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nordsb [41]
2 years ago
8

. . . . . . . . . . ..........

Engineering
1 answer:
Naddika [18.5K]2 years ago
5 0

Answer: Thx for points...

Explanation:

Heres Random Info About Enginnering (because why not): Engineers are involved in all aspects of interactive TV technology, from designing new cables, to creating new film emulsions, to engineering better sound quality. This technology allows viewers to select any program, film, or game from more than 500 channels. Engineers play an instrumental role in the theme park industry?

Hope This Helps Someday!

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(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without
steposvetlana [31]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

5 0
3 years ago
A sample of sand weighs 490 g in stock and 475 in Oven Dry (OD) condition, respectively. If absorption capability of the sand is
Ivahew [28]

The weight of the specimen in SSD condition is 373.3 cc

<u>Explanation</u>:

a) Apparent specific gravity = \frac{A}{A-C}

Where,

A = mass of oven dried test sample in air = 1034 g

B = saturated surface test sample in air = 1048.9 g

C = apparent mass of saturated test sample in water = 975.6 g

apparent specific gravity = \frac{A}{A-C}

                                         = \frac{1034}{1034-675 \cdot 6}

Apparent specific gravity = 2.88

b) Bulk specific gravity G_{B}^{O D}=\frac{A}{B-C}

G_{B}^{O D}=\frac{1034}{1048.9-675 \cdot 6}

       =  2.76

c) Bulk specific gravity (SSD):

G_{B}^{S S D}=\frac{B}{B-C}

=\frac{1048 \cdot 9}{1048 \cdot 9-675 \cdot 6}

G_{B}^{S S D} = 2.80

d) Absorption% :

=\frac{B-A}{A} \times 100 \%

=\frac{1048 \cdot 9-1034}{1034} \times 100

Absorption = 1.44 %

e) Bulk Volume :

v_{b}=\frac{\text { weight of dispaced water }}{P \omega t}

=\frac{1048 \cdot 9-675 \cdot 6}{1}

= 373.3 cc

5 0
2 years ago
Write down the types and tasks of the pressure control valves ?
Yuki888 [10]

Answer:

There are 6 types of pressure control valves and their function is to regulate the pressure below a threshold level within safe limits and to maintain and control  pressure of a particular circuit.

Explanation:

The six type of Pressure valve with their functions are given below:

a. Unloading Valve:

These type of pressure valve are used to pour fluid into the container at very low or no pressure.

b. Safety valve:

These are used when the pressure within the vessel is in excess as inside temperature is near about preset [point point then these valves are open to release the extra pressure and are closed once normal conditions are regained.

c. Pressure Reducing Valve:

These are basically used for the control of the pressure in downstream not exceeding the design limits.

d. Pressure Relief Valves:

These are basically used to limit and regulate the pressure of any system.

e. Counter Balance Valve:

These are used to develop pressure in the reverse direction at the actuator's return line in order to keep the load under control.

f. Sequence Valve:

These are used to maintain sequence or order in the operations of two parts or branches.

8 0
3 years ago
1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and
belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of  sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = \frac{wet mass - dry mass }{wet mass} = \frac{0.525 - 0.49}{0.525} = 0.07 or 7%

water content =  \frac{wet mass - dry mass}{wet mass} = 7%

8 0
3 years ago
Read 2 more answers
Which measuring tool will be used to determine the diameter of a crankshaft journal?
guapka [62]

Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

Hope it's helpful to you

pls mark me as brain list

4 0
3 years ago
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