Answer:
b)false
Explanation:
Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling is a metal forming process.
We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.
So given statement is wrong.
1 micro gram of Strontium-90 has an activity of
0.0000053 terabecquerels (TBq),
Explanation:
Given information denotes that .,one gram of Strontium-90 has an activity of 5.3 terabecquerels (TBq)
the activity of 1 micro gram is
1 gram = 1,000,000 micro gram has activities of 5.3 terabecquerels
therefore 1 micro gram has the activity of (5.3 ÷ 1,000,000 = 0.0000053 )
= 
Hence ., 1 micro gram of Strontium-90 has an activity of
0.0000053 terabecquerels (TBq),
Answer:
a) 22.5number
b) 22.22 m length
Explanation:
Given data:
Bridge length = 500 m
width of bridge = 12 m
Maximum temperature = 40 degree C
minimum temperature = - 35 degree C
Maximum expansion can be determined as
where , \alpha is expansion coefficient 
degree C
SO, 
number of minimum expansion joints is calculated as
b) length of each bridge
Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
