All categories

3 years ago

What is the best free graphic design program for windows?

Engineering

2 answers:

Lynna [10]3 years ago

8 0

ANSWER:
Gravit Designer

Is available on Windows macOS Linux and ChromeOS

kumpel [21]3 years ago

3 0

Gravit designer hope this helps :)))))))))

You might be interested in

The collapse of the magnetic field inside the ignition coil happens as a

Tanzania [10]

Primary coil

See an example below

3 0

3 years ago

The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel. Determine the magnitude of forc

Sonbull [250]

Answer:

Magnitude of force P = 25715.1517 N

Explanation:

Given - The wires each have a diameter of 12 mm, length of 0.6 m, and are made from 304 stainless steel.

To find - Determine the magnitude of force P so that the rigid beam tilts 0.015∘.

Proof -

Given that,

Diameter = 12 mm = 0.012 m

Length = 0.6 m

$\theta$ = 0.015°

Youngs modulus of elasticity of 34 stainless steel is 193 GPa

Now,

By applying the conditions of equilibrium, we have

∑fₓ = 0, ∑ $f_{y}$ = 0, ∑M = 0

If ∑ $M_{A}$ = 0

⇒ $F_{BC}$ ×0.9 - P × 0.6 = 0

⇒ $F_{BC}$ ×3 - P × 2 = 0

⇒ $F_{BC}$ = $\frac{2P}{3}$

If ∑ $M_{B}$ = 0

⇒ $F_{AD}$ ×0.9 = P × 0.3

⇒ $F_{AD}$ ×3 = P

⇒ $F_{AD}$ = $\frac{P}{3}$

Now,

Area, A = $\frac{\pi }{4} X (0.012)^{2}$ = 1.3097 × 10⁻⁴ m²

We know that,

Change in Length , $\delta$ = $\frac{P l}{A E}$

Now,

$\delta_{AD} = \frac{P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 9.1626 × 10⁻⁹ P

$\delta_{BC} = \frac{2P(0.6)}{3(1.3097)(10^{-4}) (193)(10^{9} }$ = 1.83253 × 10⁻⁸ P

Given that,

$\theta$ = 0.015°

⇒ $\theta$ = 2.618 × 10⁻⁴ rad

So,

$\theta = \frac{\delta_{BC} - \delta_{AD}}{0.9}$

⇒2.618 × 10⁻⁴ = ( 1.83253 × 10⁻⁸ P - 9.1626 × 10⁻⁹ P) / 0.9

⇒P = 25715.1517 N

∴ we get

Magnitude of force P = 25715.1517 N

6 0

3 years ago

Natural Convection Cooling of an Orange. An orange 102 mm in diameter having a surface temperature of 21.1°C is placed on an ope

natali 33 [55]

Answer:

q = 3.181 w

Explanation:

$T_w = 21.1$ degree celcius

$T_b = 4.4$ degree celcius

D = 0.102 m

Radius = 0.051 m

$\Delta T = T_w - T_b= 21.4 - 4.4 = 16.7$ degree celcius

$L^3 \Delta t = 0.051^3\times 16.7 = 2.22\times 10^{-3}$

$h = 1.37 (\frac{\Delta}{L})^{1/4}$

$= 1.37\times (\frac{16.7}{0.051})^{1/4}$

$h = 5.828 w/m^2 K$

$A =4\pi r^2$

$A = 4\times \pi 0.05^2 = 0.03268 m^2$

$q = hA\Delta t$

h =5.828 (0.03268) 16.7

q = 3.181 w

7 0

4 years ago

What is the deflection of a W12x30 16 ft long with a distributed load of 2k/ft? Include the weight of the beam. a) 0.027in b) 0.

cestrela7 [59]

Answer:

correct answer is option C)

Explanation:

W 12 x 30 section of the beam dimension

Value of E = 29000 ks i/in²

I = 238 in⁴

weight of beams per length = 30 lbs/ft = 30 × 10⁻³ ks i/ft = 2.5 × 10⁻³ ks i/in.

distribution load = 2 k/ft = 1/6 k/in

L = 16 × 12 = 192 inch.

deflection due to distributed load = $\dfrac{5}{384}\dfrac{wl^4}{EI}$

= $\dfrac{5}{384}\times \dfrac{192^4}{6\times 29000\times 238}$

= 0.427 in.

deflection due to self weight = $\dfrac{wl^4}{48EI}$

= $\dfrac{2.5 \times 10^{-3}\times 192^4}{48\times 29000\times 238}$

= 0.01025 inch

total deflection = 0.427 in.+ 0.01025 inch

total deflection = 0.4375 in

correct answer is option C)

6 0

3 years ago

4 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is

ankoles [38]

Answer:

Following are the responses to the given question:

Explanation:

Effective red duration is applied each cycle r=30 second D/D/1 queuing

In total, its approach delay is 83.33 sec vehicle per cycle

Flow rate(s) of saturated = 1,000 vehicles each hour

Total vehicle delay per cycle $= \frac{v \times 30^2}{2(1-\frac{v}{0.2778})}$

$\to \frac{v\times 30^2}{2(1-\frac{v}{0.2778})}= 83.33\\\\\to 900v=166.66-599.928v\\\\\to v=0.111 \frac{veh}{sec}\\\\$

The flow rate for such total approach is 0.111 per second.

The overall flow velocity of the approach is 400 cars per hour

The approach capacity refers to the number of arrivals per cycle.

Environmentally friendly time ratio to cycle length:

$,\frac{g}{C} \ is = \frac{400}{1000}=0.4\\\\r= c-g\\\\30\ sec =C - 0.4 C\\\\C=50 \ sec$

6 0

3 years ago