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Select the properties and typical applications for the high carbon steels.

yanalaym [24]

Answer:

Option-(A)

Explanation:

Typical applications for the high carbon steels includes the following;

It is heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.

3 0

3 years ago

Earth whose in situ weight is 105lb/cf and whose compacted weight is 122 lb/cf is placed in a fill at the rate of 260 cy/hr, mea

Stells [14]

Answer:

Number of rollers required to complete the compaction are 2

Explanation:

The solution is given in the attachments.

6 0

3 years ago

Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s

suter [353]

Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.

<h3>Wat is the max stress theory?</h3>

The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”

Stress states at various critical locations are f= 2.662.

Read more about strain:

brainly.com/question/6390757

#SPJ1

3 0

2 years ago

Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

FOR SUBSTANCE A:

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

FOR SUBSTANCE B:

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

C = 1.16 J/g

5 0

3 years ago

What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?

Tom [10]

Answer:

F=1.47 KN

Explanation:

Given that

Diameter of plate = 25 cm

Height of pool h = 3 m

We know that force can be given as

F= P x A

P=ρ x g x h

Now by putting the values

P=1000 x 10 x 3

P= 30 KPa

$A=\dfrac{\pi}{4}\times 0.25^2\ m^2$

$A=0.049\ m^2$

F= 30 x 0.049 KN

F=1.47 KN

So the force on the plate will be 1.47 KN.

4 0

3 years ago