Answer:
number of pulses produced = 162 pulses
Explanation:
give data
radius = 50 mm
encoder produces = 256 pulses per revolution
linear displacement = 200 mm
solution
first we consider here roll shaft encoder on the flat surface without any slipping
we get here now circumference that is
circumference = 2 π r .........1
circumference = 2 × π × 50
circumference = 314.16 mm
so now we get number of pulses produced
number of pulses produced = 
× No of pulses per revolution .................2
number of pulses produced = 
× 256
number of pulses produced = 162 pulses
Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
Answer:
Zero 1 = -1
Zero 2 = -3
Pole 1 = 0
Pole 2 = -2
Pole 3 = -4
Pole 4 = -6
Gain = 4
Explanation:
For any given transfer function, the general form is given as
T.F = k [N(s)] ÷ [D(s)]
where k = gain of the transfer function
N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.
D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.
k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]
it is evident that
Gain = k = 4
N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)
= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)
The zeros are -1 and -3
D(s) = s⁴ + 12s³ + 44s² + 48s
= s(s³ + 12s² + 44s + 48)
= s(s + 2)(s + 4)(s + 6)
The roots are then, 0, -2, -4 and -6.
Hope this Helps!!!
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution
