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dsp73
3 years ago
11

Yield and tensile strengths and modulus of elasticity . with increasing temperature. (increase/decrease/independent)

Engineering
1 answer:
meriva3 years ago
6 0

Answer:

Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.

Explanation:

The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.

Mathematically we can write,

E(t)=E_o[1-a\frac{T}{T_m}]

where,

E(t) is the modulus of elasticity at any temperature 'T'

E_o is the modulus of elasticity at absolute zero.

T_{m} is the mean melting point of the material

Hence we can see that with increasing temperature modulus of elasticity decreases.

In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.

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Explain all the characteristics of computer<br>​
Sedaia [141]
They are as follows-

Speed
Accuracy
Storage
Versatility
Diligence
Automation
Reliability
Power of Remembering

Let us understand each characteristic in brief:-

Speed – Speed of a computer means the time it takes to complete any given task. The computer works at a very lightning speed. For example, a computer takes a second to calculate 3 million calculations rather than a human which takes years. Hence nowadays the speed of the computer is measured in terms of microseconds, Nanoseconds, and even in Pico seconds instead of seconds or milliseconds.


Accuracy – The computers are programmed and designed in such a manner that their results are almost 100% accurate. It performs all the tasks with the same accuracy. There are very few chances of uncertainty in results and even though an error occurs, it can happen either due to wrong input data or unreliable programs by a programmer. Such errors are referred to as Garbage In Garbage Out (GIGO) which means, if you provide wrong instructions to the computer then you get wrong results.


Versatility – Along with being accurate and diligent a computer is also a versatile device. It can perform several types of tasks at a time if they are reduced to a certain set of logical steps. A computer can be used for many purposes at different places such as booking air and rail tickets, weather forecasting, listening to music, and playing games. Also at a time, one can pay his utility bills and can make a monthly budget too. It can prepare documents, data sheets and also can make models of houses, dams, etc.
Diligence – The ability of a computer to perform tasks without getting tired is known as diligence. Computers are highly reliable, they do not get fed up, exhausted, or lack concentration. They can work for hours and hours unlike humans and can give accurate or error-free results. For example, humans will start feeling tired after 2 -3 hours and will lack concentration whereas a computer will keep on working until the results are achieved.
Automation – Automation means working automatically. A computer can work on its own without the intervention of the person using it. Programs can be made for the computer to perform the task and based on the sequence it executes the tasks and gives accurate results. In case an error occurs, it gives error messages too and registers the logs.
Reliability – High reliability of a computer depends on low failure rate and easy maintenance. The results given out by the computer are reliable as the accuracy is almost 100%.
Power of Remembering – The computer has the power of storing data or information for several years. It cannot lose the data on its own. A person can retrieve the data whenever required and the data remains the same after a number of years also. It gives the freedom to the user to decide on how much data to store and how much to remove.
5 0
3 years ago
Consider the base plate of an 800-W household iron with a thickness of L = 0.6 cm, base area of A =160 cm2, and thermal conducti
ratelena [41]

Given-

Power, P = 800W

Thickness, L = 0.6cm

Area, A = 160cm²

Thermal conductivity, k = 60W/mK

The heat conduction would be

\frac{d^2T}{dx^2} + \frac{d^2T}{dy^2} + \frac{d^2T}{dz^2} + \frac{e(gen)}{k}  = \frac{1}{\alpha } \frac{dT}{dt}

Except \frac{d^2T}{dx^2} all the values are 0.

Therefore,

\frac{d^2T}{dx^2} = 0

Thus, the boundary conditions here would be

1. Q_x_=_0 = -kA \frac{dT (0)}{dx} = Q_o

2. T(L) = T_L

4 0
4 years ago
Yfchggfcvkuhgvjbjhbsdfhjkibgihdfgb
Molodets [167]

Answer:

dysurygfyubiugyunhuhfnbgjdsbnm

Explanation: Because ngubduykrtvytiudhurthgvniuonssdf

8 0
3 years ago
The high electrical conductivity of copper is an important design factor that helps improve the energy efficiency of electric mo
ludmilkaskok [199]

Answer:

B

Explanation:

This is a two sample t-test and not a matched pair t-test

null hypothesis(H0) will be that mean energy consumed by copper rotor motors is greater than or equal to mean energy consumed by aluminium rotor motors

alternate hypothesis(H1) will be that mean energy consumed by copper rotor motors is less than or equal to mean energy consumed by aluminium rotor motors.

So, option D is rejected

The hypothesis will not compare mean of differences of values of energy consumed by copper rotor motor and aluminium rotor motor.

Option A and C are also rejected

5 0
3 years ago
1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bot
liberstina [14]

Answer:

T_{f} = 90.07998 ° C

Explanation:

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

                 Qc = -Qb

                M c_{e_Te} (T₁ -T_{f}) = m c_{e_air} (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

   ρ = m / V  

   V = 1000 -100 = 900 ml  

   V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³  

   V_air = 0.100 l = 0.1 10⁻³ m³  

Tea Mass  

     M = ρ V_te  

     M = 1100 0.9 10⁻³  

     M = 0.990 kg  

Air mass  

     m = ρ _air V_air  

     m = 1.225 0.1 10⁻³  

     m = 0.1225 10⁻³ kg  

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀  

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})  

Let's calculate  

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)  

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)  

T_{f} = 98097.11 / 1089.0002  

T_{f} = 90.07998 ° C  

This temperature decrease is very small and cannot be measured

3 0
3 years ago
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