Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Answer:
<u><em>note:</em></u>
<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>
Answer and Explanation:
It is True that packet size increases the transmission time of the packets. This is because packet transmission would be impacted depending on the packet size, whether it is large or small. Here comes the concept of bandwidth of the medium. For a particular bandwidth of the medium, if the packet is small then many such packets can pass through the medium at a given time, rather than less packets at a given time when the packet sizes are larger. Transmission time of a packet = Packet size / Bandwidth of the medium.
If the bandwidth is high, the transmission time will be less. Also, if the packet size increases then, as we saw above the transmission time will increase.
Every packet has to have a header information in which there will be several information and data about the packet and the protocol it follows, also what is the source and destination of the packet. The information contained in the header of a packet may vary in case of different protocols. The address portion of the packet is a overhead but us actually necessary when the packet needs to be transferred from a source to a destination using a protocol. The size of the header varies from protocol to protocol and is an overhead and has impact on data transmission rates since size of the packet increases because of the header.
Answer:
44.59°c
Explanation:
Given data :
Total pressure = 105 kpa
complete combustion
A) Determine air-fuel ratio
A-F = 
N = number of mole
m = molar mass
A-F = 
= 22.2 kg air/fuel
hence the ratio of Fuel-air = 1 : 22.2
B) Determine the temperature at which water vapor in the products start condensing
First we determine the partial pressure of water vapor before using the steam table to determine the corresponding saturation temp
partial pressure of water vapor
Pv = 
N watervapor ( number of mole of water vapor ) = 3
N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol
Pro = 105
hence Pv = ( 3/33.53 ) * 105 = 9.39kPa
from the steam pressure table the corresponding saturation temperature to 9.39kPa = 44.59° c
Temperature at which condensing will start = 44.59°c
An equation showing the products of propylene with their mole numbers is attached below
