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blsea [12.9K]
3 years ago
15

Which type of engineering design uses an already existing design

Engineering
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

Reverse engineering

Explanation:

Reverse Engineering is the remaking of already made products following the deconstruction and examination of the product to make known the product design, code and architecture features, gain knowledge of the composition and construction in a scientific research approach

Reverse engineering is also known as back engineering and consists of three main stages

1) Recovery implementation

2) Design recovery

3) Recovery analysis.

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Select the creative imaging fields that require knowledge of programming.
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Video game designer for sure
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3 years ago
There are two methods to create simple robots. First, you can construct them by purchasing various individual components and ass
Vlad1618 [11]

Answer:

second way

Explanation:

robotic kits are simple and easy to make

5 0
3 years ago
A force of 50N acts on a roller of mass 15 kg initially at rest on a frictionless surface. The roller travels 12 m while the for
kenny6666 [7]

Answer: 600 J

Explanation:

By definition, the work is the process due to which an applied force causes a mass to move, and can be calculated as the product of the component of the force applied along the direction of the movement, times the distance traveled.

In this case, is W= F*d= 50 N*12 m= 600 J

It can be showed, that the work done on the mass (in absence of friction ) is equal to the change in the kinetic  energy of the mass.

If the roller is initially at rest, this means the following:

ΔK= 1/2 m * vf²

Now, applying Newton's 2nd Law, F=m*a ⇒ a= F/m= 3.33 m/s²

If vo=0, we can write the following:

vf²=2*a*x= 2*3.33 m/s²*12m=79.98 m²/s²

Replacing in ΔK, we obtain

ΔK= 1/2*15 kg* 79.98 m²/s²≈ 600 J

6 0
3 years ago
Miriam is doing a measurement with her multimeter and the LCD is showing Hz. What's she measuring?
-BARSIC- [3]

Answer:

voltage

Explanation:

because it is used with a moving pointer to display readings.

7 0
3 years ago
A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, deter
uysha [10]

Answer:

4.83m/s^{2}

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by a_{radial}=w^{2}r

Applying values we get  a_{radial}=(2t)^{2}X0.3m

Thus a_{radial}=1.2t^{2}

At time = 2seconds a_{radial}= 4.8m/s^{2}

The tangential acceleration is given by a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}

a_{tangential}=\frac{d(2tr)}{dt}

a_{tangential}= 2r

a_{tangential}=0.6m/s^{2}

Thus the resultant acceleration is given by

a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}

a_{res} =\sqrt{4.8^{2}+0.6^{2}  } =4.83m/s^{2}

8 0
4 years ago
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