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Natalka [10]
3 years ago
12

Example for strengthening mechanism in single-phase material A. Dispersion strengthening B. Precipitation hardening C. Fiber str

engthening D. Strain hardening
Engineering
2 answers:
Inessa05 [86]3 years ago
5 0
Precipitation hardening
Lena [83]3 years ago
3 0

Answer:

Example for strengthening mechanism in single-phase material: Strain hardening- D.

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To purchase a new car, you borrow $20,000. The bank offers a 6-year loan at an interest rate of 3.25% compounded annually. If yo
Natalka [10]

Answer:

SPCA factor

Single payment compound amount factor.

Total amount pay A = $24,230.95 (Approx)

Interest paid = $4,230.95  (Approx)

Explanation:

Given:

P = $20,000

n = 6 year

r = 3.25%

Find:

Total amount pay A

Computation:

A=p(1+r)ⁿ

A=20,000[1+3.5%]⁶

A=20,000[(1.0325)⁶]

Total amount pay A = $24,230.95 (Approx)

Interest paid = $24,230.95 - 20,000

Interest paid = $4,230.95  (Approx)

6 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 875 K to a cold reservoir at 300 K. Calculate the rate at which th
blsea [12.9K]

Answer:

The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.

Explanation:

Solution

From the question given we recall that,

The transferred heat rate is = 2kW

A reservoir cold at = 300K

The next step is to find the rate  at which the entropy of the two reservoirs changes is kW/K

Given that:

Δs =  Q/T This is the entropy formula,

Thus

Δs₁ = 2/ 300 = 0.006667 kW/K

Δs₂ = 2 / 875 =0.002285

Therefore,

Δs = 0.006667 - 0.002285

= 0.004382 kW/K

Yes, the second law is satisfied, because it is seen as positive.

8 0
4 years ago
A material with a yield strength of 40 kpsi and an endurance strength of 20 kpsi is intended for an application with a von Mises
wlad13 [49]

To solve this problem we will apply the concepts related to Soderberg's relation, which will allow us to find the safety factor based on yield stress, endurance strength, the mean and average stress. The mean and average stress values can be found through the alternating Stress previously given. We will proceed by defining the known values, that is

Yield stress \sigma_y = 40ksi

Endurance strength \sigma_e = 20ksi

The two values for alternating stress are

\sigma_1 = 17ksi

\sigma_2 = 3ksi

We know that mean stress is

\sigma_m = \frac{\sigma_1+\sigma_2}{2}

\sigma_m = \frac{17+3}{2} = 10ksi

And the average stress is

\sigma_v = \frac{\sigma_1-\sigma_2}{2}

\sigma_v = \frac{17-3}{2} = 7ksi

According to Soderberg relation

\frac{1}{F.s} = \frac{\sigma_m}{\sigma_y}+\frac{\sigma_v}{\sigma_e}

\frac{1}{F.s} = \frac{10}{40}+\frac{7}{20}

\frac{1}{F.s} = 0.6

F.s = 1.66 \approx 2

Therefore the factor of safety is 2

3 0
3 years ago
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