The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole
of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)
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Answer:
Explanation:
Regardless of the initial velocity of the pebble, the acceleration of the pebble is equal to the gravitational acceleration which is equal to 9.8 m/s2 towards downwards direction.
This can be shown by Newton's Second Law. According to the law, the net force applied on an object is equal to mass times acceleration of that object.
During the downward motion, the only force acting on the pebble is the gravitational force, hence its acceleration is equal to gravitational acceleration.