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3 years ago

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true or false incident reports, such as situation reports and status reports enhance situational awareness and ensure that perso

nnel can access needed information.

1 answer:

Degger [83]3 years ago

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True becauseidbajqkfkmxzmbakwlgof

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The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re

Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 % , Irreversible

b) W_cycle = 600 KW , n_th = 100 % , Impossible

c) W_cycle = 400 KW , n_th = 66.67 % , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

TL = 400 K

TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

n_th < n_max ...... irreversible

n_th = n_max ...... reversible

n_th > n_max ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And, n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 400 = 200 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 200 / 600 = 33.33 %

- The type of process according to second Law of thermodynamics:

n_th = 33.333 % n_max = 66.67 %

n_th < n_max

Hence, Irreversible Process

b) QH = 600 kW, QC = 0 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 0 = 600 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 600 / 600 = 100 %

- The type of process according to second Law of thermodynamics:

n_th = 100 % n_max = 66.67 %

n_th > n_max

Hence, Impossible Process

c) QH = 600 kW, QC = 200 kW

- The work done by cycle according to First Law is:

W_cycle = 600 - 200 = 400 KW

- The thermal efficiency of the cycle is given by n_th:

n_th = W_cycle / QH

n_th = 400 / 600 = 66.67 %

- The type of process according to second Law of thermodynamics:

n_th = 66.67 % n_max = 66.67 %

n_th = n_max

Hence, Reversible Process

7 0

3 years ago

Which of the following best distinguishes between superficial design improvements and functional

Contact [7]

Answer:

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

Explanation:

As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.

I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)

In short ...

<em>Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.</em>

8 0

2 years ago

Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br

vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation $v = v_o + a_ct$ to determine the velocity of car B.

where;

$v_o =$ initial velocity

$a_c$ = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

$S = d + v_ot + \dfrac{1}{2}at^2$

Then:

$v_B = 60-12t$

The distance traveled by car B in the given time (t) is expressed as:

$S_B = d + 60 t - \dfrac{1}{2}(12t^2)$

For car A, the needed time (t) to come to rest is:

$v_A = 60 - 18(t-0.75)$

Also, the distance traveled by car A in the given time (t) is expressed as:

$S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

Relating both velocities:

$v_B = v_A$

$60-12t = 60 - 18(t-0.75)$

$60-12t =73.5 - 18t$

$60- 73.5 = - 18t+ 12t$

$-13.5 =-6t$

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

$S_B = S_A$

$d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2$

$d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2$

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0

3 years ago

Gfci stands for ground fault circuit interference.

dybincka [34]

Answer:

The ground-fault circuit interrupter, or GFCI, is a fast-acting circuit breaker designed to shut off electric power in the event of a ground-fault within as little as 1/40 of a second.

Explanation:

4 0

2 years ago

Please help me I need to answer this problem

Norma-Jean [14]

Answer:

f

Explanation:

7 0

2 years ago