Answer:
The table is attached as a picture.
a)
Select VENDOR_CONTACT_LAST_NAME || ', ' || VENDOR_CONTACT_FIRST_NAME "full_name" from VENDORS where VENDOR_CONTACT_LAST_NAME like 'A%' or VENDOR_CONTACT_LAST_NAME like 'E%' order by VENDOR_CONTACT_LAST_NAME,VENDOR_CONTACT_FIRST_NAME;
concatenation operator || is used . Also LIKE is used for pattern matching. full_name is alias for the concatenated column
b) As sample data is not given ,Please test the query for the data given in table
Explanation:
Answer:
see explaination for all the answers and full working.
Explanation:
deflection=8P*DN/Gd^4
G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2
for outer spring,
deflection=8*3*50^3*5/(70*9^4)=32.66mm
for inner spring
deflection=8*3*30^3*10/(70*5^4)=148.11mm
max stress=k*8*P*C/(3.14*d^2)
for outer spring
c=50/9=5.55
k=(4c-1/4c-4)+.615/c=1.2768
max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2
for inner spring
c=6
k=1.2525
max stress=2.29KN/mm^2
Answer:
33.4
Explanation:
Step 1:
\sumMo=0 (moment about the origin)
Fb(15)-Fc(15)=0
Fb=Fc
Step 2:
\sumFx=0
-Fb-Fccos\theta+Ncsin\theta=0
Fc=0.3Nc=Fb
-0.3Nc-0.3Nccos\theta+Ncsin\theta=0
(-0.3-cos\theta+sin\theta)Nc=0----(1)
\sumFy=0
Nccos\theta+Fcsin\theta-Nb=0
Nccos\theta+0.3Ncsin\theta-Nc=0
Nc[cos\theta+0.3sin\theta-1]=0--------(2)
Solving eq (1) and eq (2)
\theta=33.4
Step 3:
As the roller is a two force member
2(90-\phi)+\theta=180
\phi=\theta/2
\phi=Tan(\muN/N)-1
\phi=16.7
\theta=2x16.7=33.4
Answer:
w ( mass flow rate of air ) = 3.16 kg/s
Explanation:
<u>Determine the mass flow rate of air </u>
mass flow rate of water = 1.5 kg/s
Height at which air enters the cooling tower = 1m
velocity of air entering at 1 m = 20 m/s
Height at which air leaves the cooling tower = 7 m
attached below is a detailed solution of the problem
Answer:
Explanation:
30 we know that radius is 18 and the circumference is 36pi and the time to go around is is 36pi/30=1.2pi≈3.76991118
