Multiply. Write the answer in simplest form. 1 3/10×1/8

Consider a continuous-time LTI system whose frequency response is I x sin(4w) H(jw) = -x h(t)e-jwtdt = w If the input to this sy

Helga [31]

Answer:

I believe this is your question.

See the hand worked solution attached.

8 0

3 years ago

In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W

GrogVix [38]

Answer:

The shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

Explanation:

Given:

Temperature $T = 1273.15$ K

Initial Pressure $P_{1} = 1.8$ MPa

Final pressure $P_{2} = 0.1$ MPa

From the table superheated,

$h_{i} = 4635$ $\frac{K J}{Kg}$ and $h_{f} = 2706.54$ $\frac{K J}{Kg}$

Work done by shaft is,

$W = h_{f} - h_{i}$

$W = 2706.54 - 4635$

$W = -1928.46 \frac{kJ}{kg}$

But here efficiency is 0.56,

So work generated per kg is,

Work = $0.56 \times(- 1928.46)$

Work = $-1.3$ $\frac{MJ}{kg}$

Therefore, the shaft work generated per kilogram is $-1.3 \frac{MJ}{kg}$

6 0

3 years ago

Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa a

Elena L [17]

Answer:

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

The entropy change of the carbon dioxide is -0.1104 kJ/kg.K

Explanation:

We are given that carbon dioxide undergoes a process in a closed system.

We are asked to find the entropy change of the carbon dioxide with the assumption that the specific heats are constant.

The entropy change of the carbon dioxide is given by

$S_2 - S_1 = C_p \ln (\frac{T_2}{T_1}) - R\ln (\frac{P_2}{P_1})$

Where Cp is the specific heat constant

Cp = 0.846 kJ/kg.K

R is the universal gas constant

R = 0.1889 kJ/kg.K

T₁ and T₂ is the initial and final temperature of carbon dioxide.

P₁ and P₂ is the initial and final pressure of carbon dioxide.

$S_2 - S_1 = 0.846 \ln (\frac{800}{400}) - 0.1889\ln (\frac{2000}{50})$

$S_2 - S_1 = 0.846(0.69315) - 0.1889(3.6888)$

$S_2 - S_1 = 0.5864 - 0.6968$

$S_2 - S_1 = -0.1104 \: \: kJ/kg.K$

Therefore, the entropy change of the carbon dioxide is -0.1104 kJ/kg.K

6 0

2 years ago

Consider each statement below and determine which are correct concerning dietary fiber. Select all that apply. View Available Hi

PIT_PIT [208]

Answer:

Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. This is correct if you are consuming less than 25-30gms of fiber per day. exceding this limit won't be beneficial.

Consuming a high-fiber diet most likely promotes the health of the digestive system. This is correct. Fibers are important for the digestive system´s health, especially for intestines and colon.

Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. This is Incorrect. Fiber is absorbed and digested at a much slower rate than sugar or starch.

Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. This is incorrect. Consuming a diet high in dietary fiber would decrease the LDL.

Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day. This is incorrect. The data obtained by the University of California San Francisco said that currently the amount of fiber intake by Americans adults is about 15g a day, which is half the recommended amount.

7 0

3 years ago

Read 2 more answers

You get invited to a party at a 1,200 sf apartment, where there are a total of 23 people. The apartment has ceilings that are 8

yawa3891 [41]

Answer:

concentration of air = 0.0000366 ppm is not sufficient to cause any eye irritation

Explanation:

Average amount of aldehyde generated by a cigarette = 1 mg = 10⁻⁶g

Room volume = Height * Base area

Room volume = 8ft * 1200 sq.ft = 9600 cubic feet

number of people in the room = 23

If 1 person smokes 2 cigarette in 1 hour,

23 people will smoke 23*2 cigarettes = 46 cigarettes in 1 hr

Mass of aldehyde generated by 1 cigarette = 10⁻⁶g

Mass of aldehyde generated by 46 cigarettes = 46 * 10⁻⁶g

Based on the law of mass conservation,

Mass of aldehyde coming in = mass of aldehyde going out = Rate of accumulation of aldehyde

Rate of accumulation of aldehyde in the steady state = 0

m*C*V= 0

m = 46 * 10⁻⁶g

C = concentration in the steady state

V = Volumetic flow of the fresh air

V = 600 ft³/min = 600 *60

V = 36000 ft³/hr

Density = Mass/volume

Density = 46 * 10⁻⁶g /36*10³

Density = 1.28 * 10⁻⁹ g/ft³

If the density of air is taken as 36 g/ft³

concentration = (1.28 * 10⁻⁹ g/ft³)/36 g/ft³

concentration = 3.66 * 10⁻¹¹g/g

1*10⁻⁶g/g of air = 1ppm(part per million)

concentration of air =( 3.66 * 10⁻¹¹)(/1*10⁻⁶)

concentration of air = 3.66 * 10⁻⁵ppm = 0.0000366 ppm

concentration of air = 0.0000366 ppm is not sufficient to cause any eye irritation

7 0

2 years ago