Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following a
nalog signals without causing aliasing error. (a) f1(t) = f(t) sin(4000πt) (b) f2(t) = f(t) ∗ sin(4000πt) (c) f3(t) = f(t) ∗ f(sample the following analog signals without causing aliasing error
1 answer:
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Answer:
fracture will occur since ( 31.8 Mpa√m ) is greater than the of the material ( 30 Mpa√m )
Explanation:
Given the data in the question;
To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.
On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.
so we first of all solve for the parameter Y in the condition where fracture occurred.
= 30 Mpa√m
σ = 237 MPa
2α = 2.78 mm = 2.78 × 10⁻³ m
so
Y = / σ√πα
we substitute
Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )
Y = (30 Mpa) / (237)( 0.06608187 )
Y = 30 / 15.6614
Y = 1.9155
Next we solve for Yσ√πα for the second case;
σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m
so
Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )
= 1.9155 × 355 × 0.0467269
= 31.8 Mpa√m
so
( 31.8 Mpa√m ) > ( 30 Mpa√m )
Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the of the material ( 30 Mpa√m )
Answer:
Condition A
Heat flux is 1400 W/M^2
Condition B
Heat flux is 12800 w/m^2
Explanation:
Given that:
is given as 30 degree celcius
condition A
Air temperature = - 5 degree c
convection coefficient h = 40 w/m^2. k
condition A
water temperature = 10 degree c
convection coefficient = 800 w/m^2.k