Answer:
F = 0.0022N
Explanation:
Given:
Surface area (A) = 4,000mm² = 0.004m²
Viscosity = µ = 0.55 N.s/m²
u = (5y-0.5y²) mm/s
Assume y = 4
Computation:
F/A = µ(du/dy)
F = µA(du/dy)
F = µA[(d/dy)(5y-0.5y²)]
F = (0.55)(0.004)[(5-1(4))]
F = 0.0022N
Answer:
2.46 * 10⁵ W/m³
Explanation:
See attached pictures for detailed explanation.
Answer:
The percentage ductility is 35.5%.
Explanation:
Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.
Step1
Given:
Diameter of shaft is 10.2 mm.
Final area of the shaft is 52.7 mm².
Calculation:
Step2
Initial area is calculated as follows:
A = 81.713 mm².
Step3
Percentage ductility is calculated as follows:
D = 35.5%.
Thus, the percentage ductility is 35.5%.
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
