1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Musya8 [376]
4 years ago
14

Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach

ining tool.
Engineering
1 answer:
STatiana [176]4 years ago
7 0

Answer:

Explained

Explanation:

Cutting tools:

 1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

You might be interested in
I need unseen passage
monitta

Answer:

this unseen passage__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

8 0
3 years ago
Read 2 more answers
Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid,
77julia77 [94]

Answer:

Explanation:

The image attached to the question is shown in the first diagram below.

From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.

IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :

From the diagram; we can determine the length of BC by using pyhtagoras theorem;

SO;

L_{BC}^2 =  L_{AB}^2 + L_{AC}^2

L_{BC}^2 = (3.5+2.5)^2+ 4^2

L_{BC}= \sqrt{(6)^2+ 4^2}

L_{BC}= \sqrt{36+ 16}

L_{BC}= \sqrt{52}

L_{BC}= 7.2111 \ m

The cross -sectional of the cable is calculated by the formula :

A = \dfrac{\pi}{4}d^2

where d = 4mm

A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2

A = 1.26 × 10⁻⁵ m²

However, looking at the maximum deflection  in length \delta ; we can calculate for the force F_{BC by using the formula:

\delta = \dfrac{F_{BC}L_{BC}}{AE}

F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}

where ;

E = modulus elasticity

L_{BC} = length of the cable

Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for L_{BC} and 0.006 m for \delta ; we have:

F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}

F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN     ---- (1)

Similarly; we can determine the force F_{BC} using the allowable  maximum stress; we have the following relation,

\sigma = \dfrac{F_{BC}}{A}

{F_{BC}}= {A}*\sigma

where;

\sigma = maximum allowable stress

Replacing 190 × 10⁶ Pa for \sigma ; we have :

{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \  kN     ------ (2)

Comparing (1) and  (2)

The magnitude of the force F_{BC} = 2.09676 \ kN since the elongation of the cable should not exceed 6mm

Finally applying the moment equilibrium condition about point A

\sum M_A = 0

3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0

3.5 P - 3.328 F_{BC} = 0

3.5 P = 3.328 F_{BC}

3.5 P = 3.328 *2.09676 \  kN

P =\dfrac{ 3.328 *2.09676 \  kN}{3.5 }

P = 1.9937 kN

Hence; the maximum load P that can be applied is 1.9937 kN

4 0
3 years ago
Motion is defined as a change in an object's position when compared to other objects around it. Mary Ann was watching a slug cra
lara [203]

Answer:

Explanation:

Do it yourself bum

4 0
4 years ago
What is the angle of the input
olga55 [171]
Absolute positions — latitudes and longitudes
Relative positions — azimuths, bearings, and elevation angles
Spherical distances between point locations
3 0
3 years ago
explain the four functional blocks on an oscilloscope and describe the major controls within each block
steposvetlana [31]

Answer:

The cathode ray oscilloscope (CRO) consists of a set of blocks. Those are vertical amplifier, delay line, trip circuit, time base generator, horizontal amplifier, cathode ray tube (CRT) and power supply. The CRO block diagram is shown in attached figure.

The function of each CRO block is mentioned below,

Vertical amplifier amplifies the input signal, which will be displayed on the CRT screen.

Delay line provides a certain amount of delay to the signal, which is obtained at the output of the vertical amplifier. This delayed signal is then applied to the CRT vertical deflection plates.

Trigger circuit produces a trigger signal to synchronize the horizontal and vertical deviations of the electron beam.

Time base generator produces a sawtooth signal, which is useful for horizontal deviation of the electron beam.

Horizontal amplifier amplifies the sawtooth signal and then connects it to the CRT horizontal deflection plates.

Power supply produces high and low voltages. The high negative voltage and the low positive voltage apply to CRT and other circuits respectively.

Cathode ray tube (CRT)

it is the main important block of CRO and consists mainly of four parts. Those are electronic guns, vertical deflection plates, horizontal deflection plates and fluorescent display.

The electron beam, which is produced by an electron gun, is deflected both vertically and horizontally by a pair of vertical deflection plates and a pair of horizontal deflection plates, respectively. Finally, the deflected beam will appear as a point on the fluorescent screen.

In this way, CRO will display the input signal applied on the CRT screen. So, we can analyze the signals in the time domain using CRO.

Explanation:

The oscilloscopes which is widely used for analysis purpose of circuits is divided into four main groups: the horizontal and vertical controls, the input controls and the activation controls.

Found in the front panel section marked Horizontal, the oscilloscope's horizontal controls allow users to adjust the horizontal scale of the screen. This section includes the control of the horizontal delay (displacement), as well as the control that indicates the time per division on the x-axis. The first control allows users to scan through a time range, while the latter allows users to approach a particular time range by decreasing the time per division.

Meanwhile, the oscilloscope's vertical controls are usually found in a section specifically marked as Vertical. The controls found in this section allow users to adjust the vertical appearance of the screen and include the control that indicates the number of volts per division on the axis and the grid of the screen. Also in this section is the control of the vertical displacement of the waveform, which translates the waveform up or down on the screen.

Signal activation helps provide a usable and stable display and allows users to synchronize the oscilloscope acquisition in the waveform of interest. The oscilloscope trigger controls allow users to choose the vertical trigger level, as well as the desired trigger capability. Common types of activation include fault activation, edge activation and pulse width activation.

Useful for identifying random errors or failures, the activation of faults allows users to fire at a pulse or event whose width is less than or greater than a specific period of time. This activation mode allows users to capture errors or technical problems that do not occur very frequently, which makes them very difficult to see.

The most famous trigger mode, edge tripping occurs when the voltage exceeds a set threshold value. This mode allows users to choose between shooting on a falling or rising edge.

Although pulse width activation is comparable to fault activation when users search for pulse width, it is, however, more general since it allows users to fire pulses of specified width. Users can also select the polarity of the pulses to be activated and set the horizontal position of the trigger. This allows users to see what happened during pre-shot or post-shot.

The input panels of an oscilloscope usually include two or four analog channels. They are usually numbered and have a button associated with each channel that allows users to activate and deactivate them. This section may also include a selection that allows users to specify the DC or AC coupling. Selecting the DC coupling implies that the entire signal will be input. The AC pairing, on the other hand, blocks the DC component and focuses the waveform around zero volts. Operators can also identify the probe impedance of the channels through a selection button. In adding, the input panels permit users to select the type of sampling to be used.

5 0
3 years ago
Other questions:
  • 3. An isolated system has an initial temperature of 30oC. It is then placed on top of a Bunsen burner whose temperature is 70oC
    10·1 answer
  • How does a vacuum carpet cleaner work?
    10·2 answers
  • Show that a chirped Gaussian pulse is compressed initially inside a single-mode fiber when ftC < 0. Derive expressions for th
    12·1 answer
  • To increase fault-tolerance, the security administrator for Corp has installed an active/passive firewall cluster where the seco
    11·1 answer
  • A level loop began and closed on BM_A (elevation = 823.368 ft). The plus and minus sights were kept approximately equal. Reading
    11·1 answer
  • A steam turbine has an inlet of 3 kg/s water at 1200 kPa, 350 °C and velocity 15 m/s. The exit is at 100 kPa, 150 °C and very lo
    13·1 answer
  • Whats a person who enters the electrical trade
    7·2 answers
  • Which of the following activities could lead to injuries?
    6·1 answer
  • Technician A says mismatching tires of the same size on a heavy vehicle will generally not affect ABS operation. Technician B sa
    13·1 answer
  • How can you exercise without hurting yourself.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!