List 3 specific things you can do with a screw gun (not just screw in screws)

Answers

Answer 1

Answer:

1: Attaching a fork and wip eggs/batter/ ect.

2: You can entertain a cat by putting yarn on the gun/drill

3: If you get yarn everywere, you can spin the yarn up on the drill/gun

Explanation:

I hope this helps you out somehow


Related Questions

the fixed scale specifies 10.5 millimeters and the thimble specifies 30 hundredths of a mm. what is the total value shown by the micrometer when you add these readings together

Answers

Answer:

10.80 mm

Explanation:

Given that the reading of the fixed scale is : 10.5 mm

The thimble scale specifies : 0.30 mm

The micrometer reading will be : 10.5 + 0.30 =10.80 mm

A business owned by one people is a

Answers

Answer:

A sole proprietorship, also known as the sole trader, individual entrepreneurship or proprietorship, is a type of enterprise that is owned and run by one person and in which there is no legal distinction between the owner and the business entity.

Explanation:

Answer:

A business owned by one people is a-Sole proprietorship

Explanation:

This is correct i did it for my electives class which was career reaserch and decision making.

brainliest?

A plane stress state (2D) of material is defined by: [σ] = [800300300−400] (MPa). Define the principal stresses σ1, σ2 using:

• Determinant Method
• Mohr’s Circle (determine also ????max)

Answers

Answer:

Explanation:

Given that:

[tex][\sigma] = \left[\begin{array}{cc}800&300\\ \\300&-400\end{array}\right] (MPa)[/tex]

[tex][\sigma] = \left[\begin{array}{cc}\sigma_{xx}&\sigma_{xy}\\ \\\sigma_{yx}&\sigma_{yy}\end{array}\right][/tex]

Using Determinant method

The principal stress is the maximum or minimum normal stress acting on any plane. For the 2D stress system, the 2-principle plane always carries zero shear stress.

For principal stress [tex]( \sigma_1, \sigma_2)[/tex]

[tex]\sigma_{1,2} = \dfrac{\sigma_x+ \sigma_y}{2} \pm \sqrt{( \dfrac{\sigma_x - \sigma_y}{2} )^2 + \sigma^2_{xy}}[/tex]

[tex]\sigma_{1} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]

[tex]\sigma_{1} = \dfrac{400}{2} \pm \sqrt{ (600)^2 + (300)^2}[/tex]

[tex]\sigma_{1} = 200+ 670.82 \\ \\ \sigma_{1} = 870.82 \ MPa[/tex]

[tex]\sigma_{2} = \dfrac{800+(-400)}{2} \pm \sqrt{( \dfrac{800 -(-400)}{2} )^2 + (300)^2}[/tex]

[tex]\sigma_{2} = 200- 670.82 \\ \\ \sigma_{1} = - 470.82 \ MPa[/tex]

According to Mohr's circle;

Mohr's circle is the locus provided that the position of the normal stress and the shear stress is acting on any plane.

Center = (a,0)

[tex]a = \dfrac{\sigma_{x}+\sigma_{y}}{2}[/tex]

[tex]a = \dfrac{800+(-400)}{2}[/tex]

a = 200 MPa

radius (r) = [tex]\sqrt{ (\dfrac{\sigma_{x}-\sigma_{y}}{2})^2 + \sigma^2 _{xy}}[/tex]

[tex]=\sqrt{ (\dfrac{800-(-400)}{2})^2 + (300)^2}[/tex]

[tex]=\sqrt{ (600)^2 + (300)^2}[/tex]

r = 670.82 MPa

[tex]\sigma_1 = a +r \\ \\ \sigma_1 = 200 + 670.82 \\ \\ \sigma_1 = 870.82 \ MPa[/tex]

[tex]\sigma_2 =-(r-a)[/tex]      (it is negative because of the negative x-axis)

[tex]\sigma_2 =670.82 - 200 \\ \\ \sigma_1 = 470.82 \ MPa[/tex]    

[tex]\tau_{max} = radius \ of \ Mohr's \ circle[/tex]

[tex]\tau_{max} = 670.82 \ MPa[/tex]

Alex loves to build things and wants to study robotics. He needs to get a job over the summer but the only one he can find is at a fast food restaurant. Does taking the job mean that Alex cannot pursue a career in robotics? Explain your answer.

Answers

Answer:

He can take the job at the fast food restaurant to make money in order to pursue his career in robotics it doesnt mean he cant pursue it just he needs to make money for it first.

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