Consider the 'scope trace showing the input signal to an rc filter (the one having the larger amplitude) and the output of the f
1 answer:
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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:
a) 740 W b) 6.2 A c) 8.1%
Explanation:
We need first to get the total energy spent during the 30 days, that can be calculated as follows:
1 month = 30 days = 720 hr
If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:
80 $/mo = 0.15 $/Kwh*x Kwh/mo
Solving for the total energy spent in the month:
Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:
b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:
c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:
P = 740 W - 60 W = 680 W
The new value of the energy spent during the entire month will be as follows:
E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo
The reduction in percentage regarding the total energy spent can be calculated as follows:
ΔE =
⇒ ΔE(%) = 8.1%