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elena-s [515]
3 years ago
6

A family member who hasn’t worked with computers before has decided to change jobs. You’ve been asked to explain some of the bas

ics. You begin by explaining the basic idea of word processing software, spreadsheets, databases, and presentation software. You sense that the following question is coming: “Why so many different things? Why can’t there just be one software that you could use for everything?” How would you answer?
Engineering
2 answers:
-BARSIC- [3]3 years ago
6 0
That due to the specific tasks that needs to be accomplished by each program to make an all encompassing program would be inefficient and full of bugs
Svetllana [295]3 years ago
6 0
It will be full of bugs with the software being in that state.
You might be interested in
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo
bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0
3 years ago
Read 2 more answers
6. What types of injuries can occur in an electronics lab and how can they be prevented?
marysya [2.9K]

Answer:

The most common injuries in a chemistry lab is making a fire, heat burns, chemical burns, cuts and scrapes, contamination, inhalation, and spills and breaks.

1.) You can prevent making a fire by making sure you close and seal flammable materials.

2.) You can prevent heat burns by teaching the students how to properly use tongs,water baths, and other cooling equipment. 

3.) You can prevent chemical burns by treating the chemicals with caution, measure carefully, and use the approved containers.

4.) You can prevent cuts and scrapes by telling the students how to use the blades safely, and also when they are disposing broken or sharp items they should know how to wrap them up so no one else will get hurt. 

5.) You can prevent contamination by washing your hands, protect their clothing and skin with a lab coat or a lab apron, gloves and glasses, and cleaning your area where the germs of the chemicals were so no one will become.

6.) You can prevent inhalation by opening up windows, using ventilation fans, and using an equipment that measures the amount of gas emission in a room.

7.) Finally, you can prevent spills and breaks by telling the students what will happen if anything spills, and tell them to clean up.  

8 0
3 years ago
Describe how you could engineer the situation to produce even more friction and heat
lana66690 [7]
True the use many abstract power
6 0
2 years ago
A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0
3 years ago
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