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3 years ago

I need help with my autos

Engineering

1 answer:

oksano4ka [1.4K]3 years ago

6 0

Answer:

what is wrong with it and what is the question

Explanation:

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The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr

djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

6 0

4 years ago

Marshall's ruling in Marbury v. Madison is a testament to his strategic political skills for which of the following reasons? Sel

gladu [14]

Answer:

The following reason is letter D.

Explanation:

<em>Even though the real correct answer should be: It did not immediately strenghten the Court's power in its relations with the President or Congress.</em>

6 0

4 years ago

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha

alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α = 15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ + β - α = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

$r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}$

$r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}$

r=0.88

New coefficient of friction ,μ' = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' + β' - α = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

$r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}$

$r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}$

r'=0.70

Percentage change

$\Delta r=\dfrac{r-r'}{r}\times 100$

$\Delta r=\dfrac{0.88-0.70}{0.88}\times 100$

Δr=20.45 %

8 0

3 years ago

Is a steam power plant a heat engine?

Novosadov [1.4K]

Answer:

Yes

Explanation:

Yes steam power plant is heat engine.

As we know that a heat engine takes from source heat and produce some amount of work and reject the heat to the sink.

So the steam power plant fulfill the all requirement like it take heat from boiler and produce some amount work and then reject the heat by using condenser.So we can say that steam power plant is an example of heat engine.

7 0

4 years ago

City A and city B are located at the same latitude and elevation. City A is a coastal city, and city B is located far inland. A

jenyasd209 [6]

Answer:

Option 4

Explanation:

8 0

3 years ago

Read 2 more answers