The mass of 2 kg from problem (12) is released from rest at a distance xo to the right of the equilibrium position. Find the displacement as a function of time, where t = 0 is the instant when the mass is released. B = 42 N-s/m H wwww k = 392 N/m 2 kg
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Q: Needs Complete solution with 100 % accuracy don't use chat gpt or ai plz plz plz plz plz.
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- A skier is standing on a 30o hill. Draw the hill and add idealized force vectors that represent the force of the skier (FS), the normal force (FN), and the force of friction (FF).PART A) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0cm^2? (This is approximately the cross-sectional area of a tibia, or shin bone, at its narrowest point.) Express your answer using two significant figures. PART B) Estimate the maximum height from which a 70kg man could jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be 0.006s, and assume that the stress is distributed equally between his legs. Express your answer using two significant figures. Thank you!!!Needs Complete typed solution with 100 % accuracy. Don't use chat gpt please.
- Consider the motion described by the figure below. (a) Find the displacement at t = 12.4 s(b) Find the displacement at t = 12.9 s(c) Find the displacement at t = 16.5 sFigure x Part A - Moment due to a force specified by magnitude and endpoints F As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) VE ΑΣΦ ↓↑ vec Mo 0.013,97.9, - 196 Submit The force has magnitude F = 140 N and is directed as shown. The dimensions are x₁ = 0.450 m, x₂ = 1.70 m, y₁ = 2.60 m, and z₁ = 1.30 m. B What is the moment about the origin due to the applied force F? Express the individual components of the Cartesian vector to three significant figures, separated by commas. ► View Available Hint(s) Previous Answers X Incorrect; Try Again; 4 attempts remaining 1 of 3 ? i, j, k] N · mThe velocity of a particle is given by v = 29t²-88t - 196, where v is in feet per second and t is in seconds. Plot the velocity v and acceleration a versus time for the first 4 seconds of motion. After you have the plots, answer the questions as a check on your work. -1 Questions: + 0 + 1 When t = 0.5 sec, V = 2 i When t = 3.3 sec, V = i + 3 +s, ft or m When the acceleration is zero, the velocity is i ft/sec, a = ft/sec, a = i i ft/sec. ft/sec² ft/sec²