Answer:
Procurement Process
Explanation:
Procurement Process describes the series of activities that an organization partakes in to get products or services in order to achieve their goals. The choice of the procurement process is very important for the success of a construction project.
So during a bidding process, the procurement process is section where the organization will need to get water and other utilities, sanitation equipment or storage needed for the success of a construction project.
Answer: D. All of the choice A, B and C are correct.
Answer: 12.4 feet
Explanation:
If there is a smooth transition and there is no change in slopes, energy considerations can be used
The cube has a kinetic energy of
ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when there is a gain in potential energy
pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
If there is no loss in energies,
pe = ke
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
h = 2000 /322 = 6.211 (ft)
= h / sin(30) = 12.4 ft
Answer:
Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)
Explanation:
The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by
V(t) = V₀ [1 - e⁻ᵏᵗ]
where k = (1/time constant)
when V(t) = V₀/2
(1/2) = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = - 0.693
-kt = - 0.693
kt = 0.693
t = (0.693/k)
Recall that k = (1/time constant)
Time to charge to half of max voltage = T(1/2)
T(1/2) = 0.693 (Time constant)
when V(t) = 0.75
0.75 = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
In e⁻ᵏᵗ = In 0.25 = -1.386
-kt = - 1.386
kt = 1.386
t = 1.386(time constant) = 2 × 0.693(time constant)
Recall, T(1/2) = 0.693 (Time constant)
t = 2 × T(1/2)
Hope this Helps!!!
Answer:
0.2 kcal/mol is the value of 
for this reaction.
Explanation:
The formula used for is:
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol
![=-1.987 cal/K mol\times 298 K\ln [35]+1.987 cal/K mol\times 298K\times \ln [46]](https://tex.z-dn.net/?f=%3D-1.987%20cal%2FK%20mol%5Ctimes%20298%20K%5Cln%20%5B35%5D%2B1.987%20cal%2FK%20mol%5Ctimes%20298K%5Ctimes%20%5Cln%20%5B46%5D)
1 cal = 0.001 kcal
0.2 kcal/mol is the value of for this reaction.
