $

Cost = $2527.2 per month.

7 0

4 years ago

The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 4840

Assoli18 [71]

Answer:

70,900

Explanation:

Given :

True stress (psi) _____ True strain (psi)

48400 ______________ 0.11

60400 ______________ 0.19

Using ratio simplification :

Let :

s = True stress ; t = true strain

s1 = 48400

s2 = 60400

t1 = 0.11

t2 = 0.19

True stress, s0 ; needed to produce a True plastic strain, tp = 0.26

(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)

(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)

(s0 - 48400)/12000 = 0.15/0.08

Cross multiply :

0.08(s0 - 48400) = 0.15 * 12000

0.08s0 - 3872 = 1800

0.08s0 = 1800 + 3872

0.08s0 = 5672

s0 = 5762 / 0.08

s0 = 70,900

The true stress required to produce a true plastic strain of 0.26 is 70,900

5 0

3 years ago

Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is already cache

aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

8 0

3 years ago

The propeller shaft of the submarine experiences both torsional and axial loads. Draw Mohr's Circle for a stress element on the

Alina [70]

Answer: Attached below is the missing detail and Mohr's circle.

i) б1 = 9.6 Ksi

б2 = -10.7 ksi

ii) 10.2 Ksi

iii) -0.51Ksi

Explanation:

First step :

direct compressive stress on shaft

бd = P / π/4 * d^2

= -20 / 0.785 * 5^2 = -1.09 Ksi

shear stress at the outer surface due to torsion

ζ = 16*T / πd^3

= (16 * 250 ) / π * 5^3 = 010.19 Ksi

Calculate the Principal stress, maximum in-plane shear stress and average normal stress

Using Mohr's circle ( attached below )

i) principal stresses:

б1 = 4.8 cm * 2 = 9.6 Ksi

б2 = -5.35 cm * 2 = -10.7 ksi

ii) maximum in-plane shear stress

ζ = radius of Mohr's circle

= 5.1 cm = 10.2 Ksi ( Given that ; 1 cm = 2Ksi )

iii) average normal stress

= 9.6 + ( - 10.7 ) / 2

= -0.51Ksi

8 0

3 years ago