Answer:
D
Explanation:
To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.
![\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D%26Cost%5C%20per%5C%20day%5C%20%28%5C%24%29%26Time%5C%20to%5C%20complete%5C%20%28days%29%26Total%5C%20cost%5C%20%28%5C%24%29%5C%5CZoe%26500%268%264000%5C%5CGreg%26650%2610%266500%5C%5COrion%26400%2612%264800%5C%5CJin%26700%265%263500%5Cend%7Barray%7D%5Cright%5D)
As you can see, Greg is the least cost-effective because he charges the most for the project.
Answer:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Explanation:
Calculation to estimate the upper and lower bounds of the modulus of this composite.
First step is to calculate the maximum modulus for the combined material using this formula
Modulus of Elasticity for mixture
E= EcuVcu+EwVw
Let pug in the formula
E =( 110 x 0.40)+ (407 x 0.60)
E=44+244.2 GPa
E=288.2GPa
Second step is to calculate the combined specific gravity using this formula
p= pcuVcu+pwTw
Let plug in the formula
p = (19.3 x 0.40) + (8.9 x 0.60)
p=7.72+5.34
p=13.06
Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness
UPPER BOUNDS
Using this formula
Upper bounds=E/p
Let plug in the formula
Upper bounds=288.2/13.06
Upper bounds=22.07 GPa
LOWER BOUNDS
Using this formula
Lower bounds=EcuVcu/pcu+EwVw/pw
Let plug in the formula
Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3
Lower bounds=(44/8.9)+(244.2/19.3)
Lower bounds=4.94+12.65
Lower bounds=17.59 GPa
Therefore the Estimated upper and lower bounds of the modulus of this composite will be:
Upper bounds 22.07 GPa
Lower bounds 17.59 GPa
Answer:
minimum flow rate provided by pump is 0.02513 m^3/s
Explanation:
Given data:
Exit velocity of nozzle = 20m/s
Exit diameter = 40 mm
We know that flow rate Q is given as
where A is Area
minimum flow rate provided by pump is 0.02513 m^3/s
Answer:
Fuel efficiency for highway = 114.08 miles/gallon
Fuel efficiency for city = 98.79 miles/gallon
Explanation:
1 gallon = 3.7854 litres
1 mile = 1.6093 km
Let's first convert the efficiency to km/gallon:
48.5 km/litre = (48.5 * 3.7854) km/gallon
48.5 km/litre = 183.5919 km/gallon (highway)
42.0 km/litre = (42.0 * 3.7854) km/gallon
42.0 km/litre = 158.9868 km/gallon (city)
Next, we convert these to miles/gallon:
183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon
183.5919 km/gallon = 114.08 miles/gallon (highway)
158.9868 km/gallon = (158.9868 /1.6093) miles/gallon
158.9868 km/gallon = 98.79 miles/gallon (city)
