Answered: A system has the characteristic… | bartleby

Related questions

Question

A system has the characteristic equation
q(s) = s + 4KS² + (5+K)s + 10 = 0
The range of K for a stable system is:
O 0 < K < 0.46
O K< 0.46
O K> 0.46
O Unstable for all values of K

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

Physics 1 523 er % (9/12) on 100% (4/4) Work & Energy OPEN Sanal Sin My Drive TurboTax co c to Probability Unversity Physics V equating 31 WEBSplach Search A shown in the picture, a 223 g mass slides down a curved incline then collides with a spring. The spring constant of the spring is 164 N/m. Assume that friction is negligble in these problems. Iam finished (A) If the mass starts from rest at a height of 1.07 cm, what is the final compression of the spring when the mass comes to rest? Assume the spring is initially uncompressed. 1.69 cm (B) If the mass initially compresses the spring 1.50 cm, what is the maximum height the mass rises to on the incline? Assume the mass is released from rest. 0.95 cm (C) If the mass is released from rest at a height of 1.07 cm, what is the compression of the spring when the mass has a speed of 33 cm/s? 1.17 cm 2:00 PM //2023
Given the internal energy U and entropy S of N weakly interacting particles in a closed system with fixed volume V. U = NkgT² (27In 2) U S = Nkg lnz + T (a) Prove the Helmholtz free energy (b) Prove the Pressure of the system is F = -NkBT ln z P = Nk Tln z) ( T
Use the e-S definition of infinite limits to prove the statement. lim -00 %3D X - X→5 f(x) 1 is defined for all x # Let N 0 such that f(x) ---Select--- V < ---Select--- v Thus lim = -0, |x - 5| х — 5 |x – 5| X→5- X
Find C please step by step .Minimal use of the calculator please
Problem 6: A uniform wooden meter stick has a mass of m = 763 g. A clamp can be attached to the measuring stick at any point P along the stick so that the stuck can rotate freely about point P, which is at a distance d from the zero-end of the stick as shown. Randomized Variables m = 763 g 1 2 3045 6 7 12 13 14 15 16 17 18 10 11 Otheexpertta.com
How to solve this question
Suppose that ak > 0 for all k e N and E ak < 0. For each of the following, prove that the given series converges. ak ( a ) ΣΕΙ 1+ k3ak (b) Lk=1 1+ ak Vak (c) Ek=1 k < 0o.
Which of the following is/are correct for the equation y(x) dx defined for a particle whose state function is y(x) (11) (iii) This equation gives the probability of the particle with the range x to X₂. This equation applies to the particle moving in any dimension. This equation defines relation between the state function and the probability with the range x; to x₂- (a) Only (1) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (ii)
The following set up represents an isolated skeletal muscle with an attached 0.5g weight. The muscle is stimulated once with 8.5V, and the resulting distance the weight moves over time is recorded. Which of the following can you conclude based ONLY on these results? E Simulated Ruler (mm) 5g Distance (mm) 17 N 0 20 40 The muscle contracted isometrically The muscle experienced fatigue after 80 msec The muscle generated maximal tetanic tension at 40 msec The muscle contracted isotonically 100 120 140 160 180 200 Time (msec)
Four different particles are trapped in one dimensional wells with infinite potential at the walls and zero potential inside. The masses of the particles and their energy level, n , are given by: 1. mass=m, n=1 2. mass=2m, n=2 3. mass =3m, n=3 4. mass=4m, n=2 All these wells have the same length L. Rank the kinetic energy of the particles in order of size, smallest to largest. Group of answer choices E3 <E1<E2<E4 E1 = E4 < E2 < E3 E4 <E3 <E2 = E1 E1 < E2 = E4 < E3 E1 < E2 = E3 <E4
Region 1 is x 0 with &2 = 460. If E2 = 6a - 10a, + 8a: V/m, (a) find P1, and P2, (b) calculate the energy densities in both regions.
24. Consider a modified box potential with V(x) = V₁x, Vi(ar), x a Use the orthogonal trial function = c₁f₁+c₂f₂ with f₁ = √√sin (H) and f2 = √√ √√sin sin (2) to determine the upper bound to ground state energy.