Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s
emester I-2018? List the names of the students who have taken at least one course that Professor Collins is qualified to teach. List the names of the students who took at least one course with ""Syst"" in its name during the semester I-2018.
1 answer:
Answer:
See explanations
Explanation:
Consider the following Venn diagram to retrieve the name of course that student Altvater took during semester I-2015.
• In above Venn diagram inner Ellipse represent the subquery part, this subquery part select the Student ID from STUDENT table.
• Second sub query is used to determine the SectionNo of all student whose studentID retrived in the first subquery
• Finally the main query displays the CourseName from COURSE table.
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Answer:
F = 9.11 x 10³ N = 9.11 KN
Explanation:
The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:
ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper
ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)
where,
ΔL = Equivalent Change in Length = ?
ΔT = Change in Temperature = 25°C - 12°C = 13°C
Ls = Length of Steel Segment = 300 mm = 0.3 m
Lb = Length of Brass Segment = 200 mm = 0.2 m
Lc = Length of Copper Segment = 100 mm = 0.1 m
Therefore,
ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)
ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m
ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)
Now, we calculate this deflection in terms of an applied force (F):
ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)
ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)
ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)
ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)
Comparing equation (1) and equation (2):
123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)
F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)
<u>F = 9.11 x 10³ N = 9.11 KN</u>
Answer:
The correct answer is "1341.288 W/m".
Explanation:
Given that:
T₁ = 300 K
T₂ = 500 K
Diameter,
d = 0.2 m
Length,
l = 1 m
As we know,
The shape factor will be:
⇒
By putting the value, we get
⇒
⇒
hence,
The heat loss will be:
⇒