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rjkz [21]
3 years ago
8

How is engine power expressed?

Engineering
2 answers:
Kobotan [32]3 years ago
6 0
Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.
mars1129 [50]3 years ago
5 0

Answer: Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.

Explanation: (I hope this helped!! ^^)

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Why is it important for engineers to consider both short and long term implications of their work?
polet [3.4K]
The right answer should be A).
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3 years ago
The Review_c object has a lookup relationship up to the Job_Application_c object. The job_Application_c object has a master-deta
fomenos

Answer:

Utilize the Standard Controller for Position_c and a Controller Extension to query for Review_c data.

Explanation:

Data types such as  individual, pedigree, sample, and marker can be displayed depending on the type of information needed, and the database fields will be recorded based on their levels with the appropriate data type placed in each field. The relationship that will be gotten will depend on the auto-populated defaults which can changed to suit our requirements.

6 0
3 years ago
A 0.01 m3 piston and cylinder contains an ideal gas at 5.92154 atm, 290 K . A constant pressure process produces 54 kJ of work o
Debora [2.8K]

Answer:

Final Volume of the gas = 0.1 m^3

Explanation:

Check the attached file for the calculations involved in this solution

4 0
3 years ago
A mercury thermometer has a cylindrical capillary tube with an internal diameter of 0.2 mm. If the volume of the thermometer and
vova2212 [387]

To solve this problem we will proceed to calculate the specific volume from the area of the cylinder and the sensitivity. Later we will calculate the volumetric coefficient of thermal expansion and finally we will be able to calculate the volume through the relation of the two terms mentioned above. Our values are

\text{Sensitivity}= 2mm/\°C

\text{Internal diameter } d= 0.2mm

\text{Differential expansion of Hg } \lambda_L = 1.82*10^{-4}/\°C

Let's start by calculating the specific volume which is given by

v = \pi (\frac{d}{2})^2 \gamma

Here,

d = Diameter

\gamma = Sensitivity

Replacing our values we have

v = (\frac{\pi}{4})(0.2mm)^2(2mm/\°C)

v = 0.0628mm^3 /\°C

Now we will obtain the value of the volumetric coefficient of thermal expansion of mercury through the differential expansion coefficient of Hg whic is three times, then

\lambda_V = 3\lambda_L

\lambda_V = 3(1.82*10^{-4}/\°C)

\lambda_V = 5.46*10^{-4}/\°C

Finally the relation to calculate the volume the bulb must is

\text{Specific volume} = \text{Bulb Volume} \times \text{Volumetric Coefficient}

v = v_B \times \lambda_V

v_B = \frac{v}{\lambda_V}

v_B = \frac{0.0628mm^3/\°C}{5.46*10^{-4}/\°C}

v_B = 115mm^3

Therefore the volume that the bulb must have is 115mm^3

5 0
3 years ago
Select the correct answer.
Ulleksa [173]
The answer is A. Immediately inform her colleague
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3 years ago
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