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Rama09 [41]
2 years ago
6

1. Band saw lower wheel does not require a guard *

Engineering
1 answer:
soldi70 [24.7K]2 years ago
8 0

Answer:

1. false

2. true

3. true

Explanation:

thats all hope I help let me know if you need something eles

branliest pls

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Which of the following terms is defined as small bumps and slashes within a fluid power system?
-BARSIC- [3]

Answer:

friction

Explanation:

''.''

7 0
3 years ago
Read 2 more answers
Mention the types of water demend
amm1812

Answer:

domestic, public, commercial, and industrial uses.

8 0
2 years ago
1 . How are encoders used in the measurement of speed? Explain the encoder with a neat diagram.​
rusak2 [61]

The most common use for encoders is to measure angular or linear distance, but encoders can also be used to perform speed or velocity measurements. In other words, as the encoder rotates faster, the pulse frequency increases at the same rate

4 0
2 years ago
g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima
nikdorinn [45]

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

6 0
3 years ago
P1.30 shows a gas contained in a vertical piston– cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is a
iogann1982 [59]

Answer:

1.737 kJ

Explanation:

Thinking process:

Step 1

Data:

Area of the shaft = 0.8 cm²

Combined mass of shaft and piston  = (24.5 + 0.5) kg

                                                             = 25 kg

Piston diameter                                   = 0.1 m

External atmospheric pressure          = 1 bar = 101.3 kPa

Pressure inside the gas cylinder      = 3 bar = 3 × 101.3 kPa

g                                                           = 9.81 m/s²

Step 2

Draw a free body diagram

Step 3: calculations

area of the piston = 0.0314 m²

Change in the elevation of the piston, \deltaz

\deltaz = \frac{PE}{mg}

    = \frac{0.2*10x^{3} }{25*9.81}

    = 0.82 m

Next, we evaluate the work done by the shaft:

W_{s} = F_{s} Z

     = (1668) ( 0.082)

     = 1. 37 kJ

Net area for work done = A (piston) - Area of shaft

                                       = \pi*(0.1)^{2}  - 0.8 cm^{2}

                                        = 77.7 cm²

                                        = 0.007774 m²

Work done in overcoming atmospheric pressure:

 Wₐ = PAZ

       =101.3 kPa * 0.007774 * 0.82

      =  0.637 kJ

total work = work done by shaft + work to overcome atmospheric pressure = 0.367 + 1.37

= 1.737 kJ Ans

6 0
3 years ago
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