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1 year ago

Question 9

1 answer:

5 0

Answer:
1. Make sure the regulator adjustment screws are completely backed out and loose.
2. Open the main cylinder valves slowly.
3. Set the Oxygen and Acetylene working pressures.

4. Light and adjust the torch.

Explanation:

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A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep

VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

Determine if the fluid is satisfied

guapka [62]

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8 0

3 years ago

Represent each of the following units as a combination of primitive

zimovet [89]

Answer:

a. unit of length: [L]

b. unit of volume: [ $L^3$ ]

c. unit of pressure: $P=\frac{F}{A} \equiv\frac{[MLT^{-2}]}{[L^2]}$ $[ML^{-1}T^{-2}]$

d. unit of power: $N.m.s^{-1}\equiv [ML^2T^{-3}]$

e. unit of force: $[kg.m/s^2]\equiv [MLT^{-2}]$

f. unit of power: $N.m.s^{-1}\equiv [ML^2T^{-3}]$

Force: $F=m.a=m.\frac{v}{t}=m.\frac{x}{t}\div t$

Power: $P=\frac{W}{t}=\frac{F.x}{t}$

where:

F = force

A = area

W = work

t = time

a = acceleration

v = velocity

x = displacement

3 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 106 GPa and an original cross-sectional diameter of 3.9

kiruha [24]

Answer:

L= 312.75 mm

Explanation:

given data

elastic modulus E = 106 GPa

cross-sectional diameter d = 3.9 mm

tensile load F = 1660 N

maximum allowable elongation ΔL = 0.41 mm

to find out

maximum length of the specimen before deformation

solution

we will apply here allowable elongation equation that is express as

ΔL = $\dfrac{FL}{AE}$ ....................1

put here value and we get L

L = $\dfrac{0.41\times 10^{-3}\times \dfrac{\pi}{4}\times (3.9\times 10^{-3})^2\times 106\times 10^9}{1660}$

solve it we get

L = 0.312752 m

L= 312.75 mm

8 0

4 years ago

(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng

igor_vitrenko [27]

Answer:

a) $\sigma = 12.2$ (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

$\sigma = \frac{I l}{V \pi r^2}$

$\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}$

$\sigma = 12.2$ (Ω-m)^{-1}[/tex]

$Resistance = \frac{l}{\sigma A}$

$= \frac{l}{ \sigma \pi r^2}$

$= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}$

Resistance = 121.4 Ω

8 0

3 years ago