What should always be done before beginning any diagnosis?

g The function below takes a single string parameter: sentence. Complete the function to return everything but the middle 10 cha

ale4655 [162]

Answer:

def processString(sentence):
middle = len(sentence) // 2
output = sentence[0: middle - 5] + sentence[middle+5:]
return output
print(processString("I have a dream"))

Explanation:

Create a function processString that take sentence as input paratemer (Line 1).

Next create a variable middle to hold the value of middle index of sentence string (Line 2)

Build the output string by slicing the input sentence from first character to character middle - 5th and from middle + 5th till the end of the string (Line 3).

Test the function using a sample sentence and we shall get the output "I am"

4 0

2 years ago

Consider a normal shock wave in air. The upstream conditions are given by M1=3, p1 = 1 atm, and r1 = 1.23 kg/m3. Calculate the d

mart [117]

Answer and Explanation:

The answer is attached below

7 0

2 years ago

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which

Westkost [7]

Answer:

Entropy generation rate of the two reservoirs is approximately zero ( $\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$ ) and system satisfies the Second Law of Thermodynamics.

Explanation:

Reversible heat pumps can be modelled by Inverse Carnot's Cycle, whose key indicator is the cooling Coefficient of Performance, which is the ratio of heat supplied to hot reservoir to input work to keep the system working. That is:

$COP_{H} = \frac{\dot Q_{H}}{\dot W}$

The following simplification can be used in the case of reversible heat pumps:

$COP_{H,rev} = \frac{T_{H}}{T_{H} - T_{L}}$

Where temperature must written at absolute scale, that is, Kelvin scale for SI Units:

$COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}$

$COP_{H, rev} = 17.479$

Then, input power needed for the heat pump is:

$\dot W = \frac{\dot Q}{COP_{H,rev}}$

$\dot W = \frac{300\,kW}{17.749}$

$\dot W = 16.902\,kW$

By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

$-\dot Q_{H} + \dot W + \dot Q_{L} = 0$

$\dot Q_{L} = \dot Q_{H} - \dot W$

$\dot Q_{L} = 300\,kW - 16.902\,kW$

$\dot Q_{L} = 283.098\,kW$

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

$\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0$

$\dot S_{gen} = \dot S_{in} - \dot S_{out}$

$\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}$

$\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}$

$\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}$

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, $\dot S_{in} = \dot S_{out}$ .

5 0

2 years ago

Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Un

Bogdan [553]

Answer:

Answer for the question:

Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Under which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?

Is given in the attachment.

Explanation: