Answer:
Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.
Explanation:
It is study of the relationships between heat, temprature, work and energy
Answer:
Paradox of Organizational Change: Engineering Organizations with Behavioral Systems Analysis. by. Maria E. Malott.
Answer:
E= 15 GPa.
Explanation:
Given that
Length ,L = 0.5 m
Tensile stress ,σ = 10.2 MPa
Elongation ,ΔL = 0.34 mm
lets take young modulus = E
We know that strain ε given as
![\varepsilon =\dfrac{\Delta L}{L}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B%5CDelta%20L%7D%7BL%7D)
![\varepsilon =\dfrac{0.34}{0.5\times 1000}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B0.34%7D%7B0.5%5Ctimes%201000%7D)
![\varepsilon =0.00068](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D0.00068)
We know that
![\sigma = \varepsilon E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cvarepsilon%20%20E%5C%5C%5C%5CE%3D%5Cdfrac%7B10.2%7D%7B0.00068%7D%5C%5CE%3D%0915000%5C%20MPa%5C%5CE%3D15%5C%20GPa)
Therefore the young's modulus will be 15 GPa.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)