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nadya68 [22]
2 years ago
6

“In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set

MINIMUM to X.” Read this STATEMENT very carefully and Answer the below Questions.
A. Will this Algorithm RUN?? Yes or No with a justification

B. Will X Replace Any Value or Not?
Engineering
1 answer:
artcher [175]2 years ago
7 0

Answer:

no i cant answer it sorry

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What is future active and future passive and future perfect active
san4es73 [151]
The future perfect tense forms are made by putting ‘will / shall + have’ before the past participle from the verb. these sentences can be changed into the passive if the active verb has an object


i hope this helps :D thanks
5 0
3 years ago
Two vehicles arrive at an uncontrolled intersection from different streets at the same time 1.The driver on the right must yield
ss7ja [257]
3. Both vehicles must stop
4 0
3 years ago
Subject: Electronics
Maslowich

Answer:

U just believe in yourself ..........

Explanation:

<em>If </em><em>there </em><em>a</em><em>r</em><em>e </em><em>more </em><em>electrons </em><em>than </em><em>protons </em><em>in </em><em>a </em><em>piece </em><em>of </em><em>matter </em><em>it </em><em>will </em><em>have </em><em>a </em><em>negative</em><em> </em><em>charge </em><em>.</em><em> </em><em>i</em><em>f</em><em> </em><em>there </em><em>are </em><em>fever </em><em>it </em><em>will </em><em>have </em><em>positive</em><em> </em><em>charge </em><em>and </em><em>if </em><em>there </em><em>are </em><em>e</em><em>qual </em><em>numbers </em><em>it </em><em>will </em><em>have </em><em>neutral</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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hope it was helpful to you.....

6 0
2 years ago
Which option identifies the next step in the following scenario?
Whitepunk [10]

Answer: The engineer will create a detailed sketch that labels all of the visual components.

Explanation:

It should be noted that the reverse engineering is required for the replacement and the modification of an existing product.

With regards to the question, the correct answer is option A "The engineer will create a detailed sketch that labels all of the visual components".

4 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at
sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

h_{f} = 173.358 \\h_{fg} = 2402.522

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

h_{2a} =  489.752\\h_{2b} =  313.2

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a}  - h_{4a}) - (h_{2a}  - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26  - 2241.448938 ) - (489.752  - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3a} -  h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4a} -  h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.06485

Part b) @ 4 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b}  - h_{4b}) - (h_{2b}  - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14  - 2405.54119 ) - (313.12  - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3b} -  h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4b} -  h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.038275

6 0
3 years ago
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