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2 years ago

بی با positions and engin drive

Engineering

1 answer:

djverab [1.8K]2 years ago

3 0

Answer:

Un motor de paso es un motor eléctrico CC sin escobillas que divide una rotación completa en varios pasos iguales. Rota una distancia incremental específica por cada paso. El número de pasos que se ejecutan controla el grado de rotación del eje del motor.

Los motores de paso tienen cierta capacidad inherente para controlar la posición, ya que tienen pasos de salida integrados. Pueden controlar con gran precisión cuán lejos y cuán rápido rotará el motor de paso. El número de pasos que ejecuta el motor es igual al número de comandos de pulsos del controlador. Un motor de paso rotará una distancia y a una velocidad proporcional al número de la frecuencia de sus comandos de pulso.

Explanation:

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Una empresa realizó en el ejercicio de compras al contado por valor

Tanzania [10]

Answer:

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Explanation:

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8 0

3 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0

2 years ago

A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.

Setler79 [48]

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered 0.5742 Kg.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

6 0

3 years ago

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy + flow work

$E = \dot m ( \frac{v^2}{2] + zg + p\nu)$

$E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})$

$57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})$

solving for flow rate

$\dot m = 99.977we know that [tex]\dot m = \rho AV$

$\dot m = 780 \frac{\pi}{4} D^2\times 16$

solving for d

$99.97 = 780 \times \frac{\pi}{4} D^2\times 16$

d = 0.090 m

so radius = 0.045 m

3 0

3 years ago

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that

iVinArrow [24]

Answer:

Explanation:

Complete question:

Fill in the blanks

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that time ,___1______the agency ends. In addition, the parties can specify that the agency is for a particular____2______ . Once that is achieved, the agency ends. Alternatively, the parties can include a specific event as a trigger for termination; once that event,_____3______ the agency ends. The parties can terminate an agency relationship prior to any of the preceding events by ______4_________agreement, or revocation_____5______ by individual party.

Answer

1) lapses

(2) purpose

(3) occurs / begins

(4) mutual

(5) either

8 0

3 years ago

بی با positions and engin drive​

بی با positions and engin drive