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
B !! Is the correct answer
Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Capillary action occurs When the adhesion to the walls stronger than dirt cohesive forces between a liquid molecules. the head towards Capillery action will take water in a uniform circular is limited by surface tension and, of course, gravity.
Answer:
(c) 5.71 V
Explanation:
The circuit can be redrawn to a Thevenin equivalent that is 6V through a 5-ohm resistor into a 100-ohm load. Then the voltage at the load is ...
(6 V)(100/(100 +5) ≈ 5.71 V