Answer:
i) 3750 veh/hr/ln
ii) 100 veh/mi/In
iii) 37.5 mph
Explanation:
number of lanes = 3
sf for both directions = 75 mph ( free mean speed )
Dj for both directions = 200 veh/mi/In
<u>Calculate the value of S0, D0 (veh/mi/ln) and maximum Vm (veh/hr)</u>
For either direction we will consider the total volume = 3 lanes
value of Dj = 3 lanes * 200 = 600 veh/mi/
i) value of SO
= ( Dj * sf ) / 4 = ( 600 * 75 ) / 4 = 11250 veh/hr = 3750 veh/hr/lane
ii) Value of DO
DO = Dj / 2 = 200 /2 = 100 veh/mi/In
iii) Value of Vm
= sf /2 = 75 / 2 = 37.5 mph
Answer:
It is important for the client and the engineering team to maintain open communications even after the deliver of the product because
1) One of doctrine of Six Sigma which is a tool used to improve processes requires the input of effort meant to attain a stable process that can be easily predicted and geared towards the reduction of variation in the process is a pre requisite for the business to succeed
2) Communication can also help in product monitoring and control as well as ensuring that the product performance requirements are met
3) Communication between the client and the engineering team can serve as a means of reporting failures and collecting data for analysis
Explanation:
Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.
Answer:
Height of the water = √(128)/147456 ft
Explanation:
Given
Radius, r = 1 ft
Height, h = 2 ft
Radius of hole = 1/32in
Acceleration of gravity, g = 32ft/s²
c = 0.6
Area of the hold = πr²
A = π(1/32)² ---- Convert to feet
A = π(1/32 * 1/12)²
A = π/147456 ft²
Area of water = πr²
A = π 1²
A = π
The differential equation is;
dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water
A1 = π/147456, A2 = π
dh/dt = (π/147456)/π √(2*32*2)
dh/dt = 1/147456 * √128
dh/dt = √128/147456 ft
Height of the water = √(128)/147456 ft
Answer:
15.4 g/cm³, 17.4 g/cm³
Explanation:
The densities can be calculated using the formula below
ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)
fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8
a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³
b) density after infiltration with silver
fraction occupied by silver = 20 / 100 = 0.2
density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³
