Explanation:
umm can you explain me what i have to answer
Answer:
Goal address: Based on the above Ethernet outline design, we get that the goal address begins from the first hex worth and is of size 12 hex values (got from question 1). In light of the bundle given, we get that the goal address is 6c40 0889 c448.
Source address: Similarly, the source address begins after the goal address and is of size 12 hex values (got from question 1). Thus, the appropriate response is f832 e4a7 fb38.
Type/Length: The sort/length begins after the source address and is of size 4 hex values (got from question 1). Thus, the appropriate response is 0806.
Data (Payload): The data(payload) begins after the sort/length and finishes not long before FCS(Checksum) which is of 4 bytes for example 4 * 2 = 8 hex qualities. So the information comprises of everything between the 'type/length' and 'CRC'. We persuade the CRC to be c0a8 01f2. Henceforth, the appropriate response is 0001 0800 0604 0002 f832 e4a7 bf38 c0a8 0101 6c40 0889 c448.
Question 4:
The Ethernet parcel type characterizes the convention utilized for sending the bundle information. We realize that type 0x0800 demonstrates the IPv4 convention, 0x0806 shows an ARP convention, and 0x86DD demonstrates an IPv6 convention. In light of the appropriate response we got in Question 3, we realize that the sort/length esteem for the given parcel is 0806, which implies that convention utilized for sending the information bundle was ARP convention.
Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows
Here temperature should be in Kelvin.
For engine A
For engine B
So from above we can say that engine B is more efficient.
Explanation:
Ohm's law is used here. V = IR, and variations. The voltage across all elements is the same in this parallel circuit. (V1 =V2 =V3)
The total supply current is the sum of the currents in each of the branches. (It = I1 +I2 +I3)
Rt = (8 V)/(8 A) = 1 Ω . . . . supply voltage divided by supply current
I3 = 8A -3A -4A = 1 A . . . . supply current not flowing through other branches
R1 = (8 V)/(3 A) = 8/3 Ω
R2 = (8 V)/(4 A) = 2 Ω
R3 = (8 V)/(I3) = (8 V)/(1 A) = 8 Ω
V1 = V2 = V3 = 8 V