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Arada [10]
3 years ago
7

Write a single statement to print: user_word,user_number. Note that there is no space between the comma and user_number. Sample

output with inputs: 'Amy' 5
Engineering
1 answer:
kykrilka [37]3 years ago
8 0

Answer:

cout<<"''<<user_word<<"' "<<user_number;

Explanation:

The above question was answered using C++ programming language.

The keyword cout represents print and it carries out print operation only.

It prints all variable in front of it.

Assume the values of user_word and user_number to be Charles and 20, respectively.

The output of the above instruction would be

'Charles' 20 just as it is in the sample output in the question.

In java programming language, it is

System.out.print("'"+user_word+"' "+user_number);

In Qbasic, it is

PRINT "'"+user_word+"' "+ user_number

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What is differentiation​
oksian1 [2.3K]

Answer:

the action or process of differentiating or distinguishing between two or more things or people.

7 0
3 years ago
A 4-stroke Diesel engine with a displacement of Vd = 2.5x10^-3m^3 produces a mean effective pressure of 6.4 bar at the speed of
yKpoI14uk [10]

Answer:

The power developed by engine is 167.55 KW

Explanation:

Given that

V_d=2.5\times 10^{-3} m^3

Mean effective pressure = 6.4 bar

Speed = 2000 rpm

We know that power is the work done per second.

So

P=6.4\times 100\times 2.5\times 10^{-3}\times \dfrac{2\pi \times2000}{120}

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.

P=167.55 KW

So the power developed by engine is 167.55 KW

4 0
3 years ago
The minimum requirements for engineering documents are enumerated in
vladimir1956 [14]

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23'  from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

8 0
3 years ago
(a) Differentiate between heat treatment of ferrous and non-ferrous alloys (b) With your understanding of material's thermal pro
liubo4ka [24]

Answer:

In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.

Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are  the alloy element of iron ans carbon.Wrought iron is the purest from of iron.

Heat treatment process for  ferrous materials :

1.Normalizing

2.Annealing

3.Quenching

4.Surface hardening

Heat treatment process for non ferrous materials :

Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.

When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.

The use of bimetallic structure -In clock ,thermometers ,engines.

7 0
3 years ago
A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
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