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3 years ago

How has drafting evolved in the 21st century

1 answer:

8 0

Drafting has been around a long time. We can safely assume that since we’ve had a tool in our hands, we’ve been describing plans and technical representations and doodling ideas. Let’s take a closer aspect at drafting and its advance from an under-the-radar part of the method to a very developed skill set.

Explanation

• 1970s – The beginning computer-aided design systems were included in the industry. Following the design engineers tried the learning curve of using CAD, their performance and productivity went through the roof. Over time, CAD software became affordable and more user-friendly, and its fame grew.

• 1990s – CAD software was expanded further to include 3-D characteristics, and quickly the technical designs of the past enhanced increasingly simulated and accessible to engineer.

• Present – The development of drafting has brought us to the present day, were using 3-D representations is the standard and the aim to generate full virtual prototypes.

You might be interested in

How much does it cost to replace a roof on a 2,200 square foot house.

Norma-Jean [14]

Answer:

An estimated amount of $8,200 to $12,000

Explanation:

~~It all really depends on what materials you have purchased for the replacement.

~~If you're to choose lower-quality materials the cost would be less but high-quality ones will be more expensive.

(The answer is if you're most likely to purchase asphalt shingles)

Hope this Helps!

7 0

2 years ago

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and a

anyanavicka [17]

Answer:

branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
branch 2: i = 21.466∠63.435°; pf = 0.447 leading
total: i = 31.693∠10.392° leading; pf = 0.984 leading

Explanation:

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

$X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915\times10^{-3}\approx j4.99984\,\Omega$

The capacitive reactance is ...

$X_C=\dfrac{1}{j\omega C}=\dfrac{-j}{100\pi\cdot 318.31\times10^{-6}F}\approx -j10.00000\,\Omega$

Branch 1

The impedance of branch 1 is ...

Z1 = 8 +j4.99984 Ω

so the current is ...

I1 = V/Z = 240/(8 +j4.99984) ≈ 25.440∠-32.005°

The power factor is cos(-32.005°) ≈ 0.848 (lagging)

Branch 2

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

I2 = 240/(5 +j10) ≈ 21.466∠63.435°

The power factor is cos(63.436°) ≈ 0.447 (leading)

Total current

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

It ≈ 31.173 +j5.717 ≈ 31.693∠10.392°

The power factor for the circuit is cos(10.392°) ≈ 0.984 (leading)

The phasor diagram of the currents is attached.

_____

Additional comment

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

3 0

2 years ago

At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide

Xelga [282]

Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0 (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If 2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0 (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

⇒ vB = 70 mm/s

d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

4 0

3 years ago

Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t

raketka [301]

Answer:

a) 53 MPa, 14.87 degree

b) 60.5 MPa

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0

3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago