Which option identifies the tool best to use in the following scenario?

(8 points) Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight

FromTheMoon [43]

Answer:

Weight of cement = 10909.08 pounds

Weight of sand = 18008.64 lb

Weight of gravel = 28051.92 lb

Weight of water = 5454.54 lb

Explanation:

given data

length = 40 feet

width = 12 feet

thickness = 6 in = 0.5 feet

mix proportion = 1:2:3

specific gravity of cement = 3.15

specific gravity of sand = 2.60

specific gravity of gravel = 2.70

water to cement ratio = 0.50

solution

first we get here total volume that is

volume = length × width × thickness ...........1

volume = 40 × 12 × 0.5

volume = 240 feet³

and

now we get here dry volume of concrete that will be

Dry volume = wet volume × 1.54 .......................2

Dry volume = 240 × 1.54

Dry volume = 360 feet³

and

we have given air content entrained is = 7.5 %

so required volume will be as

required volume = Dry volume × ( 1 - 7% ) .............3

required volume = 360 × ( 1 - 7.5 % )

required volume = 333 feet³

and

total ratio is

total ratio = 1+ 2+ 3

total ratio = 6

so volume of sand cement and gravel will be

volume of cement = $\frac{1}{6} \times 333$

volume of cement = 55.5 feet³

volume of sand = $\frac{2}{6} \times 333$

volume of sand = 111 feet³

volume of gravel = $\frac{3}{6} \times 333$

volume of gravel = 166.5 feet³

and

now we get here weight of sand cement and gravel as

weight = volume × specific gravity × density of water .............4

and now put here value and we get

Weight of cement = 55.5 × 3.15 × 62.4

Weight of cement = 10909.08 pounds

and

Weight of sand = 111 × 2.60 × 62.4

Weight of sand = 18008.64 lb

and

Weight of gravel = 166.5 × 2.7 × 62.4

Weight of gravel = 28051.92 lb

and

Weight of water = 0.5 of weight of cement

Weight of water = 0.5 × 1090.9.08

Weight of water = 5454.54 lb

8 0

2 years ago

Consider a steady-state experiment in which the observed current due to reduction of Ox to R is 85 mA/cm2. What is the concentra

Elanso [62]

Answer: Yes

Explanation:

Surface Concentration: In electrochemistry, there is an important distinction between the concentration of a species at the electrode’s surface and its concentration at some distance from the electrode’s surface (in what we call the bulk solution). Suppose we place an electrode in a solution of Fe3+ and fix the potential at 1.00 V. At this potential Fe3+ is stable—the standard state reduction potential for Fe3+ to Fe2+ is +0.771 V, the concentration of Fe3+ remains the same at all distances from the electrode’s surface.

Bulk Concentraton: If we change the electrode’s potential to +0.500 V, the concentration of Fe3+ at the electrode’s surface decreases to approximately zero. The concentration of Fe3+ increases as we move away from the electrode’s surface until it equals the concentration of Fe3+ in bulk solution. The resulting concentration gradient causes additional Fe3+ from the bulk solution to diffuse to the electrode’s surface.

5 0

2 years ago

The jib crane is supported by a pin at Cand rod AB. The rod can withstand a maximum tension of 40 kN. If the load has a mass of

lakkis [162]

Answer:

The maximum allowable distance = 5 m

Explanation:

Data:

There will be three forces on the jib. Let the forces be denoted as:

$C_{x}$ , $C_{y}$ and the force on pole AB

To find the angle AB makes with the horizontal beam:

$tan^{-1}(\frac{3}{4}) = 36.8699$

The load has a mass of 2 000 kg then, the force will be:

F = mg, where g = 9.81 m/s²

= $2000* 9.81 = 19 620 N$

Breaking AB into its x and y coordinates:

$AB_{x}=ABcos(36.8699)\\AB_{y} = ABsin(36.8699)$

Then,

∑ $M_{c}$ = 0

$0 = (4)(AB sin (36.8699) + 0.2 (AB cos (36.8699) - (5*19620)\\AB = 38 320. 3 N$

∑ $F_{x} = 0\\C_{x} = AB cos(36.8699)\\C_{x} = 30 656.2 N$

∑ $F_{y} = 0\\C_{y} + AB sin (36.8699) - 19 620 = 0\\C_{y} = - 3 372.18 N$

so the components of the forces will be 30 656.2 N and - 3 372.18 N

6 0

2 years ago

Which of the following explains why trucking is the most widely used transportation method?

True [87]

Answer:

Trucks are the fastest mode of transportation

Explanation:-Means of transport is a term used to distinguish between different ways of transportation or transporting people or goods. The different modes of transport are air, water, and land transport, which includes Rails or railways, road and off-road transport. Other modes also exist, including pipelines, cable transport, and space transport. Human-powered transport and animal-powered transport are sometimes regarded as their own mode, but never fall into the other categories. In general, transportation is used for moving of people, animals, and other goods from one place to another. The means of transport, on the other hand, refers to the (motorized) vehicles necessary for transport according to the chosen mode (car, airplane, ship, truck and rail). Each mode of transport has a fundamentally different technological solution, and some require a separate environment. Each mode has its own infrastructure, vehicles, and operations.

****†********************MD.Tashin*************************

6 0

2 years ago

Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select fro

Vinvika [58]

Answer:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

7 0

2 years ago