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lukranit [14]
3 years ago
14

What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?

Engineering
1 answer:
sasho [114]3 years ago
7 0

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

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In your opinion, what is the external opportunity cost of a successful biking company in a community
OleMash [197]

Answer:

The cost of increased rates of accidents/injuries due to road accidents associated with cycling.

Explanation:

Opportunity cost is referred to as an alternative cost which is the value of the other choose left out while settling for a better alternative. External cost will be the cost a society has to bear as a result of a private cost. A successful biking company in a community can led to increased rates of accidents due to road accidents associated with cycling. In addition, the society will have to bear the costs of the threats presented by dangerous car drivers who posse more danger to cyclists as compared to trucks/car drivers.

8 0
3 years ago
A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an
vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0
2 years ago
I'm bored. I want to talk to you humans! weird. STUPID QUARANTINE!!! So. What are yall up to?
kiruha [24]

Answer:

nothing much what class r u in

3 0
3 years ago
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Answer:k

Explanation:

6 0
2 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
2 years ago
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