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natita [175]
3 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]3 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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3 years ago
In the last 5 meters of braking, you lose ___ of your speed.
expeople1 [14]

Answer:

answer is

a)

3/4

Explanation:

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3 years ago
Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30
hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

V_z=V_z_o+I_zr_z

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

V_z_o=V_z-I_zr_z

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V

The value of I_L is given as 5 mA

Now the expression of current is as

I=I_z+I_L\\I=10mA+5mA\\I=15 mA

Now the resistance is calculated as

R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA

The load voltage is 7.485 V

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5. Consider the LTI system defined by the differential equation (a) Draw the pole-zero plot for the system. Is the system stable
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