ENGINEERING COLLEGE

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period.
b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits.
c. Do all of these samples meet the contract specifications? Explain your answer.
d. What trend do you observe based on the data? What could cause this trend?"

Answers

Answer 1

Answer:

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535, standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

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Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,
b) sketch vc and ic.

Answers

Answer:

hello your question is incomplete attached is the complete question

A) Vc = 15 ( 1 - $e^(-t/0.15s)$ ) , ic = $1.5 mAe^(-t/0.15s)$

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

$R_(th)$ = 8 k ohms || 24 k ohms = 6 k ohms

$E_(th)$ = $(20 k ohms(20 v))/(24 k ohms + 8 k ohms)$ = 15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc = $E( 1 - e^(-t/t) )$

hence Vc = 15 ( 1 - $e^(-t/0.15)$ )

ic = $(E)/(R) e^(-t/t)$ = $(15)/(10) e^(-t/t)$ = $1.5 mAe^(-t/0.15s)$

attached

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)$

$10*2=\sqrt{1+8F^(2) } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_(1) }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*√(32.2*1) =42.0833ft/s$

b) The flow rate:

$q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s$

c) The Froude number:

$v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s$

$F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345$

d) The flow energy dissipated:

$E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft$

e) The critical depth:

$d_(c) =((((q)/(L))^(2) )/(g) )^(1/3) =((((3366.664)/(80))^(2) )/(32.2) )^(1/3) =3.8030ft$

A chemical process converts molten iron (III) oxide into molten iron and carbon dioxide by using a reducing agent of carbon monoxide. The process allows 10.08 kg of iron to be produced from every 16.00 kg of iron (III) oxide in an excess of carbon monoxide. Calculate the percentage yield of iron produced in this process.

Answers

Answer:

percentage yield = 63%

Explanation:

The yield efficiency or percentage yield measure the amount of products that are formed from a given amount of reactant. For a percentage yield of 100, all the reactants are completely converted to product. Mathematically, the percentage yield is given by:

$percentage\ yield = (Actual\ yield)/(expected\ yield) * 100\nActual\ yield = 10.08kg\nExpected\ yield= 16.00kg\n\n\therefore percentage\ yield = (10.08)/(16.00) * 100 = 63 \%$

The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms in any proportion without altering the crystal structure except by expanding it slightly: this is an example of a binary solid solution series. For different compositions, the lines in the powder diffraction patterns are in slightly different positions, because of the cell expansion, but the overall pattern remains basically the same. The spacing of the lattice planes varies linearly with composition, and this can be used in a rapid and non- destructive method of analysis. a. The (062) reflection from olivine is strong and well resolved from other lines. Calculate d062 for an olivine that displays its (062) reflection at a Bragg angle of 37.21° (i.e., a diffraction angle of 74.42°) when x-rays with a wavelength of 0.1790 nm are used. b. The d062 spacing as measured accurately for synthetic materials is 0.14774 nm for Mg2SiO4 and 0.15153 nm for Fe2SiO4. What would be the approximate composition, expressed in mol.% Mg2SiO4, of an olivine material for which do62 has the value obtained in part 2.1 above?

Answers

Answer:

The answer is "0.147 nm and 99.63 mol %"

Explanation:

In point (a):

$\to nk1 = 062$

$\to \text{Bragg angle}$ $\theta =37.21^(\circ)$

$\to \text{diffraction angle}$ $2 \theta = 74.42^(\circ)$

$\to \lambda = 0.1790 nm$

find:

d(062)=?

formula:

$\to nx = 2d \sin \theta$

$\to d(062) = (1 * 0.1790^(\circ))/(2 * \sin 37.21^(\circ))\n$

$= (0.1790^(\circ))/(2 * 0.604738126)\n\n= (0.29599589)/(2)\n\n= 0.147 \n$

In point (b):

$\to Mg_2SiO_4\longleftrightarrow Fe_2SiO_4$

$d= 0.14774 \ \ \ \ \ olivine = 0.147 \ \ \ \ \ 0.15153$

formula:

$\to d=(a)/(√(n^2+k^2+i^2))\n$

that's why the composition value equal to 99.63 %

Under the right conditions, it is possible, due to surface tension,to have metal objects float on water. Consider placing a shortlength of a small diameter steel ( γ = 490 lb/ft3)rod on a surface of water. What is the maximum diameter that therod can have before it will sink? Assume that the surface tensionforces act vertically upward. Note: A standard paper cliphas a diameter of 0.036 in. Partially unfold a paper clip and seeif you can get it to float on water. Do the results of thisexperiment support your analysis?

Answers

Answer:

A) 0.0614 inches

b) The standard steel paper clip should float on water

Explanation:

The maximum diameter that the rod can have before it will sink

we can calculate this using this formula :

D = $((8\alpha )/(\pi y ) )^{(1)/(2) }$ ----- 1

∝ = value of surface tension of water at 60⁰f = 5.03×10^−3 lb/ft

y = 490 Ib/ft^3

input the given values into equation 1 above

D = $((8*(5.3*10^(-3)) )/(\pi *490 ) )^{(1)/(2) }$

= 5.11 * 10^-3 ft convert to inches

= 5.11 *10^-3 ( 12 in/ 1 ft ) = 0.0614 inches

B) The diameter of a standard paper Cliphas = 0.036 inches

and the diameter of the rod = 0.0614. Hence the standard steel paper clip should float on water

One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.

Answers

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.

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