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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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The mathematical model for finite element method in structural mechanics is the
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Explanation:

Conservation of velocity equation

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4 years ago
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The enforcement of OSHA standards is provided by federal and state
Gnoma [55]

Answer:

Explanation:

Enforcing OSHA, Occupational Safety and Health Administration, standards is not a job for electricians, lawmakers or tax collectors. The right answer is safety inspectors.

3 0
2 years ago
Water at 15°C is to be discharged from a reservoir at a rate of 18 L/s using two horizontal cast iron pipes connected in series
Zina [86]

Answer:

0.245 m^3/s

Explanation:

Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s

5 0
3 years ago
Swing arm restraints are intended to prevent a vehicle from falling off a lift.
castortr0y [4]

The question is asking whether that statement is true or false. Options are;

A) True

B) False

This is about usage of Swing arm restraints.

<em><u>B) False</u></em>

There are different safety features that people employ when a vehicle is lifted. However, for this question, we will only talk about swing arm restraints.

  • Swing arm restraints are lifting restraint devices that are used to prevent a cars arms from shifting or going out of position after that car has been lifted and mounted.

  • This swing arm restraint does not prevent a vehicle from falling off a lift as it just helps to ensure that the swing arms that are unloaded basically maintain their position.

Read more at; brainly.com/question/17972874

4 0
3 years ago
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum
adoni [48]

Answer:

5.6 mm

Explanation:

Given that:

A cylindrical tank is required to contain a:

Gage Pressure P = 560 kPa

Allowable normal stress \sigma = 150 MPa = 150000 Kpa.

The inner diameter of the tank = 3 m

In a closed cylinder  there exist both the circumferential stress and the longitudinal stress.

Circumferential stress \sigma = \dfrac{pd}{2t}

Making thickness t the subject; we have

t = \dfrac{pd}{2* \sigma}

t = \dfrac{560000*3}{2*150000000}

t = 0.0056 m

t = 5.6 mm

For longitudinal stress.

\sigma = \dfrac{pd}{4t}

t= \dfrac{pd}{4*\sigma }

t = \dfrac{560000*3}{4*150000000}

t = 0.0028  mm

t = 2.8 mm

From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value  with the maximum thickness = 5.6 mm

8 0
4 years ago
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