A reservoir rock system located between a depth of 2153m and a depth of

The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm

Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0

3 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago

A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a te

Artyom0805 [142]

Explanation:

a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures

100% (3 ratings)

A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2

3 0

3 years ago

Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app

stiks02 [169]

Answer:

a) Fb= 275.77 lb Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

Fay = 337.75 lb

Fbx = 195 lb

Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ $F_{x} = 0$ & ∑ $F_{y} = 0$

Hence,

$F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0$

$F_{A}sin(30) - F_{B} sin(45) = 0$

and

$F_{A} = 390 lb$

Inserting the value of $F_{A}$ and solving the remaining equations simultaneously yields (magnitudes),

$F_{B} = 275.77 lb\\F_{C} = 142.75 lb$

b) Summing up moments

$M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))$

$M = -779.97 lb.ft$ (i.e. 779.97 lb.ft clockwise)

c)

$F_{Ax} = 390 sin(30) = 195 lb$

$F_{Ay} = 390 cos(30) = 337.75lb\\$

$F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb$

8 0

3 years ago

Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772

slega [8]

Answer:

atomic percentage = 143 %

Explanation:

Let x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:

$\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }$

where;

the lattice parameter is given as : 4.7589 × 10⁻⁸ cm

The atomic mass of tin is 118.69 g/mol

The atomic mass of copper is 63.54 g/mol

The density is 8.772 g/cm³

$\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }$

569.32 = 118.69x + 254.16-63.54x

569.32 - 254.16 = 118.69x - 63.54 x

315.16 = 55.15x

x = 315.16/55.15

x = 5.72 atoms/cell

As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is calculated as follows :

atomic % = $\frac{no \ of \ atoms \ per \ cell \ in \ tin }{no \ of \ atoms \ per \ cell \ in \ the \ metal}*100$

atomic % = $\frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100$

atomic % = 143 %

7 0

3 years ago