The power factor of the ballast is 0.54 which is the ratio of working power to apparent power.
The power factor is a measure of energy efficiency. It is typically expressed as a percentage, with a lower percentage indicating inefficient power usage.
The power factor (PF) is the ratio of working power (in kW) to apparent power (in kilovolt amperes) (kVA).
Given that a 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts.
PF = (True power)/(Apparent power)
PF = W/VA
Here W = 22 watts, V = 120-volt, and A = 0.34 ampere
PF = 22 / (120 × 0.34)
PF = 22 / 40.8
PF = 0.5392
PF = 0.54
Thus, the power factor of the ballast is 0.54.
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Answer:
PF= .54
Explanation:
Power Factor equals working/real power (W) over apparent power (VA). 1.0 PF is an efficient equipment. PF= 22/(120*.34)
Answer:
It is important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.
Data compression is also required to reduce the space that is occupied by data during transmission, now once the presentation is added to the physical layer, data from the physical layer can be compressed at the presentation layer and sent by improving the throughput.
Explanation:
Solution
The presentation of data involves the following as shown below:
Presentation of data comprises of the task like translating between receiver and sender devices so that machines with different capabilities sets can communicate with one another.
It involves encoding and decoding of data to provide data security that is been transmitted by different machines.
Data sometimes needs to compressed for efficiency improvement for transmission.
The physical layer of the TCP/IP protocol suite is responsible or refers to the transmission of physical data over a physical medium
It is good or important to add presentation layer after the physical layer, so that the data along with it's headers can be translated, when the receiver machine is applying a set of different characters.
Data encryption at this stage is good for security instead of encrypting the data at upper/higher layers.
Hence, it is advisable to add presentation layer after the physical layer in the TCP/IP suite.
Answer:
The layer ought to be embedded between Layer 2 and 3.
Explanation:
Applications often communicate with each other. This cannot be successful if they don't see data the same way. The Presentation Layer in the Open Systems Interconnection defines how data is presented and is often processed in the TCP/IP applications.
While the Presentation Layer does not exist as a different layer in the TCP/IP protocol order of arrangement, it is important to note that the Network Layer is also known referred to as the TCP/IP’s Network Layer.
Therefore, if the presentation of the data layer will be separated, it should be between layer 2 and 3.
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Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
Answer: Fossil remains of the same land-dwelling animal.
Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.
Answer:
Fossil remains of the same land-dwelling animal
Explanation:
Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.
Answer:
891.027 lbm/ft³
Explanation:
See it in the pic
Answer:
Decrease to typical from utilizing lambda-decrease:
The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2
The of taking the terms is significant in lambda - math,
For the term, (λy, y×3)2, we can substitute the incentive to the capacity.
Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6
Presently the tem becomes, (λf λx f(f(fx)))6
The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.
Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.
In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.
Answer: hello your question is incomplete below is the missing part
question :Determine the temperature of the cooled water exiting the cooling tower
answer : T = 43.477° C
Explanation:
Temp of water at exit = 45°C
mass flow rate of cooling tower = 15,000 kg/s
Temp of makeup water = 20°C
Assuming an atmospheric pressure of = 101.3 kPa
Determine temperature of the cooled water exiting the cooling tower
Water entering cooling tower at 45°C
Given that Latent heat of water at 45°C = 43.13 KJ/mol
Cp(wet air) = 1.005+ 1.884(y1)
where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273
Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C
First step : calculate the value of Q
Q = m*Cp*(ΔT) + W(latent heat)
Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)
Q = 95935.8547 KJ/s
Given that mass rate of water = 15000 kg/s
Hence the temperature of the cooled water can be calculated using the equation below
Q = m*Cp*∆T
Cp(water) = 4.2 KJ/Kg°C
95935.8547 = (15000)*(4.2)*(45 - T)
( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C