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Vsevolod [243]
2 years ago
9

An engineer is working with archeologists to create a realistic Roman village in a museum. The plan for a balance in a marketpla

ce calls for granite stones, each weighing denarium. (The denarium was a Roman unit of mass: ). The manufacturing process for making the stones will remove of the material. If the granite to be used has a density of , what is the minimum volume of granite (in ) that the engineer should order?
Engineering
1 answer:
NeTakaya2 years ago
3 0

Answer:

The minimum volume requirement for the granite stones is 1543.64 cm³

Explanation:

1 granite stone weighs 10 denarium

100 granted stones will weigh 1000 denarium

1 denarium = 3.396g

1000 denarium = 3396g.

But we're told that 20% of material is lost during the making of these stones.

This means the mass calculated represents 80% of the original mass requirement, m.

80% of m = 3396

m = 3396/0.8 = 4425 g

This mass represents the minimum mass requirement for making the stones.

To now obtain the corresponding minimum volume requirement

Density = mass/volume

Volume = mass/density = 4425/2.75 = 1543.64 cm³

Hope this helps!!!

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A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp
iragen [17]

Answer:

Elastic modulus of steel  = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm  = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × 10^{-4} m

solution

we know that Elastic modulus is express as

Elastic modulus = \frac{stress}{strain}    ................1

here stress is

Stress = \frac{Force}{area}       .................2

Area = (0.022)²

and  

Strain = \frac{extension}{length}       .............3

so here put value in equation 1 we get

Elastic modulus = \frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }  

Elastic modulus of steel = 202.27 × 10^9 Pa

Elastic modulus of steel  = 202.27 GPa

3 0
3 years ago
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0
3 years ago
4. What are these parts commonly called?
patriot [66]

These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.

The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.

The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.

These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.

Learn more about carburetors here:

brainly.com/question/4237015

7 0
2 years ago
Technician A states that a brake lathe is used to make a used brake rotor surface "like new". Technician B states that a brake l
nikitadnepr [17]

Answer:

Both Technician A and B are correct.

Explanation: A brake lathe is a special tool used to improve or work on the surface of brake pads it helps to smoothen the surface.

Brake lathe has been found to be very effective in removing rusts in rotors and unevenness in the brake pad surfaces in order to ensure the efficiency and effectiveness of the brake system of a vehicle. Hence, a brake lathe helps to make brake rotor surface as smooth as possible.

7 0
3 years ago
Discuss typical advantages and disadvantages of an irrigation system?
trapecia [35]
Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.
4 0
3 years ago
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