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3 years ago

On a CNC lathe, which of the following axes specifies the distance of the cutting tool from the workpiece centerline?

Engineering

1 answer:

kvasek [131]3 years ago

8 0

Answer:

The axis of motion that is parallel to the spindle axis is always the Z-axis.

Explanation:

Z-Axis Which axis is which depends on the orientation of the spindle.

If the spindle is vertical (Figure 2.1), the Z-axis is vertical. Either the quill or the knee of a vertical spindle mill will move when a Z-axis command is executed.

This is the best answer I could give you, maybe you could show us a pic of the question so we see all the possible choices?

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we wish to send at a rate of 10Mbits/s over a passband channel. Assuming that an excess bandwidth of 50% is used, how much bandw

gayaneshka [121]

Answer:

QPSK: 7.5 MHz

64-QAM:2.5 MHz

64-Walsh-Hadamard: 160 MHz

Explanation:

See attached picture.

6 0

3 years ago

What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p

Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is = $55\times 25$

=1375psf

1 ft = 12 inch

pressure at bottom = $\frac{1375}{12^2}$

= 9.55 psi

so, answer will be option (d) which is 9.55 psi

3 0

3 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

This problem has been solved!

lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0

3 years ago