1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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.

You might be interested in
Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of
earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant

Since the losses are neglected thus applying this theorm between upper and lower porion we have

\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}

Now by continuity equation we have

A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s

Applying the values in the Bernoulli's equation we get

\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars

6 0
3 years ago
Answer my question I will mark brainliest
Iteru [2.4K]

Answer:

150

Explanation:

Mark me Brainliest

6 0
3 years ago
Read 2 more answers
A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

   = 270 / 0.0144

δ  = 18750 lb/in² = 19,000 Psi

<u>Part (B)</u> The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

<u>Part (C)</u> The modulus of elasticity of the steel

E = δ/σ

   = 19,000 / 0.00063

E = 30,000,000 Psi

4 0
3 years ago
What are the general rules for press fit allowances
Keith_Richards [23]

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

please rate brainliest if helps and follow

4 0
1 year ago
Which examples best demonstrate likely tasks for Health, Safety, and Environmental Assurance workers? check all that apply
Over [174]

Sam, Elijah, Joy Those are 100% correct from my human knowledge

6 0
3 years ago
Read 2 more answers
Other questions:
  • The spring has a stiffness k=200 N/m and is unstretched when the 25 kg block is at A. Determine the acceleration of the block wh
    6·1 answer
  • Write a function digits() that accepts a non-negative integer argument n and returns the number of digits in it’s decimal repres
    13·1 answer
  • A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg lid. If the local atmospheric pressure
    9·2 answers
  • A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.
    7·1 answer
  • Which of the following describes a product concept?
    15·1 answer
  • If my current directory is ‘AR’ write the path for my current directory
    5·1 answer
  • Which component found in fertilizer is a known cancer-causing agent?
    11·2 answers
  • What type of plans have to do with earth, soil, excavation, and location<br> of a house on a lot?
    12·1 answer
  • List six clues that indicates that you are approaching an intersection
    10·1 answer
  • Advanced manufacturing does NOT serve the transportation, communications, or medical industries. Is this statement TRUE or FALSE
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!