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dimulka [17.4K]
3 years ago
11

What is a solution-based process and why it is important in engineering?

Physics
1 answer:
Murrr4er [49]3 years ago
3 0

The engineering design process is a series of steps that engineers follow when they are trying to solve a problem and design a solution for something; it is a methodical approach to problem solving. This is similar to the “Scientific Method” which is taught to young scientists.


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No they say "Watch out it's the fuzz"

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Why does a crumpled paper land before a flat sheet of paper?
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The crumpled paper hits the ground first due to air resistance. The crumpled paper has less surface area than the flat paper, therefore the crumpled paper hits first.

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Read the excerpt from The Building of Manhattan.
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3 years ago
What is the x-component of a vector with a magnitude of 115 km at an angle of 22°?
ruslelena [56]

The x-component of a vector are < 106.6, 43.07 >

Depending on the angle we are provided, the x-component of a vector can either be cos or sin. Cos always corresponds to the right triangle's side that contacts the specified angle.

If a vector v with magnitude ||v|| makes an angle θ with the positive x-axis then,

v = ||v|| cos θi + ||v|| sin θj

 =  < ||v|| cos θ , ||v|| sin θ >

Magnitude p = 115 km

Angle = 22°

p = ||p|| < cos θ, sin θ >

p = 115 < cos 22°, sin 22° >

p = 115 < 0.927, 0.3746 >

p = < 106.6, 43.07 >

Therefore,  the x-component of a vector are < 106.6, 43.07 >

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4 0
2 years ago
Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direc
Ira Lisetskai [31]

Answer:

0 < r < r_exterior     B_total = \frac{\mu_o I}{2\pi  r}

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = \frac{\mu_o I}{2\pi  r}

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = \frac{\mu_o I}{2\pi  r}  

r > r_exterior

          B_total = B₁ -B₂

          B_total = 0

6 0
3 years ago
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