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RUDIKE [14]
3 years ago
5

Explain how a CO2 cartridge powers the dragster you will be building. A good website to use is How Stuff Works. (howstuffworks.c

om) Write a good paragraph or two to explain your answer in complete sentences.
Engineering
2 answers:
Sonbull [250]3 years ago
7 0
Co2 cartridges are pressed
andrezito [222]3 years ago
5 0

Answer

C02 cartridges are compressed.

Explanation:

The C02 Cartridge will be used to power our dragsters by the compression inside of the bottle. Since C02 is a compressed gas inside of a small container, The pressure when released my provide enough force to move the dragster at high speeds. An example of this would be like an air soft gun. some air soft guns are used by small C02 Cartridges, which let out a strong burst of compressed C02, which fires the projectiles at very high speeds.

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The Acme tool is aligned to the work with: A. A square B. The eye C. An Acme tool gage D. A center gage
kotegsom [21]

Answer:

(c) an acme tool gage

Explanation:

with the help of an acme tool we measure the pitch of a screw thread it is used as reference for finding the pitch of the thread it is also called an inspection tool is aligned to work with an acme tool gage it is also used for gage the internal and external threaded product. it is also used for grinding and setting tools it is mainly work 29° thread angle

8 0
4 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}

Original length L_(i}

\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)

Where d_{f} is final diameter, d_{i} is original diameter, L_{f} is final length and L_{i} is original length.

\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}

\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}

Therefore, the original length is 105.70 mm

7 0
4 years ago
Explain why the scenario below fails to illustrate an understanding of the importance of metrology. Situation: Natalie is a cali
VMariaS [17]

Answer:

Explanation:

The situation being described completely fails in regard to the importance of metrology. This is because the main importance of metrology is making sure that all of the measurements in a process are as accurate as possible. This accuracy allows an entire process to function efficiently and without errors. In a food production plant, each individual department of the plant relies on the previous function to have completed their job with the correct and accurate instructions so that they can fulfill their functions correctly and end up with a perfect product. If the oven (like in this scenario) is a couple of degrees off it can cause the product to come out burned or undercooked, which will then get transferred to the next part of production which will also fail due to the failed input (burned or undercooked product). This will ultimately lead to an unusable product at the end of the process and money wasted. Which in a large production plant means thousands of products in a single batch are thrown away.

8 0
3 years ago
Given the following MATLAB statement: ( 3 + 2 ) / 5 * 4 + 5 ^ 2 In what order will these operations be done?
Andrei [34K]

Answer:

first is the parentheses, (3+2)=5 next is the exponent 5^2=25, next is the division 5 / 5 = 1, then the multiplication 4*1=4 and then you add 4+25=29. so the answer is 29.

7 0
3 years ago
A 2 m3 insulated rigid tank contains 3 kg of nitrogen at 90 kPa. Now work is done on the system until the pressure in the tank r
Nutka1998 [239]

Answer: \Delta S = 1.47kJ/K

Explanation: <u>Entropy</u> is the measure of a system's molecular disorder, i.e, the unuseful work a system does.

The nitrogen gas in the insulated tank can be described as an ideal gas, so it can be used the related formulas.

For the entropy, the ratio of initial and final temperatures is needed and as volume is constant, we use:

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}

\frac{P_{2}}{P_{1}} =\frac{T_{2}}{T_{1}}

\frac{T_{2}}{T_{1}} =\frac{175}{90}

\frac{T_{2}}{T_{1}} =1.94

<u>Specific</u> <u>Heat</u> is the quantity of heat required to increase the temperature 1 degree of a unit mass of a substance. Specific heat of nitrogen at constant volume is c_{v}= 0.743kJ/kg.K

The change in entropy is calculated by

\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})-Rln(\frac{V_{2}}{V_{1}} )]

For the nitrogen insulated in a rigid tank:

\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})]

Substituing:

\Delta S= 3[0.743ln(1.94)]

\Delta S= 1.47

The entropy change of nitrogen in an insulated rigid tank is 1.47kJ/K

6 0
3 years ago
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