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padilas [110]
3 years ago
5

4. What are the basic scientific principles the engineers who designed the digital scales would have needed to understand to des

ign this tool? Choose all that apply. A. Electric currents B. Physical properties of metals C. Influence of gravitational force on objects D. Physical and chemical properties of materials used in building circuits
Engineering
1 answer:
tamaranim1 [39]3 years ago
5 0

Answer:

Electric currents

Physical properties of metals

Influence of gravitational force on objects

Explanation:

The engineer who designed the digital scale must have a good working knowledge of the electric current, and its principle of flow through various materials, and also, factors that can affect its flow through a material (especially metals).

The physical properties of metal would have to be known by the engineer, in order to design the digital scale. The engineer must be able to predict some of the effect of the changes in the physical properties of a metal, and then use this to get some function that he wants from the digital scale, especially the effect of these physical changes on the electrical conducting ability of a metal.

The knowledge of the effect of gravitational force on objects is a must-know for the engineer. The process of scaling here on earth is very much dependent on the gravitational pull on objects downwards to the earth. This knowledge can be used to predict how much deflection this gravity forces will induce on an object towards the earth.

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What is a Flame Front Generator?
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The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

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3 years ago
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