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s2008m [1.1K]
3 years ago
13

3. In order to obtain your commercial driver's license (CDL) you must first:

Engineering
1 answer:
Murljashka [212]3 years ago
8 0
A and C is the answer to the question. Be 15 years old & get a permit
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Who ever is here first get brainliest
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Heey

Explanation:

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Select the properties and typical applications for the high carbon steels.
yanalaym [24]

Answer:

<u>Option-(A)</u>

Explanation:

<u>Typical applications for the high carbon steels includes the following;</u>

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

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3 years ago
The application of technology results in human-made things called
Sergio039 [100]

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Internet of things

Explanation:

This is a good example where the application of technology results are applied to human made things.

Internet of things (IOT), involves the application of one technology results–the internet, embedded into devices such as refrigerator, television etc so as to send and receive data (digital instructions). Such applications of technology results has revolutionized the way we use "human made things".

8 0
4 years ago
A spacecraft is fueled using hydrazine ​(N2H4​; molecular weight of 32 grams per mole​ [g/mol]) and carries 1 comma 630 kilogram
Varvara68 [4.7K]

Answer:

attached below

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7 0
3 years ago
An ideal gas turbine operates using air coming at 355C and 350 kPa at a flow rate of 2.0 kg/s. Find the rate work output
GuDViN [60]

Answer:

The rate of work output = -396.17 kJ/s

Explanation:

Here we have the given parameters

Initial temperature, T₁ = 355°C = 628.15 K

Initial pressure, P₁ = 350 kPa

h₁ = 763.088 kJ/kg

s₁ = 4.287 kJ/(kg·K)

Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately

h₂ = 79.572 kJ/kg

The saturation temperature at the given

T₂ = 79°C

The rate of work output \dot W = \dot m×c_p×(T₂ - T₁)

Where;

c_p = The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)

\dot m =  The mass flow rate = 2.0 kg/s

Substituting the values, we have;

\dot W = 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s

\dot W = -396.17 kJ/s

7 0
4 years ago
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