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TEA [102]
3 years ago
12

5. Consider the LTI system defined by the differential equation (a) Draw the pole-zero plot for the system. Is the system stable

? (5 points) (b) Find the system’s impulse response h(t). (5 points) (c) Find the system’s output if the input signal is x(t) = cos(2t). (5 points) (d) Find the system’s output if the input is x(t) = e-t u(t). (10 points)

Engineering
1 answer:
stealth61 [152]3 years ago
8 0

Answer:

Detailed solution is attached in the images below showing step wise solution and answer for each part individually.

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A heat pump cycle whose coefficient of performance is 2.5 delivers energy by heat transfer to a dwelling at a rate of 20kW.
12345 [234]

Answer:

a) 8kW

b) $128

Explanation:

Given the coefficient of performance of the heat pump cycle to be 2.5

Energy delivered by the heat pump = 20kW

a) net power required to operate the heat pump = Energy delivered / coefficient of performance

Net power required = 20/2.5

= 8kW

b) Given the cost of electricity is $0.08 for 1kWhour

Since net power required to operate heat pump = 8kW

If the heat pump operate for 200hours, total power required for a month = 8kW×200hours = 1600kWhour

since 1kWh of electricity costs $0.08, cost of electricity used in a month when the pump operates for 200hour will be 1600kWh×$0.08 which is equivalent to $128

8 0
3 years ago
Are you?
Inessa [10]

Answer:

are you wht

didn't understand the question

6 0
2 years ago
Read 2 more answers
(a) Differentiate between heat treatment of ferrous and non-ferrous alloys (b) With your understanding of material's thermal pro
liubo4ka [24]

Answer:

In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.

Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are  the alloy element of iron ans carbon.Wrought iron is the purest from of iron.

Heat treatment process for  ferrous materials :

1.Normalizing

2.Annealing

3.Quenching

4.Surface hardening

Heat treatment process for non ferrous materials :

Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.

When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.

The use of bimetallic structure -In clock ,thermometers ,engines.

7 0
3 years ago
identify which country has an absolute advantage in production of cookies and which has the absolute advantage in production of
nirvana33 [79]

a) Question Completion:

INPUT HOURS OF LABOR

Country   Cookies      Milk

Atlantis       2 hours   1 hour

Neverland  4 hours   1 hour

Answer:

1. Atlantis has the absolute advantage in the production of cookies.

2. No country has the absolute advantage in the production of milk.

Explanation:

Absolute advantage refers to superior production capability.  It is determined when a country, for example, has the ability to produce a particular good or service at lower cost or more efficiently (i.e. with less resources) than the other country.  In the scenario above, Atlantis has an absolute advantage in the production of cookies because it can produce the same quantity of cookies using 2 labor hours that Neverland can produce using 4 labor hours.  But for the production of milk, Atlantis and Neverland share the same comparative advantage less they can use less labor hours to produce milk than they can produce cookies.

5 0
3 years ago
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin
Leni [432]

Answer:

Q=41.33 KVAR\ \\at\\\ 480 Vrms

Explanation:

From the question we are told that:

Voltage V=480/0 \textdegree V

Power P=120kW

Initial Power factor p.f_1=0.77 lagging

Final Power factor p.f_2=0.9 lagging

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

p.f_1=0.77

cos \theta_1 =0.77

\theta_1=cos^{-1}0.77

\theta_1=39.65 \textdegree

And

p.f_2=0.9

cos \theta_2 =0.9

\theta_2=cos^{-1}0.9

\theta_2=25.84 \textdegree

Therefore

Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)

Q=-41.33VAR

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

Q=41.33 KVAR\ \\at\\\ 480 Vrms

6 0
2 years ago
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