Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ = 
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ = 
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
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.
Answer:
I did not what is it about?
Answer:
def theRoundTrip(movement):
x=0
y=0
for i in movement:
if i not in ["U","L","D","R"]:
print("bad input")
return
if i=="U":
y+=1
if i=="L":
x-=1
if i=="D":
y-=1
if i=="R":
x+=1
return x==0 and y==0
