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zepelin [54]
3 years ago
10

The Emergency Stop Button icon on the Inputs toolbar can be used to press or release the Emergency Stop button on the CNC machin

e.
Select one:
True
False
Engineering
1 answer:
Tanya [424]3 years ago
4 0

The Emergency Stop Button icon on the Inputs toolbar can be used to press or release the Emergency Stop button on the CNC machine - False

<h3><u>Explanation:</u></h3>

The Emergency Stop button is practiced to stop machine operation. When actuated, machine operation terminates instantly. The execution of the part program can be hindered by touching the Control and Spacebar buttons on the computer keyboard. Unlike applying the Emergency Stop button, this way of stopping the milling center does not create the software to drop the path of the tool position.

The Inputs Toolbar is idle. It contributes information only on the state of the Emergency Stop, the safety shield, and the limit switches.  An input is active (on) when the button is depressed.

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The components of an electronic system dissipating 90 W are located in a 1-m-long circular horizontal duct of 15-cm diameter. Th
Artyom0805 [142]

Answer:

Given data

 electronic system dissipating = 90 W

diameter = 15 cm

The components in the duct are cooled by forced air which enters at 32°C at a rate of 0.65 m3 /min

the duct and the remaining = 15 %

See pictures for solution.

Explanation:

See attached pictures for detailed explanation.

4 0
3 years ago
An AC circuit has a resistor, capacitor and inductor in series with a 120 V, 60 Hz voltage source. The resistance of the resisto
aliina [53]

Answer:

(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

We have given voltage V =120 volt

Frequency f=60 Hz

Resistance R =3 ohm

Inductance L =0.01 H

Capacitance C =0.00075 farad

(i) reactance of of inductor X_L=\omega L=2\pi fL=2\times 3.14\times 60\times 0.01=3.768ohm

X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 60\times 0.00075}=3.5385ohm

(ii) Total impedance Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{3^2+(3.768-3.5385)^2}=3.008ohm

Current i=\frac{V}{Z}=\frac{120}{3.008}=39.89A

(viii) power factor cos\Phi =\frac{R}{Z}=\frac{3}{3.008}=0.9973

(VII) cos\Phi =0.9973

\Phi =4.1796^{\circ}

So power factor angle is 4.1796°

(iii) Apparent power P=VICOS\Phi =120\times 39.89\times 0.9973=4773.875W

(vi) Reactive power Q=VISIN\Phi =120\times 39.89\times SIN4.17^{\circ}=348var

5 0
3 years ago
A short-circuit experiment is conducted on the high-voltage side of a 500 kVA, 2500 V/250 V, single-phase transformer in its nom
Anarel [89]

Given Information:

Primary secondary voltage ratio = 2500/250 V

Short circuit voltage = Vsc = 100 V

Short circuit current = Isc = 110 A

Short circuit power = Psc = 3200 W

Required Information:

Series impedance = Zeq = ?

Answer:

Series impedance = 0.00264 + j0.00869 Ω

Step-by-step explanation:

Short Circuit Test:

A short circuit is performed on a transformer to find out the series parameters (Z = Req and jXeq) which in turn are used to find out the copper losses of the transformer.

The series impedance in polar form is given by

Zeq = Vsc/Isc < θ

Where θ is given by

θ = cos⁻¹(Psc/Vsc*Isc)

θ = cos⁻¹(3200/100*110)

θ = 73.08°

Therefore, series impedance in polar form is

Zeq = 100/110 < 73.08°

Zeq = 0.909 < 73.08° Ω

or in rectangular form

Zeq = 0.264 + j0.869 Ω

Where Req is the real part of Zeq  and Xeq is the imaginary part of Zeq

Req = 0.264 Ω

Xeq = j0.869 Ω

To refer the impedance of transformer to its low voltage side first find the turn ratio of the transformer.

Turn ratio = a = Vp/Vs = 2500/250 =  10

Zeq2 = Zeq/a²

Zeq2 = (0.264 + j0.869)/10²

Zeq2 = (0.264 + j0.869)/100

Zeq2 = 0.00264 + j0.00869 Ω

Therefore, Zeq2 = 0.00264 + j0.00869 Ω is the series impedance of the transformer referred to its low voltage side.

4 0
3 years ago
Tyuuyiopopiouyttrrtrffrlkl,k;;';'l.l
Viktor [21]

Answer:

Explanation:

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5 0
3 years ago
Who would most likIf you are interested in working for a specific company, what type of job site should you look at for opening?
fredd [130]

Answer:

if you are interested in working for a company then you should opt for c). Company Site

Explanation:

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3 years ago
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