If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.

Answers

Answer 1
Answer:

Answer:

critical stress required is  18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × 10^(-3) m

so we get here critical stress that is

\sigma _c = \sqrt{(2E \gamma )/(\pi a)}     ...............1

put here value we get

\sigma _c =   \sqrt{(2* 225* 10^9 * 1 )/(\pi * 0.4* 10^(-3))}

\sigma _c =  18923493.9151 N/m²

\sigma _c =   18.92 MPa


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Isothermal process also means adiabatic internal reversible process. a)-True b)-False

Answers

Answer:

(b) False

Explanation:

Isothermal process is a process in which temperature is constant ,heat can be transferred but temperature is constant and as the temperature is constant so internal energy is also constant

In other hand in adiabatic process there no transfer of heat and internal energy also changes

So the given statement is false statement  

An escalator with 35° incline is designed to have two passengers per step. Find number of persons moved per hour for the design if velocity is 50cm/mins and the step tread is 600mm.

Answers

Answer:

The escalator disposes 58 passengers each hour.

Explanation:

The velocity diagram of the escalator is shown in the attached figure

We can obtain the vertical distance that the escalator moves in 1 hour as

D_(v)=vsin(\theta )* 1hr\n\nD_(v)=50cm/min* sin(35^(o))* 60mins\n\n\therefore D_(v)=1720.73cm

Thus in 1 hour the last thread moves 1720.73 cm

Now it is given that 1 thread = 600 mm =60 cm

Thus the number of times the last thread moves equals

n=(1720.73)/(60)=28.68times

Since each time last thread moves it disposes 2 passengers thus the number of passengers disposed when the thread moves 28.68 times equals

N=28.68* 2=57.36

Thus the escalator disposes 58 passengers each hour.

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.

Answers

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.

Answers

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 * 10^(-5) kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s

total volume of air  infiltration= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s

Time period = (210)/(5.04* 10^(-3)) = 41654.08 s

Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above

Answers

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer:

v = 250[1 - {e^(-6000t)}] mV

Explanation:

The voltage across a capacitor at a time t, is given by:

v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)                 ----------------(i)

Where;

v(t) = voltage at time t

t_(0) = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3e^(-6000t) mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)    

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0

v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt            

v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt             [Solve the integral]

v = (3)/(2*10^(-6)*(-6000))  {e^(-6000t)}|_0^t

v = (-3000)/(12)  {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)}|_0^t

v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]

v = -250 {e^(-6000t)} - [-250]

v = -250 {e^(-6000t)} + 250

v = 250 -250 {e^(-6000t)}

v = 250[1 - {e^(-6000t)}]

Therefore, the voltage across the capacitor is v = 250[1 - {e^(-6000t)}] mV