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scoray [572]
3 years ago
7

Oil system cleaning products should not use solvents problem

Engineering
1 answer:
pantera1 [17]3 years ago
7 0
Sorry I don’t know this one out I need points sorryyyy
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Brazing, Soldering and Adhesive Bonding are the types of • Liquid Solid System Welding Solid State Welding • Fusion Welding . No
olga_2 [115]

Answer: Solid state welding

Explanation: Solid state welding is the welding procedure which is based on the temperatures and pressure but without any liquid or vapor as aid for welding.This process is carried mainly cohesive forces and considering forces as less important. Brazing,soldering and adhesive is the process in which material are joint which the help of solid welding agent.Thus solid state welding is the correct option.

6 0
3 years ago
A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.
Setler79 [48]

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered  0.5742 Kg.

(c) The work done is 56.52 kJ.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

6 0
3 years ago
2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks
Amiraneli [1.4K]

Answer:

  • 100N
  • 25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0
3 years ago
Which is not an affect of alcohol
Gre4nikov [31]
Pooping problems is not an affect
6 0
3 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]

b)

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
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