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ratelena [41]
1 year ago
13

List two ways you can make an informal survey

Engineering
2 answers:
solniwko [45]1 year ago
8 0

The two ways you can use to make an informal survey are:

  • make field observations
  • interview people using informal unstructured techniques

<h3>What are informal surveys?</h3>

In informal surveys can be regarded as a type of survey that can be made by the researcher by going to the field themselves and this can be done by using different methods or ways.

For instance, the researcher can go out to interview people that can give the data that is needed about the research such as informally asking them questions,  unstructured techniques can also be used to solve critical issues.

learn more about survey at: brainly.com/question/6947486

#SPJ9

nlexa [21]1 year ago
3 0

The best informal research methods include asking quality questions, drawing on prior experience and knowledge, using guesses and hunches, having conversations, conducting surveys, reading documents, searching on the internet, running mini-experiments and doing a trial run.

You might be interested in
Define Mechanism and mechanics.​
aivan3 [116]

mechanism, in mechanical construction, the means employed to transmit and modify motion in a machine or any assemblage of mechanical parts.

6 0
3 years ago
Read 2 more answers
What steps would you take to design an improved toothpaste container?
algol [13]

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

7 0
3 years ago
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
Fibonacci sequence has many applications in Computer Science. Write a program to generate Fibonacci numbers as many as desired.
VikaD [51]

Answer:

The Python Code for Fibonacci Sequence is :

# Function for nth Fibonacci number  

def Fibonacci(n):  

if n<0:  

 print("Incorrect input")  

# First Fibonacci number is 0  

elif n==0:  

 return 0

# Second Fibonacci number is 1  

elif n==1:  

 return 1

else:  

 return Fibonacci(n-1)+Fibonacci(n-2)  

# Driver Program  

print(Fibonacci(9))  

Explanation:

The Fibonacci numbers are the numbers in the following integer sequence.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……..

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation

Fn = Fn-1 + Fn-2

with seed values

F0 = 0 and F1 = 1.

8 0
3 years ago
Read 2 more answers
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
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