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elena-s [515]
3 years ago
10

Considering the two tables below called Orders and OrderLines, how many rows would be returned by this SQL query? SELECT o.Order

ID, ol.OrderLineID FROM Orders AS o LEFT JOIN OrderLines AS ol ON o.OrderID = ol.OrderID ;

Engineering
1 answer:
natima [27]3 years ago
3 0

Answer:

The query will return four (4) rows

Explanation:

(See attachment below for the table)

The query will return rows where ORDERID is the same on both tables i.e. table ORDERS and ORDERLINES.

The similar rows (on column ORDERID) are 1897 and 87; for table ORDERLINESS and table ORDERS.

But on table ORDERLINESS, there are three rows with 1897 (under ORDERID column).

In total, they have 4 rows in common which are

ORDERID

1897

1897

1897 and

87

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3 0
1 year ago
Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
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Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

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Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

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k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

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