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Greeley [361]
2 years ago
15

Which of the following would be pictured on a fan motors ladder logic diagram

Engineering
1 answer:
Ipatiy [6.2K]2 years ago
4 0

Answer:B. two Push-button control

Explanation:

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A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
Compute the fundamental natural frequency of the transverse vibration of a uniform beam of rectanqular cross section, with one e
marshall27 [118]

Answer:

The natural angular frequency of the rod is 53.56 rad/sec

Explanation:

Since the beam is free at one end and fixed at the other hence the beam is a cantilevered beam as shown in the attached figure

We know that when a unit force is placed at the end of a cantilever the displacement of the free end is given by

\Delta x=\frac{PL^3}{3EI}

Hence we can write

P=\frac{3EI\cdot \Delta x}{L^3}

Comparing with the standard spring equation F=kx we find the cantilever analogous to spring with k=\frac{3EI}{L^3}

Now the angular frequency of a spring is given by

\omega =\sqrt{\frac{k}{m}}

where

'm' is the mass of the load

Thus applying values we get

\omega _{beam}=\sqrt{\frac{\frac{3EI}{L^{3}}}{Area\times density}}

\omega _{beam}=\sqrt{\frac{\frac{3\times 20.5\times 10^{10}\times \frac{0.1\times 0.3^3}{12}}{5.9^{3}}}{0.3\times 0.1 \times 7830}}=53.56rad/sec

8 0
3 years ago
Read 2 more answers
A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 day
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D Hey will you please help me with my essay and I’ll get back to yours please ASAP
3 0
3 years ago
Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact di
Ilya [14]

Answer:

Problem 1 (10 points) In the first homework you were instructed to design the mechanical components of an oscillating compact disc reader. Since you did such a good job in your design, the company decided to work with you in their latest Blue-ray readers, as well. However, this time the task is that once the user hits eject button, the motor that spins the disc slows down from 2000 rpm to 300 rpm and at 300 rpm a passive torsional spring-damper mechanism engages to decelerate and stop the disc. Here, your task is to design this spring-damper system such that the disc comes to rest without any oscillations. The rotational inertia of the disc (J) is 2.5 x 10-5kg m² and the torsional spring constant (k) is 5 × 10¬³NM. Calculate the critical damping coefficient cc for the system. choice of the damper, bear in mind that a good engineer stays at least a factor of In your 2 away from the danger zone (i.e., oscillations in this case). Use the Runge Kutta method to simulate the time dependent angular position of the disc, using the value of damping coefficient (c) that calculated. you Figure 1: Blue-ray disc and torsional spring-damper system.

5 0
3 years ago
A unidirectional E-Glass fiber-epoxy composite material contains 61% by volume E-Glass fibers stressed under isostrain condition
zalisa [80]

Answer:

The total load carried by the fiber will be "98%".

Explanation:

The given values are:

V_{f}=0.61

V_{m}=1-V_{f}

     =1-0.61

     =0.39

E_{f}=10 \ Mpa

\sigma_{f}=0.35 \ Mpa

E_{m}=0.45 \ Mpa , \sigma_{m}=9\times 10^{-3} \ Mpa

As we know,

⇒  E_{e}=fE_{f}+mE_{m}

On putting the estimated values, we get

⇒       =0.61\times 10+0.39\times 0.95

⇒       =6.27 \ Mpa

Now,

⇒  \sigma_{c}=f\sigma_{f}+m\sigma_{m}

On putting the estimated values, we get

⇒       =0.61\times 0.35+0.39\times 0.009

⇒       =0.217 \ Mpa

Therefore,

The load carried by fiber,

=\frac{f\sigma_{f}}{\sigma_{c}}

=\frac{0.35\times 0.61}{0.217}

=0.98 i.e., 98%

4 0
4 years ago
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