Answer:
The minimum diameter for each cable should be 0.65 inches.
Explanation:
Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,
(1/2)(Weight/Cross Sectional Area) = Allowable Stress
Here,
Weight = 1000 lb
Cross-sectional area = πr²
where, r = minimum radius for each cable
(1/2)(1000 lb/πr²) = 1500 psi
500 lb/1500π psi = r²
r = √1.061 in²
r = 0.325 in
Now, for diameter:
Diameter = 2(radius) = 2r
Diameter = 2(0.325 in)
<u>Diameter = 0.65 in</u>
Answer:
The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)
Explanation:
Since the given volume flow rate is gallons per minute.
We know that 1 gallon = 3.785 liters and
1 minute = 60 seconds
Let the flow rate be 
Now replacing the gallon and the minute by the above values we get
Thus 
Now since we know that 1 liter = 
Using this in above relation we get
From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.
The response to whether the statements made by both technicians are correct is that;
D: Neither Technician A nor Technician B are correct.
<h3>Radio Antennas</h3>
In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.
Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.
Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.
Read more about Antennas at; brainly.com/question/25789224
Answer:
5,4,1, this is a explication
Answer:
Machine 2 has a higher process capability index, it would be best considered for purchase.
Explanation:
Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]
For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7
Cpk= [0.952;0.952]= 0.952
For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3
Cpk= [1.111;3.333]= 1.111
It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.
Since machine 2 has a higher process capability index, it would be best considered for purchase.
