Answer:
Mechanical dissection
Explanation:
Taking apart old things and innovating to improve upon them is mechanical dissection.
Answer:
touching
Explanation:
The backrest of the seat should be tilted back ever so slightly, and when turning the steering wheel your shoulders should remain in contact with the seat – rather than hunched forward.
Explanation:
Engineering drawing abbreviations and symbols are used to communicate and detail the characteristics of an engineering drawing. There are many abbreviations common to the vocabulary of people who work with engineering drawings in the manufacture and inspection of parts and assemblies.
Some examples of these are:
BOT: Bottom
B.O.W.: Bottom of Wall
BP: Blue Print (or B/P)
BV: Butterfly Valve
CAD: Cadmium or Computer-Aided Drafting
CBORE: Counterbore
C.C.: Center to Center
C.D.: Construction Document
C.F.M.: Cubic Feet per Minute
CFS: Cubic Feet per Second
C.I.: Cast Iron
E.F.: Exhaust Fan
EQ: Equal, or Equally
E.W.: Each Way
Ext.: Exterior
FACP: Fire Alarm Control Panel
FAO: Finish All Over
Answer:
Aluminum cross sectional area = 1.99 * 10^-5 mm^2
Steel cross sectional area = 9.95* 10^-6 mm^2
Explanation:
Given data:
Tensile strength of Grunerite = 3.5 * 10^2 kg/mm^2 = 3.5 * 10^-4 kg/<em>u</em>m^2
Tensile strength of Aluminum = 2.5 × 10^4 lb/in2 = 2.5 × 10^4 * 703.07 kg/m^2
Tensile strength of Steel number 5137 = 5.0 × 10^4 lb/in2 = 5.0 × 10^4*703.07 kg/m^2
<u>Calculating the cross sectional area of the wires of aluminum and steel No5137</u>
first we will determine the cross sectional area of the aluminum wire ( A ) by equating tensile strength of aluminum with the tensile strength
of Grunerite
(2.5 × 10^4 * 703.07) * A = 3.5 * 10^-4 kg
Hence ; A = 1.99 * 10^-5 mm^2
Next we calculate the cross sectional area of steel
5.0 × 10^4*703.07 kg/m^2 * A" = 3.5 * 10^-4 kg
Hence ; A" = 9.95* 10^-6 mm^2
Answer:
0.19s
Explanation:
Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.
Queueing delay =(N-1) L /2R
where N = no of packet =93
L = size of packet = 4MB
R = bandwidth = 1.4Gbps = 1×10⁹ bps
4 MB = 4194304 Bytes
(93 - 1)4194304 / 2× 10⁹
queueing delay =192937984 ×10⁻⁹
=0.19s