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natali 33 [55]
4 years ago
5

The author claims that engineers should always have the final say and last word when it comes to setting safety standards. Selec

t one True False
Engineering
2 answers:
Lisa [10]4 years ago
4 0

Answer:

I select false because engineers are not the only thing we have, we have scientists doctors mathematicians and much more to give safety standards

defon4 years ago
4 0

Answer:

False.

Explanation:

Even though engineers are designing their buildings and products to meet safety expectations, things can go wrong, and engineers should never be allowed to have final say. That is the main reason why we have safety inspectors, and are able to live our lives safer today. If there were no second checks on different things, if one were to fail, we would know it for certain to be the engineer, and he does not want to take the blame, (and possibly a lawsuit), that may appear after the accident. If we want maximum safety, we ensure that there are many more people who prepare and examine products closely from time to time.

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A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?
blagie [28]

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

4 0
3 years ago
What is the effect of the workpiece specific cutting energy on the cutting forces, and why?
ella [17]

Explanation:

Specific cutting energy:

   It the ratio of power required to cut the material to metal removal rate of material.If we take the force required to cut the material is F and velocity of cutting tool is V then cutting power will be the product of force and the cutting tool velocity.

Power P = F x V

Lets take the metal removal rate =MRR

Then the specific energy will be

    sp=\dfrac{F\times V}{MRR}

If we consider that metal removal rate and cutting tool velocity is constant then when we increases the cutting force then specific energy will also increase.

8 0
4 years ago
During his military campaign in what is now Germany, Julius Caesar lead his army of 40,000 soldiers to the western bank of the R
CaHeK987 [17]

Answer:

identifying a problem

Explanation:

its right

5 0
3 years ago
A system consists of a disk rotating on a frictionless axle
kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

  1. Li = Lf where;
  2. li is the preliminary second of inertia
  3. If is the very last second of inertia
  4. wi is the preliminary angular velocity
  5. wf is the very last angular velocity
  6. Li is the preliminary angular momentum
  7. Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.

Read more about the frictionless :

brainly.com/question/13539944

#SPJ4

8 0
2 years ago
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26
scZoUnD [109]

Answer:

The strength coefficient is 625 and the strain-hardening exponent is 0.435

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  (K) and the (n).

\sigma =K\epsilon^n

We will plug the values in the formula.

250=K\times (0.12)^n\\350=K\times (0.26)^n

We will solve these equation.

K=\frac{250}{(0.12)^n} plug this value in 350=K\times (0.26)^n

350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\  \\1.4=(2.17)^n

Taking a natural log both sides we get.

ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435

Now, we will find value of K

K=\frac{250}{(0.12)^n}

K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625

So, the strength coefficient is 625 and the strain-hardening exponent is 0.435.

5 0
3 years ago
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