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solniwko [45]
3 years ago
14

Based on these statements:

Engineering
1 answer:
Evgen [1.6K]3 years ago
7 0

Answer:

the third statement is true

Explanation:

given data

Lenovos cost more than Dells

Lenovos cost less than Apples

solution

we have given 1st statement that is express as

cost (Lenovo) > cost (Dell)     ..................1

and

2nd statement that is express as

cost (Lenovo) < cost (Apple)

so we can say it as

cost (Apple) > cost (Lenovo)       ......................2

and

now above Both equation 1 and 2 can be written as

cost (Apple) > cost (Lenovo) > cost (Dell)      .........................3

so we can say cost of Apples is more than the cost of Lenovos and the cost of Dells

so as that given 3rd statement is true

You might be interested in
____ grinders are used to grind diameters, shoulders, and faces much like the lathe is used for turning, facing, and boring oper
skelet666 [1.2K]

Answer:

Cylindrical

Explanation:

<em>A cylindrical grinder </em><em>is a tool for shaping the exterior of an item. Although cylindrical grinders may produce a wide range of forms, the item must have a central axis of rotation. Shapes such as cylinders, ellipses, cams, and crankshafts are examples of this.</em><em> Cylindrical grinding</em><em> machines are specialized grinding machines that are used to process cylinders, rods, and similar workpieces. The cylinders revolve in one direction between two centers, while the grinding wheel or wheels are close together and rotate in the other direction.</em>

8 0
2 years ago
Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl
Ulleksa [173]

Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio \frac{v_1}{v_2}  = 10

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

\frac{p_1}{p_2}  = (\frac{v_2}{v_1} )^n Therefore p₂ = p₁ ÷ (\frac{v_2}{v_1} )^n = (1 bar) ÷ (\frac{1}{10} )^{1.3} = 19.953 bar

Similarly for a polytropic process we have

\frac{T_1}{T_2}  = (\frac{v_2}{v_1} )^{n-1} or T₂ = T₁ ÷ (\frac{v_2}{v_1} )^{n-1} = \frac{310}{0.1^{0.3}} = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = \frac{8.3145}{28.9628} = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R =  1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = \frac{R(T_2-T_1)}{n-1} = \frac{0.287(618.531-310)}{1.3 - 1}= 295.16 kJ/kg

Q =(\frac{n-\gamma}{\gamma - 1} )W = (\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = \frac{R(T_4-T_3)}{n-1} = Where T₄ is given by  \frac{T_4}{T_3}  = (\frac{v_3}{v_4} )^{n-1} or T₄ = T₃ ×0.1^{0.3}

= 2200 ×0.1^{0.3}  T₄ = 1102.611 K

W =  \frac{0.287(1102.611-2200)}{1.3 - 1}= -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

\eta = 1-\frac{T_4-T_1}{T_3-T_2} =1-\frac{1102.611-310}{2200-618.531} = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

p_{m}  = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}  where r = compression ratio and r_p = \frac{p_3}{p_2}

However p₃ = \frac{p_2T_3}{T_2} =\frac{(19.953)(2200)}{618.531} =70.97 atm

r_p = \frac{p_3}{p_2} = \frac{70.97}{19.953}  = 3.56

Therefore p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9} = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

8 0
3 years ago
Scientific research techniques are used to analyze the effectiveness of political advertising. False True
miv72 [106K]

Answer:

correct me if i'm wrong but i think it's false

Explanation:

5 0
3 years ago
Explain the four criteria for proving the correctness of a logical pretest loop construct of the form "while B do S end". And pr
evablogger [386]

Answer:

Check the explanation

Explanation:

The loop invariant has to satisfy some amount of requirements to be of good use. Another complex factor as to why a loop is the question of loop termination. A loop that doesn’t terminate can’t invariably be correct, and in fact the computation in whatever form amounts to nothing. The total axiomatic description of a while construct will have to involve all of the following to be true, in which I is the loop invariant:

P => I

{I and B} S {I}

(I and (not B)) => Q

Then the loop terminates

6 0
3 years ago
Describe ICP/OES in detail.
alex41 [277]

Answer:

ICP -OES stand for inductively coupled plasma optical emission spectroscopy

Explanation:

It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.

The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.

7 0
3 years ago
Read 2 more answers
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