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:
N = 38546.82 rpm
Explanation:
= 150 mm

= 17671.45 
= 250 mm

= 49087.78 
The centrifugal force acting on the flywheel is fiven by
F = M (
-
) x
------------(1)
Here F = ( -UTS x
+ UCS x
)
Since density, 





∴
-
= 50 mm
∴ F = 
F = 33618968.38 N --------(2)
Now comparing (1) and (2)

∴ ω = 4036.61
We know


∴ N = 38546.82 rpm
Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
extension lines,sketches,leader lines,dimensions describes all illustrations created by freehand.
Answer: change the tires
Explanation: you can’t drive on a flat tire