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Artemon [7]
3 years ago
12

7. The binary addition 1 + 1 + 1 gives ​

Engineering
2 answers:
scoray [572]3 years ago
7 0
1+1+1=3 should be the answer right?
kenny6666 [7]3 years ago
4 0

11 [2-bit]

011 [3-bit]

0011 [4-bit]

________

1 + 1 + 1 = 3

________

3 = 2 + 1

2¹ 2⁰

3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)

3 = ..011

Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.

[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]

[ 128 64 32 16 8 4 2 1 ]

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Read the passage.
Archy [21]

The claim being made in in the above passage is that " It makes financial sense to stop using the penny." (Option B)

<h3>What textual evidence backs up the above claim?</h3>

The textual evidence that supports the above claim is "Not only does it make financial sense to take the penny out of circulation, but it also makes environmental sense." [Para. 2]

Textual evidence is evidence related to a text which supports claims made in such a text.

Learn more about claims at:
brainly.com/question/2748145
#SPJ1

6 0
2 years ago
In homes today, what is behind the reason for flashover fires occurring much more rapidly than in the past generations?
garik1379 [7]

Answer:

One of the reasons why flashover fires are more prevalent today than it was in the past is that homes and furniture today are made from materials that are far more combustible than those of previous years.

Explanation:

A flashover fire is the rapid ignition and combustion of all flammable materials in an enclosed vicinity in a very short period of time.

Thirty years ago, the average escape time from a house that was on fire is about sixteen and fifty seconds...that would be approximately seventeen minutes. Presently that figure is down to four minutes.

One of the reasons identified is that the internal and external appurtenances especially furniture in use today are more combustible than those of previous years. That is, as they burn, they produce more heat and disintegrate faster.

The reason identified for this is, old houses were made of more natural materials such as real wood etc whilst the furniture and curtains in modern houses are mostly from synthetic materials.

Cheers

6 0
3 years ago
Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

6 0
4 years ago
What process is used to remove collodal and dissolved organic matter in waste water ​
Juli2301 [7.4K]

Answer:

Aerobic biological treatment process

Explanation:

Aerobic biological treatment process in which micro-organisms, in the presence of oxygen, metabolize organic waste matter in the water, thereby producing more micro-organisms and inorganic waste matter like CO₂, NH₃ and H₂O.

3 0
3 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

4 0
3 years ago
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