Calculate the average kinetic energy for 1. V(x) = x(5 – x) where a € [0,5]
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- m, A particle of mass 2.00 × 10-10 kg is confined in a hollow cubical three-dimensional box, each edge of which has a length, 2.00 × 10-10 and for which the potential energy function is zero inside, and infinite outside, the box. The total energy of the particle is 2.47 × 10-37 J. FindProblem 2: A student pushes a baseball of m = 0.18 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.14 meters from its equilibrium length. The spring constant of the spring is k = 740 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring. Randomized Variables m = 0.18 kg k = 740 N/m d = 0.14 mThe potential energy function for either one of the two atoms in a diatomic molecule is often approximated by U(x) = −a/x12 − b/x6 where x is the distance between the atoms. (a) At what distance of seperation does the potential energy have a local minimum (not at x = ∞)? (b) What is the force on an atom at this separation? (c) How does the force vary with the separation distance?
- Simplify the Boolean sum of products using the Karnaugh map below:Estimate the penetration distance Ax of a small ball of radius r= 10-² m and density p 2650 kg/m³, thrown at a potential barrier at velocity v= 10 m/s. The height of the potential barrier is twice the kinetic energy of the ball. 10-28 m 10-30 m 10-32 m 10-34 m 10-36 m 10-90 m 10-40 m(a) Check all of the following that are correct statements, where E stands for ymc². Read each statement very carefully to make sure that it is exactly correct. O At speeds close to the speed of light, kinetic energy is approximately equal to (1/2)mv². O The energy principle can be written AEsys Wsurr. %3D O The definition of work is W = |I. %3D O The definition K = E - mc² is valid even for speeds near the speed of light. O The definition of work is W = . O The definition of work is W = FxAx + F,AY + F;Az. %3D O The energy principle can be written Esys,f = Esys,i + Wsurr (b) An object with mass 120 kg moved in outer space. When it was at location its speed was 18 m/s. A single constant force N acted on the object while the object moved to location m. What is the speed of the object at this final location? final speed m/s %3D
- Problem 10: A student pushes a baseball of m-8.16 kg down onto the top of a vertical spring that has its lower end fixed a table, compressing the spring a distance of d-0.12 meters from its original equilibrium point. The spring constant of the spring is k-640 N/m. Let the gravitational potential energbe zero at the position of the baseball in the compressed spring. Randomized Variables m-0.16 kg k-640 N/m d-0.12 m. V05 Part (a) The ball is then released. What is its speed, v, in meters per second, just after the ball leaves the spring? Part (b) What is the maximam beight, A, in meters, that the ball reaches above the equilibrium point? Part (c) What is the ball's velocity, in meters per second, at half of the maximum height relative to the equilibrium point?Needs Complete typed solution with 100 % accuracy.(3) The speed of light in vacuum, c, is 3 X 105 km/s and 1 kilometer = 1000 meters. How many Joules of energy are released if 2 kg of matter are converted into energy at 0.1% efficiency? First convert c to units of meters/second: c = 3 x 105 km/s x 1000 m/km = 3 x 108 m/s. Then express efficiency (eff) in decimal terms: eff = 0.001. In this type of problem, the mass/ energy equivalence equation is written as: E = (eff) mc2 Joules,