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oee [108]
3 years ago
13

1. Select the punch that is used to make a mark in metal

Engineering
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

pin punch

Explanation:

  1. <em><u>Pin</u></em><em><u> </u></em><em><u>Punch</u></em><em><u> </u></em>
  2. <em><u> </u></em><em><u>pin</u></em><em><u> </u></em><em><u>punch</u></em>
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Give the general layout of centrifugal pump.​
Nadusha1986 [10]

Answer:

A centrifugal pump is a rotating machine that pumps liquid by forcing it through a paddle wheel or propeller called an impeller. It is a most common type of industrial pump. By the effect of the rotation of the impeller, the pumped fluid is drawn axially into the pump, then accelerated radially, and finally discharged tangentially.

Explanation:

Suppliers generally offer charts in the plan, which present the various things at the nominal operating point using two main arrangements: the inductors and the balancing pants.

7 0
3 years ago
What type of control would describe training, inspection, and housekeeping?
zubka84 [21]

Answer:

Administrative controls,

Explanation:

such as training, inspection, housekeeping and so on can be used to limit exposure.

8 0
3 years ago
Water flows near a flat surface and some measurements of the water velocity, u, parallel to the surface, at different heights y
BabaBlast [244]

Answer:

a) since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b) The velocity according to the equation at y=0 is equal to 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary, this value is wrong hence the equation is NOT CORRECT

Explanation:

Given that;

range ⇒ 0 < y < 0

1ft is given by the equation u = 0.81 + 9.2y + (4.1 × 10³y³)

so u=velocity of water at different layers

y= height of the layer

a)

consider BG system of units

u(ft/s) = 0.81 + 9.2y + (4.1 × 10³y³)

and consider y=0.05 ft

u = 0.81 + 9.2(0.5) + (4.1 × 10³(0.5³)

u = 0.81 + 0.46 + 0.5125

u = 1.7825 ft/s lets say this is equation 1

now consider the SI system units

u(m/s) = 0.81 + 9.2y + (4.1 × 10³y³)

also consider y=0.05ft

1ft = 3.048×10⁻¹ (from conversion table)

so 0.05ft = 0.01524m

we substitute

u(m/s) = 0.81 + 9.2(0.01524m) + (4.1 × 10³(0.01524m)³)

u = 0.81 + 0.1402 + 1.4512×10⁻²

u = 0.9647 m/s

1m/s = 3.281 ft per seconds ( conversion table)

so

0.9647 m/s = 0.9647(3.281)

u = 3.165 ft/s lets say this is equation 2

now since u from equation1 and u from equation2 are not the same ( 1.7825 ft/s ≠ 3.165 ft/s ) then this equation is not valid for any system of units.

b)

we know that the velocity of water at the surface contact is zero

u=0

so from the equation

u = 0.81 + 9.2y + (4.1 × 10³y³)

at y = 0

u = 0.81 + 9.2(0) + (4.1 × 10³(0)³)

u = 0.81 ft/s

The velocity according to the above equation at y=0 is 0.81 ft/sec but since the fluid is flowing on flat surface which is stationary this value is wrong hence the equation is NOT CORRECT

5 0
4 years ago
What software do you sue for design and the rapid prototyping machine?
mash [69]

Answer:

Computer Aided Design (CAD)

Explanation:

For design purpose and rapid prototyping, we use a software known as Computer Aided Design abbreviated as CAD.

Several production formates that are used in CAD are .stp, .stl, .igs, .step where formats like .stp, .igs and .step are used for rapid prototyping and format like .stl is used for design.

The information depending on the extensions used can be modeled in the 3-D files. The most frequent software used for prototyping are: AutoCAD, SolidWorks, SketchUP, etc

This also allows the spatial manipulation, dimension measurement and the visualization of the inner and outer parts.

3 0
4 years ago
Using the background information provided and your answers to questions 1 &amp; 2, look at the hex packet below and identify the
yanalaym [24]

Answer:

Goal address: Based on the above Ethernet outline design, we get that the goal address begins from the first hex worth and is of size 12 hex values (got from question 1). In light of the bundle given, we get that the goal address is 6c40 0889 c448.  

Source address: Similarly, the source address begins after the goal address and is of size 12 hex values (got from question 1). Thus, the appropriate response is f832 e4a7 fb38.  

Type/Length: The sort/length begins after the source address and is of size 4 hex values (got from question 1). Thus, the appropriate response is 0806.  

Data (Payload): The data(payload) begins after the sort/length and finishes not long before FCS(Checksum) which is of 4 bytes for example 4 * 2 = 8 hex qualities. So the information comprises of everything between the 'type/length' and 'CRC'. We persuade the CRC to be c0a8 01f2. Henceforth, the appropriate response is 0001 0800 0604 0002 f832 e4a7 bf38 c0a8 0101 6c40 0889 c448.  

Question 4:  

The Ethernet parcel type characterizes the convention utilized for sending the bundle information. We realize that type 0x0800 demonstrates the IPv4 convention, 0x0806 shows an ARP convention, and 0x86DD demonstrates an IPv6 convention. In light of the appropriate response we got in Question 3, we realize that the sort/length esteem for the given parcel is 0806, which implies that convention utilized for sending the information bundle was ARP convention.

8 0
3 years ago
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