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:
(d) Spheroidizing
Explanation:
Spheroidizing
This is the heat treatment process for steel which having carbon percentage more than 0.8 %.As we know that a hard and brittle material is having carbon percentage more than 0.8 %.That is why this process is suitable for the hard materials.
In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.
In this process grain become spheroidal and these grains are ductile.
Answer:
Hello there, Please follow the step by step explanations for answer.
Explanation:
Hello there, Please follow the step by step explanations for answer.
Ex1 ( A):
it depends on n
=> O(n)
2.)
O(n/2) which is equal to O(n)
3.)O(n^2)
4.)O(2n) ==> O(n)
5.)O(n^3)
Also, see file attachment on this question for more clarity. Thanks and all the best.
Answer:
Technician A is wrong
Technician B is right
Explanation:
voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.
A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.
Answer:
motion ------> electrical. winds push the turbines which generate a magnetic fields which in turn, generates electricity
