Answer:
A) while (num >= 0)
Explanation:
To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.
A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.
B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0
C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result
D) The same problem than C
E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result
The demand curve is the graphical representation of the relationship between the price of a good and the quantity demanded for a given period of time.
<h3>What is a demand schedule?</h3>
A demand schedule is a table which shows the quantity demanded of a good or service at different price levels.
A demand schedule can be graphed as a continuous demand curve on a chart where the Y-axis represents the price and the X-axis represents quantity.
Here, a typical representation, the price will appear on the left vertical axis, the quantity demanded on the horizontal axis.
Note that the complete information wasn't found and an overview was given.
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Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.
<h3>What is the
angular momentum?</h3>
The angular momentum of the device earlier than and after the clay sticks can be the same.
Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.
- Li = Lf where;
- li is the preliminary second of inertia
- If is the very last second of inertia
- wi is the preliminary angular velocity
- wf is the very last angular velocity
- Li is the preliminary angular momentum
- Lf is the very last angular momentum
Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.
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