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iogann1982 [59]

3 years ago

12

Are spheroidized steels considered as composite? If so, what is the dispersed phase a)- No b)- Yes, Chromium Carbides c)- Yes, I

ron Carbides d)- Yes, Intermetallic Compounds

1 answer:

Kruka [31]3 years ago

3 0

Answer: c)-Yes, spheroidized steel are considered as composite.the dispersed phase is iron carbide.

Explanation: Spheroidized steel are the alloy that have iron as the basic part that have been heat treated to increase their ductility and malleability property .They are considered as composite because they are made up of iron alloys. The heat treatment is usually for the carbon steel and so the dispersed phase that is obtained is iron carbide.

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There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.

sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

$C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})$

$C_{1}=1.2606\times10^{-5}$ $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

$C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})$

$C_{2}=3.334\times10^{-5}$ $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

$C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})$

$C_{3}=1.66\times10^{-5}$ $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

5 0

2 years ago

A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.

EleoNora [17]

Answer:

The temperature attains equilibrium with the surroundings.

Explanation:

When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.

Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.

Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.

6 0

3 years ago

A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

7 0

2 years ago

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as

katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:

A = as

$=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}$

Calculating the kinematics equation:

$\to v^2 = v^2_{o} + 2as\\\\$

$=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}$

Calculating the value of acceleration:

$\to a= \frac{dv}{dt}$

$=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}$

$\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\$

$=\frac{0.092}{s}$

3 0

2 years ago

a) find the state-space representation of the system. b) is the system controllable? why? c) is the system observable? why

mart [117]

If a controlled input can transfer (alter) the control system's initial states to some other desired states in a finite amount of time, the control system is said to be controllable.

Using Kalman's test, we can determine whether a control system is controllable. The evolution model for the state variables (time-varying unknowns) and the observation model, which connects the observations to the state variables, make up the state space representation of a dynamical system. The capacity to move a system about in its full configuration space using just specific permitted actions is generally referred to as controllability. The precise definition changes slightly depending on the model type or framework used.

Learn more about control here-

brainly.com/question/28540307

#SPJ4

5 0

1 year ago