Answer:
a) 4.7 kΩ, +/- 5%
b) 2.0 MΩ, +/- 20%
Explanation:
a) If the resistor has the following combination of color bands:
1) Yellow = 1st digit = 4
2) Violet = 2nd digit = 7
3) Red = multiplier = 10e2
4) Gold = tolerance = +/- 5%
this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.
b) Repeating the process for the following combination of color bands:
1) Red = 1st digit = 2
2) Black = 2nd digit = 0
3) Green = multiplier = 10e5
4) Nothing = tolerance = +/- 20%
This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.
Answer:
The maximum power that can be generated is 127.788 kW
Explanation:
Using the steam table
Enthalpy at 20 bar = 2799 kJ/kg
Enthalpy at 2 bar = 2707 kJ/kg
Change in enthalpy = 2799 - 2707 = 92 kJ/kg
Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s
Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW
Answer:
4/5
Explanation:
She is not wearing white t-shirt on the first day so she is wearing the other 4 t-shirt
Answer:
At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.
Answer:
$7,778.35
Explanation:
At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.
First, calculate the uniform payments A:
A = 12000(A/P, 4%, 5)
= 12000(0.2246) = 2695.2 (from the calculator)
Then take the last three payments as its own cash flow.
To calculate the new P:
P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35
Therefore, the final payment is $7,778.35
