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natita [175]
2 years ago
13

The following laboratory tests are performed on aggregate samples:a. Specific gravity and absorptionb. Soundnessc. Sieve analysi

s test.What are the significance and use of each of these tests (1 point each)?
Engineering
1 answer:
telo118 [61]2 years ago
7 0

Answer:

Explanation:

A- Specific gravity and Absorption Test: Specific gravity is a measure of a material’s density as compared to the density of water at 73.4°F (23°C). Therefore, by definition, water at a temperature of 73.4°F (23°C) has a specific gravity of 1. Absorption is also determined by the same test procedure and it is a measure of the amount of water that an aggregate can absorb into its pore structure.

Specific gravity is used in a number of applications including Superpave mix design, deleterious particle identification and separation and material property change identification while

B- Soundness Test : This determines an aggregate's resistance to disintegration by weathering and in particular, freeze-thaw cycles. Aggregates that are durable (resistant to weathering) are less likely to degrade in the field and cause premature HMA pavement distress and potentially failure.It is used to identify the excess amount of lime in cement.

C - Sieve analysis Test: is a practice or procedure used to assess the particle size distribution (also called gradation) of a granular material by allowing the material to pass through a series of sieves of progressively smaller mesh size and weighing the amount of material that is stopped by each sieve as a fraction of the whole mass. This test is used to describe the properties of the aggregate and to see if it is appropriate for various civil engineering purposes such as selecting the appropriate aggregate for concrete mixes and asphalt mixes as well as sizing of water production well screens.

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2 years ago
(a) Aluminum foil used for storing food weighs about 0.3 grams per square inch. How many atoms of aluminum are contained in one
atroni [7]

Answer:

note:

solution is attached due to error in mathematical equation. please find the attachment

3 0
2 years ago
Technician A says that a radio may be able to receive AM signals, but not FM signals if the antenna is defective. Technician B s
DIA [1.3K]

The response to whether the statements made by both technicians are correct is that;

D: Neither Technician A nor Technician B are correct.

<h3>Radio Antennas</h3>

In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.

Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.

Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.

Read more about Antennas at; brainly.com/question/25789224

3 0
2 years ago
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26
scZoUnD [109]

Answer:

The strength coefficient is 625 and the strain-hardening exponent is 0.435

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  (K) and the (n).

\sigma =K\epsilon^n

We will plug the values in the formula.

250=K\times (0.12)^n\\350=K\times (0.26)^n

We will solve these equation.

K=\frac{250}{(0.12)^n} plug this value in 350=K\times (0.26)^n

350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\  \\1.4=(2.17)^n

Taking a natural log both sides we get.

ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435

Now, we will find value of K

K=\frac{250}{(0.12)^n}

K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625

So, the strength coefficient is 625 and the strain-hardening exponent is 0.435.

5 0
3 years ago
(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5
bezimeni [28]

Answer:

a.) 1.453MW/m2,  b.)  2,477,933.33 BTU/hr  c.) 22,733.33 BTU/hr  d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

                        50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, Q = \frac{109*0.5*100}{0.0075} = 726,666.667W

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, Q=\frac{1*0.5*100}{0.0075} = 6,666.67W

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, Q=\frac{109*0.5*100}{0.015} =363,333.33W

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0
2 years ago
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