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The trapeze/lanyard air drop
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- Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of 34 m. During the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 6.0 ms. (Assume that neither the passenger nor the cab rebounds.) (a) What is the magnitude of the impulse on the passenger during the collision? N's (b) What is the magnitude of the average force on the passenger during the collision? N Suppose the passenger were to jump upward with a speed of 6.5 m/s relative to the cab floor just before the cab hits the bottom of the shaft. Assume the same stopping time as above. (c) What is the magnitude of the impulse on the passenger during the collision? N•s (d) What is the magnitude of the average force on the passenger during the collision? Narrow_forwardA 35 g steel ball is held by a ceiling-mounted electromagnet 3.4 m above the floor. A compressed-air cannon sits on the floor, 5.0 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a 25 g plastic ball. The two balls collide 1.2 m above the floor. What was the launch speed of the plastic ball?arrow_forwardNote: Give a clear, step-by-step handwritten calculation (without any explanations), completed entirely without using AI. The solution should be of expert quality, as I will evaluate and rate it based on its precision and overall quality. Be sure to carefully double-check all work for accuracy before submission. Question: In Figure 5, a 0.1-kg projectile is launched into a 10-kg block, initially at rest with an uncompressed spring secured at both ends. Calculate the maximum horizontal displacement, X, of the spring and the resulting oscillation period of the block and the embedded projectile.arrow_forward
- Please show me the steps for a,b,c,darrow_forwardHi! Please help me answer Problem 1, Questions 4 and 5. Question 1-3 have been answered by another tutor. Problem 1, Questions 4 and 5. Problem 1, Questions 4 and 5. Problem 1, Questions 4 and 5.arrow_forwardOblique Impact To analyze an oblique impact using the conservation of momentum and coefficient of restitution. When an oblique impact occurs between two smooth particles, the particles move away from each other with velocity vectors that have unknown directions and unknown magnitudes. If the y axis is within the plane of contact and the x axis is the line of impact, the impulsive forces of deformation and restitution act only along the line of impact (the x axis). Momentum of the system is conserved along the line of impact (the x axis): ∑m(vx)1=∑m(vx)2 The coefficient of restitution, e, relates the relative-velocity components of the particles along the line of impact (the x axis): e=(vBx)2−(vAx)2(vAx)1−(vBx)1 The momenta of both particles A and B are conserved in the plane of contact (the y axis) because no impulse acts on either particle in this plane. Therefore, the y component of the velocities before and after the collisions remains unchanged: (vy)1=(vy)2 As shown,…arrow_forward
- Oblique Impact To analyze an oblique impact using the conservation of momentum and coefficient of restitution. When an oblique impact occurs between two smooth particles, the particles move away from each other with velocity vectors that have unknown directions and unknown magnitudes. If the y axis is within the plane of contact and the x axis is the line of impact, the impulsive forces of deformation and restitution act only along the line of impact (the x axis). Momentum of the system is conserved along the line of impact (the x axis): ∑m(vx)1=∑m(vx)2 The coefficient of restitution, e, relates the relative-velocity components of the particles along the line of impact (the x axis): e=(vBx)2−(vAx)2(vAx)1−(vBx)1 The momenta of both particles A and B are conserved in the plane of contact (the y axis) because no impulse acts on either particle in this plane. Therefore, the y component of the velocities before and after the collisions remains unchanged: Immediately after the collision,…arrow_forwardIf either a simple or a compound pendulum is used to determine experimentally the acceleration of gravity g, difficulties are encountered. In the case of the simple pendulum, the string is not truly weightless, while in the case of the compound pendulum, the exact location of the mass center is difficult to establish. In the case of a compound pendulum, the difficulty can be eliminated by using a reversible, or Kater, pendulum. Two knife edges A and B are placed so that they are obviously not at the same distance from the mass center G, and the distance l is measured with great precision. The position of a counterweight D is then adjusted so that the period of oscillation τ obtained is equal to that of a true simple pendulum of length l and that g 4π2 l/τ.arrow_forwardA small block (block 1) of mass m, attached to an ideal string of length L, is initially held so that the string is taut and horizontal at height Labove a frictionless table, as shown in Fig.2. A second small block (block 2) of mass 3mis placed on the table right under the point of attachment of the string holding block 1. The tabletop is at height 2L above the floor. Block 1 is then released from rest and collides elastically with block 2 at time t=0. The acceleration due to gravity has magnitude g and air resistance is negligible. a) Determine the horizontal distance traveled by block 2as a projectile. b)Determine the speed and direction of motion of block 1 right after the collision. c) Determine the maximum angular displacement (measured from the vertical direction) of block 1 after the collision.arrow_forward
- A small block (block 1) of mass m, attached to an ideal string of length L, is initially held so that the string is taut and horizontal at height Labove a frictionless table, as shown in Fig.2. A second small block (block 2) of mass 3mis placed on the table right under the point of attachment of the string holding block 1. The tabletop is at height 2L above the floor. Block 1 is then released from rest and collides elastically with block 2 at time t=0. The acceleration due to gravity has magnitude g and air resistance is negligible. a)Determine the time t2at which block 2 hits the floor. b) Determine the horizontal distance traveled by block 2as a projectile. c)Determine the speed and direction of motion of block 1 right after the collision.arrow_forward4. When only conservative forces are acting on a system, we can use the conservation of energy theorem to solve a problem. Recall from particle kinetics: T₁ + V₁ = T2 + V2, where V is potential energy (due to both gravitational and elastic forces: V₂ = Wyc and Ve= + ½ ks², respectively). The 30-kg rod is released from rest when 0 = 45°. The spring is unstretched when 0 = 45°. Using conservation of energy theorem, determine the angular velocity of the rod when 0 = 0°. B 1.5 m k = 300 N/m Aarrow_forwardChoose the correct answer of the following questions: 1. If no external impressed force acts on the system, the total momentum (G) of a system ... c. remains constant a. increases b. decreases d. none of the these 2. Which of the following cases momentum is conserved? (1 M) a. Perfectly elastic impact d. Momentum is always conserved c. Perfectly inelastic impact b. plastic impact with 0arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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