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Evgen [1.6K]
3 years ago
7

Technician A says you should place the air ratchet setting to

Engineering
1 answer:
Bas_tet [7]3 years ago
6 0
I think it’s b sry if I’m wrong tho
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Please help<br> describe the impact that a toy robot has had or could have on its intended audience
GuDViN [60]
Depending on the age the toy is made for it could teach younger children things such as letters and numbers and for a older kid it could teach them how different things are put in the robot to help it work
3 0
3 years ago
Read 2 more answers
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri
ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Here

  • \sigma_x is the normal stress which is 1750 psf
  • \sigma_y is 0
  • \tau_{xy} is the shear stress which is 800 psf

So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

5 0
3 years ago
Which of the following shot pellets has the smallest diameter?
vitfil [10]
I believe i got this correct when i had an assignment with this question. It should be 7 1/2. B!
5 0
3 years ago
An engineer who invested very well plans to retire now because she has $2,000,000 in her ORP account. How long will she be able
Ede4ka [16]

Answer:

20 years

Explanation:

You do 2,000,000 ÷ 100,000 but you can simplify that to 20 ÷ 1 = 20

During those 20 years, the profits you earn will be 80,000 since when you do 2,000,000 x 0.04 but you can simplify that to 20,000 x 4 getting 80,000 and that quite doesn't reach 100,000 dollars

8 0
3 years ago
A steel cylindrical sample was subjected to a tensile test. The yield load was 2100N. The maximum load was 3400N and the failure
Misha Larkins [42]

Answer:

initial diameter of the sample is 2.95 mm

Explanation:

given data

yield load = 2100 N

maximum load = 3400 N

failure load = 2350 N

ultimate engineering stress = 497.4 MPa = 497 × 10^{6} N/m²

to find out

What was the initial diameter of the sample in mm

solution

we will apply here ultimate engineering stress formula that is express as

ultimate engineering stress = \frac{Pmax}{A}    ...............1

here A is area and P max is maximum load applied

so area = \frac{\pi }{4} d^2

here d is initial diameter

so put all value in equation 1

497 × 10^{6}  = \frac{3400}{\frac{\pi }{4} d^2}

solve it we get d

d = 2.95 × 10^{-3} m

so initial diameter of the sample is 2.95 mm

7 0
4 years ago
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