Problem 22.77The 20-kg block is subjected to the action of the harmonic force F = (90 cos 6t) N,where t is in seconds. (Figure 1)Figurek = 400 N/mWWWk = 400 N/mO 020 kgOOOOF90 cos 6trCc = 125 N.s/m1 of 1Part AWrite the equation which describes the steady-state motion.Express your answer in terms of t. Express your answer using three significant figures. Express the phase in radians and the final result in meters.x =Submit15. ΑΣΦΑProvide FeedbackRequest Answervec?m

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A solid sphere of mass M = 2.95 kg is mounted on a frictionless axle, which goes through its center. The moment of inertia of a solid sphere is given by I = 2MR²/5. A block of mass m = 1.05 kg hangs from a string which is wrapped around the edge of the sphere. The block is released from rest, and it begins to fall. Find the magnitude of the linear acceleration of the falling block, in m/s?. axle M a = ? m
Please explain how was the displacement of the system after 2 seconds was solved to be 5.11 m. (Please explain how was it solved in the picture given.) Question:A water tank on a column has a mass of 4000 kg. The supporting column is assumed to be massless and has a stiffness of 30,000 N/m. If the system is under free vibration where initial displacement is 15 mm and the initial velocity is 20 mm/s, Determine the following: Displacement of the system after 2 seconds
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A:YV 36 ll 46.ll Test 3.pdf F1 F2 a F3 A d. C a F4 E Consider the following values: - F5 a = 12 m; b = 10 m; c = 10 m; d = 8 m; F1 = 6 kN; F2 = 7 kN; F3 = 8 kN; F4 = 9 kN; F5 = 10 kN; a = 30° 0 = 45° B = 60° 1] What is the resultant moment of the five forces acting on the rod about point A? a) - 111.6 kN.m b) 112.6 kN.m c) - 184.6 kN.m d) - 109.6 kN.m e) - 104.1 kN.m f) None of them 2] What is the resultant moment of the five forces acting on the rod about point B? a) -95.8 kN.m b) 42.7 kN.m c) 80.9 kN.m d) 10.9 kN.m e) 51.8 kN.m f) None of them 3] What is the resultant moment of the five forces acting on the rod about point C? a) 32.6 kN.m b) - 21.1 kN.m c) 17.4 kN.m d) - 16.5 kN.m e) 15.6 kN.m f) None of them 4] What is the resultant moment of the five forces acting on the rod about point D? a) 37.6 kN.m b) 73.8 kN.m c) 71.6 kN.m d) 52.1 kN.m e) 21.1 kN.m f) None of them 5] What is the moment of the force F2 about point E? a) 90.7 kN.m b) - 88.1 kN.m c) 54.6 kN.m d) 103.1 kN.m e)…
A spring is stretched 3 cm by a force of 6 N. Thus, according to Hooke's Law, the spring DO = 200 N. Assume as usual, that r = 0 is the equilibrium position. constant is k 0.03 A mass of 8 kg is now attached to the end of the spring, and it is set in motion with initial position xo = 0 and with initial velocity 10 (m) to the left. ( So, v(0) = -10(m).)
For each of the following situations: a. Choose an initial and final state (states A and B) that will be appropriate (interesting) for analysis. b. If there is not already a physical diagram of the situation and there should be one for proper analysis, include one. c. Construct a system schema of the situation. Do not forget to include the system boundary for the system specified by the problem. Choose your system wisely. d. Construct an energy bar chart and system flow diagram (LOL plots) for each problem. e. You are allowed to add other energy storage modes. You do not have to use the energy storage modes that are printed in each LOL plot. f. Find the unknown value.
The 20-kg block is subjected to the action of the harmonic force F = (90 cos 6t) N, where t is in seconds. (Figure 1) Figure k-400 N/m Sw 20 kg k=400 N/m 00 F-90 cos 6r c-125 N-s/m < 1 of 1 Part A Write the equation which describes the steady-state motion. Express your answer in terms of t. Express your answer using three significant figures. Express the phase radians and the final result in meters. ΠΙΑΣΦ 4 ? m
C. Two masses are sliding along a rough surface as shown in the figure. The mass of each box is ma = mB = 10 kg. The masses are connected via a rope wrapped around a (frictionless, massless) pulley so that at the time shown the magnitude of the acceleration of each box is aA = AB = 0.5 m/s². KD: DO NOT SET-UP THE EQUATIONS OR SOLVE THE PROBLEM. KD: mB i. Your friend has chosen to use the rate form of the COLM to solve this problem. Neatly and completely label the system diagrams below to help your friend determine the coefficient of friction between the boxes and the rough surface. ii. List the unique unknowns in these diagrams: Block A 0 = 30° Block B FBD: pulley FBD: MA Block A 7 Block B
weighs 60 N on Earth. (3). A system consists of a pulley (vertical wheel) that is only allowed to rotate (spin). A cord is passed over the pulley and at each end there is a mass attached to the end of the cord. The mass attached on the left is m, =0.6 kg and at the right the mass is m2 = 0.8 kg. a) Calculate both the Tension in the cord and the acceleration of the system. b) What are the accelerations of both masses, if you were to cut the cord? Neglect the friction between cord and wheel,
I need the answer as soon as possible
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Solve it correctly please.I will rate accordingly. Pl

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