Did not engineer cables factoring wind shear
Answer:
0.023 Pa*s
Explanation:
The surface area of the side of the inner cylinder is:
A = π*d*l
A = π*0.15*0.75 = 0.35 m^2
At 200 rpm the inner cylinder has a tangential speed of:
u = w * r
u = w * d/2
w = 200 rpm * 2π / 60 = 20.9 rad/s
u = 20.9 * 0.15 / 2 = 1.57 m/s
The torque is of 0.8 N*m, this means that the force is:
T = F * r
F = T / r
F = 2*T / d
For Newtoninan fluids with two plates moving respect of each other with a fluid between the viscous friction force would be:
F = μ*A*u / y
Where
μ: viscocity
y: separation between pates
A: surface area of the plates
Then:
2*T / d = μ*A*u/y
Rearranging:
μ = 2*T*y / (d*A*u)
μ = 2*0.8*0.0012 / (0.15*0.35*1.57) = 0.023 Pa*s
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=