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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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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

4 0
3 years ago
Ai r is compressed by a 30-kW compressor from P1 to P2. The air t emperature i s maintained constant at 25oC during thi s proces
RUDIKE [14]

Answer:

The rate of entropy change of the air is -0.10067kW/K

Explanation:

We'll assume the following

1. It is a steady-flow process;

2. The changes in the kinetic energy and the potential energy are negligible;

3. Lastly, the air is an ideal gas

Energy balance will be required to calculate heat loss;

mh1 + W = mh2 + Q where W = Q.

Also note that the rate of entropy change of the air is calculated by calculating the rate of heat transfer and temperature of the air, as follows;

Rate of Entropy Change = -Q/T

Where Q = 30Kw

T = Temperature of air = 25°C = 298K

Rate = -30/298

Rate = -0.100671140939597 KW/K

Rate = -0.10067kW/K

Hence, the rate of entropy change of the air is -0.10067kW/K

3 0
3 years ago
WHICH COMPONENT IN A BLOWER MOTOR
bagirrra123 [75]
I think it would be the resistor?
5 0
3 years ago
Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at 573 oC (this temperature is below the m
inn [45]

Answer:

2.9\times 10^{-6}

Explanation:

Q_s = Energy for defect formation = 1.86 eV

T = Temperature = 573^{\circ}\text{C}=573+273.15=846.15\ \text{K}

k = Boltzmann constant = 8.62\times 10^{-5}\ \text{eV/K}

The fraction of lattice sites that are Schottky defects is given by

\dfrac{N_s}{N}=e^{-\dfrac{Q_s}{2kt}}\\\Rightarrow \dfrac{N_s}{N}=e^{-\dfrac{1.86}{2\times 8.62\times 10^{-5}\times 846.15}}\\\Rightarrow \dfrac{N_s}{N}=2.9\times 10^{-6}

The required ratio is 2.9\times 10^{-6}.

6 0
2 years ago
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Crank

Answer:

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7 0
3 years ago
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