A12.0-g plastic ball is dropped from a height of 2.50 m. just as it strikes the floor, it is moving at a speed of 3.20 m/s. how much mechanical energy did the ball lose during its fall?

0.233 j

0.256 j

0.128 j

0.512 j

0·233 J

Explanation:

Given

Mass of the ball = 0·012 kg

Initially the ball is at a height of 2·5 m

As initially the ball is dropped, it's initial velocity will be equal to 0

Therefore initially it has zero kinetic energy and has only potential energy

∴ Initially total mechanical energy of the ball = potential energy of the ball

Initial potential energy of the ball = m × g × h

where

m is the mass of the ball

g is the acceleration due to gravity

h is the height of the ball

∴ Potential energy = 0·012 × 9·8 × 2·5 = 0·294 J

Velocity of the ball after striking the floor = 3·2 m/s

After striking the floor, the total mechanical energy = kinetic energy just after striking the floor

Kinetic energy = 0·5 × m × v²

where m is the mass of the ball

v is the velocity of the ball

∴ Kinetic energy of the ball = 0·5 × 0·012 × 3·2² = 0·061 J

Mechanical energy that is lost = 0·294 - 0·061 = 0·233 J

∴ Mechanical energy that the ball lost during its fall = 0·233 J

25% unchanged and 75% stable

Explanation:

The half-life of a radioactive sample is the time it takes for half of the sample to decay and become stable.

In this situation, we have:

- Initial sample: 60 g

- After 1 half-life: half of the sample decays --> 30 g decays and becomes stable, while 30 g of the sample are unchanged

- After 2 half-life: half of the unchanged sample (30 g) decays --> so, 15 g decays and becomes stable, so the new stable sample is 15 g + 30 g = 45 g, while the unchanged sample is 15 g

So, after 2 half-lives we have:

- Unchanged sample:

- Stable sample:

So correct answer is

25% unchanged and 75% stable