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Yuliya22 [10]
3 years ago
9

An electronic device, such as a power transistor mounted on a finned heat sink, can be modeled as a spatially isothermal object

with internal heat generation and an external convection resistance. (a) Consider such system of mas M, specific heat c, and surface area As, which is initially in equilibrium with the environment at To. Suddenly, the electronic device is energized such that a constant heat generation Eg(W) occurs. Show that the temperature response of the device is: θ = exp RC where θ Ξ T-T(w): T(oo) is the steady-state temperature corresponding to t → oo; 0Ti T(; T: initial temperature of the device; R: thermal resistance 1/hAs; and C: thermal capacitance Mc. (b) An electronic device, which generates 60W of heat, is mounted on an aluminum heat sink weighing 0.31 kg and reaches a temperature of 100°C in ambient air at 20°C under steady- after the power is switched on?
Engineering
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

a) attached below

b) 63.71°c

Explanation:

A) Given data

The system has ; mass ( M ) , specific heat capacity , and surface area

attached below is a detailed solution

B) Given data

power generated = 60w

weight of heat sink = 0.31 kg

temperature given = 100°c

temp in ambient air = 20°c

lets take

specific heat capacity = 918.18 J /kg-k

density = 2702 kg/m^2

attached below is the detailed solution

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Determine (with justification) whether the following systems are (i) memoryless, (ii) causal, (iii) invertible, (iv) stable, and
lina2011 [118]

Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is \frac{1}{x[n] . x[n-1] x[n+1]}

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2\pi , y[n] = cos(2\pi) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

3 0
3 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
As a construction manager, Alex found a major issue with the client's design demands. He found out that the building design was
kipiarov [429]

Answer:

Which human resource skill helped Alex convince the client? its nagotiation

Explanation:

got it right on the test.

6 0
3 years ago
A ring-shaped seal, made from a viscoelastic material, is used to seal a joint between two rigid pipes. When incorporated in the
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I’m not sure what this question is talking about. I will get back to you later on this let me rethink it. 2.22578
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3 years ago
2. Consider Dekker’s algorithm written for an arbitrary number of processes by changing the statement executed when leaving the
neonofarm [45]

Answer:

Algorith does not work.

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One of the ways to obtain the Dekker Algorithm is through a change in the declaration, that is, a declaration that can be executed at the exact moment it leaves the critical section. This way it is possible that the statement,

turn = 1-i / * P0 sets turn to 1 and P1 sets turn 0 * /

It can be changed to,

turn = (turn +1) \% n / * n = number or processes * /

The result will allow to define if it works or not, that is, if it is greater than 2 the algorithm will not be able to work.

Given this consideration we can say that,

<em>- The dead lock does not occur, because the mutual is imposed (if a resource unit has been assigned to a process, then no other process can access that resource).</em>

<em>- There is the possibility of starving if the shift is established in a non-contentious process.</em>

Directly it can be concluded that there is a possibility of starvation so the algorithm could not work, despite the fact that mutual exclusion guarantees that a dead block does not occur.

4 0
3 years ago
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