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<span>Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m.
</span>
I think the answer is "She can use constructive interference to generate a wave with an amplitude of 1.5 m."
Answer:
Part a)

Part b)

Explanation:
Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

here we also use angular momentum conservation
so we have

also we know that the collision is elastic collision so we have

so we have

also we know

also we know

so we have


now we have


Part b)
Now we know that speed of the ball after collision is given as

so it is given as

A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?
Competition in the Los Angeles Flower District results in better quality flowers.
<h3>Why quality is the standard in flower competition?</h3>
Competition results in better quality flowers because in the competition, best quality of flowers will be selected as a winner so the competitors produces best quality of flowers in order to claim the prize so we can conclude that Competition in the Los Angeles Flower District results in better quality flowers
Learn more about competition here: brainly.com/question/25605883