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For a p-n-p BJT with NE 7 NB 7 NC, show the dominant current components, with proper arrows, for directions in the normal active mode. If IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA, calculate the base transport factor, emitter injection efficiency, common-base current gain, common-emitter current gain, and ICBO. If the minority stored base charge is 4.9 * 10-11 C, calculate the base transit time and lifetime.

Answers

Answer 1

Answer:

=> base transport factor = 0.98.

=> emitter injection efficiency = 0.99.

=> common-base current gain = 0.97.

=> common-emitter current gain = 32.34.

=> ICBO = 1 × 10^-6 A.

=> base transit time = 0.325.

=> lifetime = 1.875.

Explanation:

(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).

The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.

(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.

(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.

(3).common-base current gain = 0.98 × 0.99 = 0.9702.

(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.

(5). Icbo = Ico = 1 × 10^-6 A.

(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.

(7).lifetime;

= > 2 = √0.325 + √ lifetime.

= Lifetime = 2.875.

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The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-kg block B. If the unstrectched spring has a stiffness k = 1000 N/m, determine the maximum compression of the spring due to the collision. Assume the collision is perfectly plastic. Take e=0.6 .

Answers

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

List the RTL (Register Transfer Language) sequence of micro-operations needed to execute the instruction STORE X

from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.

You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.

Answers

Answer:

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

Explanation:

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

What are two advantages of forging when compared to machining a part from a billet?

Answers

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

A conical funnel of half-angle θ = 30 drains through a small hole of diameter d = 6:25 mm at the vertex. The speed of the liquid leaving the funnel is V= √ 2gy where y is the height of the liquid free surface above the hole. The funnel initially is filled to height y0 = 300 mm. Obtain an expression for the time, t, for the funnel to completely drain, and evaluate. Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm), and from 150 mm to completely empty (also a change in depth of 150 mm). Can you explain the discrepancy in these times?

Answers

Answer:

$3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

t = 12.03

t = 81.473

velocity of the fluid decreases with level of fluid. Hence, in later stages the time taken is greater than in earlier stages

Explanation:

Given:

- The half angle θ = 30°

- The diameter of the small hole d = 6.25 mm

- The flow rate out of the funnel Q = A*√ 2gy

- The volume of frustum of cone is given by:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Where,

D: Is the larger diameter of the frustum

d: Is the smaller diameter of the frustum

y: The height of the liquid free surface from small diameter d base.

Find:

- Obtain an expression for the time, t, for the funnel to completely drain, and evaluate.

- Find the time to drain from 300 mm to 150 mm (a change in depth of 150 mm) and from 150 mm to completely empty (also a change in depth of 150 mm)

- Can you explain the discrepancy in these times?

Solution:

- We will use rate of change analysis by considering the rate of change of volume, and then apply the chain rule.

- The Volume of the frustum is a function of d , D , y. V = f ( d , D , y ). In our case the diameter of base d remains constant. Then we are left with:

V = f ( D , y )

- Determine a relationship between y and d. We will use half angle θ to determine the relationship between D and y by applying trigonometric ratios:

tan ( θ ) = D / 2*y

D = 2*y*tan ( θ )

- We developed a relationship between D and y in terms of half angle which remains constant. So the Volume is now only a function of one variable y:

V = f ( y )

- The volume of frustum of the cone can be written as:

$V = (\pi )/(12)*y*(D^2 + d^2 + d*D})$

Substituting the relationship for D in terms of y we have:

$V = (\pi )/(12)*y*(4*y^2*tan(Q) ^2 + d^2 + 2*d*y*tan(Q)})$

- Now by rate of change of Volume analysis we have:

dV / dt = [dV / dy] * [dy / dt]

- Computing dV / dy, where V = f(y) only:

$V = (\pi )/(12)*(4*y^3*tan(Q) ^2 + y*d^2 + 2*d*y^2*tan(Q)})\n\n(dV)/(dy) = (\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)})\n\n$

- Where, dV/dt = Volume flow rate:

$(dV)/(dt) = - Q\n(dV)/(dt) = - A*V\n(dV)/(dt) = - (\pi*d^2 )/(4) *√(2*g*y)$

- Then from Chain rule we have:

[dy / dt] = [dV / dt] / [dV / dy]

$(dy)/(dt) = (- (\pi*d^2 )/(4) *√(2*g*y))/((\pi )/(12)*(12*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n} \n(dy)/(dt) = (-d^2 *√(2*g*y))/((4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q))\n\n}$

- Separate variables:

$\frac {(4*y^2*tan(Q) ^2 + d^2 + 4*d*y*tan(Q)} {√(2*g*y)} .dy = {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n$

- Integrate both sides:

$\frac { 1 } { √(2*g) }*(4*y^1^.^5*tan(Q) ^2 + y^-^0^.^5d^2 + 4*d*y^0^.^5*tan(Q))= {-d^2} .dt\n\n\frac { 1 } { √(2*g) }*((8)/(5)*y^2^.^5*tan(Q) ^2 + 2*d^2*y^0^.^5 + (8)/(3)*d*y^1^.^5*tan(Q)) = -d^2*t + C\n\n\frac { 1 } { √(2*y*g) }*((8)/(5)*y^3*tan(Q) ^2 + 2*d^2*y + (8)/(3)*d*y^2*tan(Q)) = -d^2*t + C\n\n0.1204*y^3 + 0.000017638*y + 0.00217*y^2 = -0.0000390625*t + C\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = -t + C$

- Evaluate @ t = 0 , y = 0.3 m

$3175.424*(0.15)^3 + 0.45153*(0.15) + 55.552*(0.15)^2 = 0 + C\n\nC = 12.0347055\n\n3175.424*y^3 + 0.45153*y + 55.552*y^2 = t + 12.0347055$

- Time taken from y = 300 to 150 mm:

$3175.424*0.15^3 + 0.45153*0.15 + 55.552*0.15^2 = -t + 90.871587\n\nt = 0 - -12.0347055\n\nt = 12.0347055 s\n$

- Time taken from y = 150 to 0 mm:

t = t_(0-300) - ( t_(0-150) = 90.871587 - 12.0347055 = 81.4721 s

- The discrepancy is time can be explained by the velocity of fluid coming out of bottom is function of y. The velocity of the fluid decreases as the level of fluid decreases hence the time taken from y =150 mm to 0 mm is larger than y = 300 mm to y = 150 mm.

An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?

Answers

Find the given attachment

what is an example of an innovative solution to an engineering problem? Explain briefly why you chose this answer.

Answers

Answer:

robotic technology

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.

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