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:
P = 80.922 KW
Explanation:
Given data;
Length of load arm is 900 mm = 0.9 m
Spring balanced read 16 N
Applied weight is 500 N
Rotational speed is 1774 rpm
we know that power is given as
T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm
angular speed 
Therefore Power is
P = 80.922 KW
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Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
