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:
Mass, in physics, quantitative measure of inertia, a fundamental property of all matter.
Explanation:
Mass is the matter that makes up objects
Answer:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
