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anyanavicka [17]
2 years ago
11

The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d

t squared end fraction plus sin (3 y )fraction numerator d y over denominator d t end fraction plus y equals t fraction numerator d f over denominator d t end fraction plus f. Then, the system is:
Engineering
1 answer:
Colt1911 [192]2 years ago
5 0

Answer:

Explanation:

The given equation is :

\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f

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A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a
prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds  

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

                          = 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

                              = 2622.6 / 25  = 104.904 ft^3/sec

8 0
3 years ago
One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150×10^{-3} m^{3}

m = 1 Kg

P_{1} = 2 MPa

T_{2}  = 40°C

The waters specific volume is calculated:

v_{1} = V/m

Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

v_{1} = 150×10^{-3} m^{3}/1

v_{1} = 0.15 m^{3}/ Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 m^{3}/ Kg and 0.13 m^{3}/ Kg.

T_{1}= 350+(400-350) \frac{0.15-0.13}{0.1522-0.1386}

T_{1} = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

v_{2} = v_{1}= 0.15 m^{3}/ Kg

In saturated water labels for T_{2}  = 40°C.

v_{f} = 0.001008 m^{3}/ Kg

v_{g} = 19.515 m^{3}/ Kg

The final state is two phase region v_{f} < v_{2} < v_{g}.

In saturated water labels for T_{2}  = 40°C.

P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

7 0
3 years ago
Which of the following sensors is used to provide suspension control module with feedback regarding vehicle cornering​ forces?
Readme [11.4K]

Answer:

A

Explanation:

4 0
2 years ago
La extensión de un resumen debe estar en un rango de _________ del texto inicial. a)25 a 35 % b)10 a 20 % c)15 a 20 % d)20 a 25
mestny [16]

Answer:

b)

Explanation:

because it is correct

3 0
3 years ago
Which of the following is typically wom when working in an atmosphere containing dust?
alukav5142 [94]

Answer:

Either D or C

Both of these masks are used for dust, but since half masks are generally cheaper and easier to use, I'd go with C.

If this is correct, I'd appreciate a brainliest.

3 0
2 years ago
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