Three capillary tubes with different radius (r1=1.0mm, r2=0.1mm, r3=0.01mm) are inserted into the same cup of water. The surface
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Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
Answer:
Part A:
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
CPI reduced by =80%
Part C:
New Execution Time=
Increase in speed=
Explanation:
FP Instructions=50*106=5300
INT Instructions=110*106=11660
L/S Instructions=80*106=8480
Branch Instructions=16*106=1696
Calculating Execution Time:
Execution Time=
Execution Time=
Execution Time=
Part A:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI reduced by =80%
Part C:
New Execution Time=
New Execution Time=
Increase in speed=