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Hatshy [7]
3 years ago
15

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

ine rod's percent ductility
Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

A=\frac{\pi d^{2}}{4}

A=\frac{\pi\times(10.2)^{2}}{4}

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

D=\frac{A_{i}-A_{f}}{A_{i}}\times100

D=\frac{81.713-52.7}{81.713}\times100

D = 35.5%.

Thus, the percentage ductility is 35.5%.

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On highways, the far left lane is usually the _____. A. emergency lane B. merge lane C. slowest D. fastest
slamgirl [31]

On highways, the far left lane is usually the<u> fastest</u> moving traffic.

Answer: Option D.

<u>Explanation:</u>

For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.

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7 0
3 years ago
Read 2 more answers
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
What are the top 4 solar inventions, how they are used, and how they are better than the original way of powering them
OlgaM077 [116]
Yes I will answer soon
7 0
3 years ago
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to =  4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × 10^{6} + 4  × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × 10^{6} - 4  × 10³

lower side frequencies =  3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies  - lower side frequencies

total width bandwidth = 8kHz

6 0
3 years ago
The minimum waste compaction ratio reported by the manufacturer of a high-pressure compaction machine is 8:1. What is the corres
Crank

Answer:

12.5%

Explanation:

Compaction ratio= Volume before reduction/volume after reduction

Compaction ratio= 8/1

% reduction in volume= Volume after reduction/Volume before reduction× 100= 1/compaction ratio × 100

% reduction in volume= 1/(8/1) × 100

                                    =12.5

6 0
3 years ago
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