We've seen that stout tendons in the legs of hopping kangaroos store energy. When a kangaroo lands, much of the kinetic energy of motion is converted to elastic energy as the tendons stretch, returning to kinetic energy when the kangaroo again leaves the ground. If a hopping kangaroo increases its speed, it spends more time in the air with each bounce, but the contact time with the ground stays approximately the same. Explain why you would expect this to be the case. Drag the terms on the left to the appropriate blanks on the right to complete the sentences._ stays the same) [doesn't depend on the amplitude] [decreases) is in inverse proportion with the amplitude is in direct proportion with the amplitude [increases) When in contact with the ground it is like a spring in simple harmonic motion. When kangaroo is hopping faster, the amplitude of the oscillation , while the period because it , hence the time in contact with the ground

Explain vector addition, triangle method and parallelogram method

A23 kg log of wood begins from rest, 300 m up a sluice (a water track used to transport logs, think of it as an inclined plane with negligible friction) inclined at 20° to the horizontal. after it reaches the flat waterway at the bottom it collides elastically with a 100 kg block of wood initially at rest near the base of the incline. (a) how long does it take the 23 kg log to travel down the incline? (b) what is the speed of the 23 kg log at the bottom of the incline? (c) what are the velocities of both blocks of wood after the collision? (d) what is the total kinetic energy before the collision? (e) what is the kinetic energy of the 23 kg log after the collision? (f) what is the kinetic energy of the 100 kg block of wood after the collision? (g) what is the total kinetic energy after the collision? (h) compare the total initial and total final kinetic energies. is this consistent with what you would expect for elastic collisions? explain!

In a good thermo flask the main cause of heat loss is