Answer:
evaluative criteria.
Explanation:
Evaluative criteria is a term in marketing that Is used when a customer decides to go for a different kind of product that is different from the one they originally had in mind because of certain attributes. These attributes could be objective in nature such as power and fuel economy or it could subjective when things like prestige and convenience are being looked at.
Answer:
4/5
Explanation:
She is not wearing white t-shirt on the first day so she is wearing the other 4 t-shirt
Answer:
Diameter will be 27394.76 m
Explanation:
Power P = 0.5 kW = 500 W
Time t required for grinding = 1 hr = 3600 sec
Energy required E = P x t
E = 500 x 3600 = 1800000 J
Flow rate of water Q = 400 ltr/min
We convert to m3/sec
400 ltr/min = 400/(1000 x 60) m3/ses
Q = 0.0067 m3/sec
Energy provided by flow will be
E = pgQd
Where p = density of water = 1000 kg/m3
g = acceleration due to gravity 9.81 m/s2
d = diameter of wheel.
Equating both energy, we have,
1800000 = 1000 x 9.81 x 0.0067 x d
1800000 = 65.73d
d = 1800000/65.73
d = 27394.76 m
Answer:
Two reasons that justify the requirement for security clearance for aerospace engineers working for government agencies are;
1) Such engineers have access to data regarding the blueprint, components, method of construction, functionality status, new systems design, future systems design, inventory of systems and aeronautical systems database which are sensitive information that are of high importance to the federal government
2) Such engineers take part in the testing of aeronautic equipment, and will require security clearance to be able to input data results into the data base of the aeronautic equipment
Explanation:
Answer:
Yes, Otto cycle is more efficient than a Diesel cycle with the same compression ratio.
Explanation:
An ideal Otto cycle models the behavior of an explosion engine. This cycle consists of six steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression:
otto performance= 1 - ( 1/(r^(б-1) ) )
r = compression ratio equal to the ratio between the volume at the beginning of the compression cycle and at the end of it.
б=1.4 exponent
A diesel engine can be modeled with the ideal cycle consisting of six reversible steps, as indicated in the figure. Prove that the performance of this cycle is given by the expression:
diesel performance= 1 - ( 1/(r^(б-1) ) ) × ( (r^б)-1 )/(б×(r-1)) )
The ideal Diesel cycle is distinguished from the ideal Otto in the combustion phase, which in the Otto cycle is assumed at constant volume and in the Diesel at constant pressure. Therefore the performance is different.
We see that the efficiency of a Diesel cycle differs from that of an Otto cycle by the factor in parentheses. This factor is always greater than the unit, therefore, for the same compression reasons r diesel performance is less than ottoperformance.