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jonny [76]
3 years ago
5

What are the risks of biohacking? Do you think the risks of biohacking outweigh the advantages? Why or why not?

Engineering
1 answer:
maksim [4K]3 years ago
5 0

Explanation:

Despite its futuristic outlook and results biohacking remains a largely unsupervised, dangerous if not illegal movement. Though the research and results produced by biohackers is ground breaking, it also breaks several ethical codes on human experimentation and endangering live of several humans.

I agree with the fact that risks outweigh advantages, but you can do things like:

- 3D Printing Body Parts

- Finding Cures

- Correcting Deformites.

The only issue is that it's illegal, unfortunately.

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What is the weight of a steel plate in the shape of a circle with a diameter of 10'? The steel weighs 14 Ib per Ft2
S_A_V [24]
Width * Length * Thickness * Density = Weight.
48″ * 96″ * . 1875″ * 0.284 lb/in3 = 245 lb.
3 0
3 years ago
Aerospace engineers who work for certain government agencies are often required to have security clearance. Explain two reasons
yuradex [85]

Answer:

Two reasons that justify the requirement for security clearance for aerospace engineers working for government agencies are;

1) Such engineers have access to data regarding the blueprint, components, method of construction, functionality status, new systems design, future systems design, inventory of systems and aeronautical systems database which are sensitive information that are of high importance to the federal government

2) Such engineers take part in the testing of aeronautic equipment, and will require security clearance to be able to input data results into the data base of the aeronautic equipment

Explanation:

3 0
3 years ago
Scientists learn more about the natural world through investigation. This involves identifying a problem, researching related in
mars1129 [50]

Answer:

This shows that the technological designs are similar to the scientific investigation processes.

Explanation:

Scientists carry out scientific investigations and discoveries. They learn from their environment by investigating the natural world. The scientist investigates by problem identification, research the related information, designing and conceptualizing, investigating and analyzing the results and deriving out conclusions.

Similarly the engineers follow the same step and provide solutions to the problems of the society through their products and technological designs. The engineers first identify the need of the product, design and implement the design to provide solution and then evaluate the solution. These are the similar steps that an engineer follow in a technological design of a product or a solution.

3 0
3 years ago
On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch
rewona [7]

Answer:

ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

Explanation:

Distance of earth from sun = R_{2} = 147.4 \times 106 Km

Spacecraft perihelion = R_{2} = 120\times106Km

gravitational parameters are now given as

\mu_{sun} = 132.7\times 10^{9}

\mu_{earth} = 398600

radius of earth  = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

   V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }

V_{D}^{v} =28.43\frac{km}{s}

Heliocentric velocity of earth = v_{earth} = 30.06\frac{km}{sec}

V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}

assume

r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}

geometric space craft velocity in its circular parking orbit

v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } }  = 7.784 \frac{km}{s}

              ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

7 0
4 years ago
Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d
Tems11 [23]

Complete Question

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the  point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is P= 210 N

and the moment M is M = -48.3N \cdot m

The horizontal force that the tong teeth exerts is F_T =89.67N

Explanation:

First let denote the dimension to corresponding to the diagram

      a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.

Next looking at the diagram let us consider the vertical direction

At equilibrium

               \sum F_{vertical} =0

This mean that

               P+ W = 0

Since they are acting in opposite direction the equation becomes

               P - W = 0

=>           P= W

=>            P= 210 N

At Equilibrium  Moment about F gives

               \sum M_f  = 0

=> F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0

=> M = -W *d

=> M = -210 * 0.230

=> M = -48.3N \cdot m

Here F_T is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

  Now the angle \theta is mathematically given as

             tan \theta = \frac{g+h}{a}

=>        \theta = arctan \frac{g+h}{a}

                = arctan (\frac{10+35}{35} )

               =52.125^o

Now at equilibrium the moment about A is

                \sum M_A = 0

          =>  P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0

                210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)* (0.035 +0.010) =0

    =>   10.29 +F_{BC} (0.02763+0.03552) =0

    =>     F_{BC} =\frac{10.29}{0.06315}

    =>      F_{BC} = - 162.925 N

Looking at the forces acting on the teeth as shown on the third uploaded image

 At Equilibrium the moment about D is

      \sum M_D = 0

=>  \frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0

=>   \frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0

=>    34.081  = F_T(0.345 +0.035)

=>   F_T =89.67N

         

   

         

7 0
4 years ago
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