Answer:
Weight of oil will be 8.1 N
So option (b) is correct
Explanation:
We have given specific gravity of oil = 0.83
We know that specific gravity is given by
So
Density of oil
Volume is given as 1 liter =
We know that mass
Weight of oil is given by
So option (b) is correct
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:
In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .
Explanation:
.
.
#hope it helps you ..
(◕ᴗ◕)
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:
For the pipe 1, the flow velocity is:
Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m
The Reynold´s number is:
Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:
For the pipe 2, the flow velocity is:
The Reynold´s number is:
The head of pipe 1 is:
The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:
The required pumping power is: