Answer: The overhead percentage is 7.7%.
Explanation:
We call overhead, to all those bytes that are delivered to the physical layer, that don't carry real data.
We are told that we have 700 bytes of application data, so all the other bytes are simply overhead, i.e. , 58 bytes composed by the transport layer header, the network layer header, the 14 byte header at the data link layer and the 4 byte trailer at the data link layer.
So, in order to assess the overhead percentage, we divide the overhead bytes between the total quantity of bytes sent to the physical layer, as follows:
OH % = (58 / 758) * 100 = 7.7 %
Answer:
Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.
Explanation:
Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.
The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.
Answer:
13.95
Explanation:
Given :
Vector A polar coordinates = ( 7, 70° )
Vector B polar coordinates = ( 4, 130° )
To find A . B we will
A ( r , ∅ ) = ( 7, 70 )
A = rcos∅ + rsin∅
therefore ; A = 2.394i + 6.57j
B ( r , ∅ ) = ( 4, 130° )
B = rcos∅ + rsin∅
therefore ; B = -2.57i + 3.06j
Hence ; A .B
( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95
Answer:
The stress in the rod is 39.11 psi.
Explanation:
The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:
Replacing the diameter the area results:
Therefore the the stress results:
Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.
Explanation:
Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.
