Berkeley Review Physics Section 5 Passage VI question 42

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So, does anyone have any idea why in section V, passage VI p 259, number 42, v=wavelength*freq. is used instead of the equation give for v in the passage? B/c assuming that the wavelength is 50 (1/3 of 150), 50/4= 12-ish if the general velocity equation is used. Using the one from the passage v=sqrt(g*wavelength/8pi) sqrt(50*10/24)= sqrt(500/25)=sqrt(20)= between 4 and 5.

Does this question have something to do between the differentiation between speed of a wave and speed of waves (like speed of a single wave crashing, versus speed of wave propagation?) How would one be able to pick that up from the question/passage (without extremely detailed analysis?)

 

[anon4895]

So, does anyone have any idea why in section V, passage VI p 259, number 42, v=wavelength*freq. is used instead of the equation give for v in the passage? B/c assuming that the wavelength is 50 (1/3 of 150), 50/4= 12-ish if the general velocity equation is used. Using the one from the passage v=sqrt(g*wavelength/8pi) sqrt(50*10/24)= sqrt(500/25)=sqrt(20)= between 4 and 5.

Does this question have something to do between the differentiation between speed of a wave and speed of waves (like speed of a single wave crashing, versus speed of wave propagation?) How would one be able to pick that up from the question/passage (without extremely detailed analysis?)

Any way you can post the passage? At the very least can you just post the equation it gives (I assume the equation you wrote is after you already started filling in values).

 

[donnnnay]

Hey, not to thread jack, but in the same book:

Passage 3, Question 20:

If the raindrop freezes into a round hailstone, how does its new terminal velocity compare with the terminal velocity it would have had if it had remained a raindrop?

A. vt stays the same
B. vt gets smaller
C. vt gets bigger
D. The relation depends upon how m changes

I put C and my thought was that as the raindrop freezes its density decreases (H2O liquid --> H2O solid), therefore a decrease in density is an increase in vt. Book says B, I'm confused. If anyone can help in the same book while answering OP I'd appreciate it.

 

[nashaiy]

Hey, not to thread jack, but in the same book:

Passage 3, Question 20:

If the raindrop freezes into a round hailstone, how does its new terminal velocity compare with the terminal velocity it would have had if it had remained a raindrop?

A. vt stays the same
B. vt gets smaller
C. vt gets bigger
D. The relation depends upon how m changes

I put C and my thought was that as the raindrop freezes its density decreases (H2O liquid --> H2O solid), therefore a decrease in density is an increase in vt. Book says B, I'm confused. If anyone can help in the same book while answering OP I'd appreciate it.

Hmm, I'll take a stab at it. You're right in that the density of the droplet decreases but why? The mass stays the same so the volume must increase when it freezes. If the volume has increased, the surface area of the raindrop has also increased since it is assumed to be perfectly spherical. Since we are not told to ignore air resistance, the air resistance on a drop with a larger surface area is higher than the air resistance on a drop with a smaller surface area. Since force of air resistance is higher for the hail, the terminal velocity is smaller since F(net) = ma = mg - F(air resistance). I hope that makes sense.

 

[anon4895]

Hey, not to thread jack, but in the same book:

Passage 3, Question 20:

If the raindrop freezes into a round hailstone, how does its new terminal velocity compare with the terminal velocity it would have had if it had remained a raindrop?

A. vt stays the same
B. vt gets smaller
C. vt gets bigger
D. The relation depends upon how m changes

I put C and my thought was that as the raindrop freezes its density decreases (H2O liquid --> H2O solid), therefore a decrease in density is an increase in vt. Book says B, I'm confused. If anyone can help in the same book while answering OP I'd appreciate it.

It's the density of the air that's important. If the air is less dense then vt will increase. The density of the raindrop/hailstone will not effect things.

When the rain drop freezes it will expand causing a larger cross sectional area causing higher drag causing vt to get smaller.

 

[BerkReviewTeach]

So, does anyone have any idea why in section V, passage VI p 259, number 42, v=wavelength*freq. is used instead of the equation give for v in the passage? B/c assuming that the wavelength is 50 (1/3 of 150), 50/4= 12-ish if the general velocity equation is used. Using the one from the passage v=sqrt(g*wavelength/8pi) sqrt(50*10/24)= sqrt(500/25)=sqrt(20)= between 4 and 5.

Does this question have something to do between the differentiation between speed of a wave and speed of waves (like speed of a single wave crashing, versus speed of wave propagation?) How would one be able to pick that up from the question/passage (without extremely detailed analysis?)

This is a typical bait-and-switch question. The passage gives you an equation for deep-ocean waves and then asks you about a boat at the top of the ocean, where the simple equation applies, and not the complex one given in the passage. It's a very common physical sciences trick to give you excessive information in the passage followed by a few simple questions where you don't need the passage.

 

[BerkReviewTeach]

Hey, not to thread jack, but in the same book:

Passage 3, Question 20:

If the raindrop freezes into a round hailstone, how does its new terminal velocity compare with the terminal velocity it would have had if it had remained a raindrop?

A. vt stays the same
B. vt gets smaller
C. vt gets bigger
D. The relation depends upon how m changes

I put C and my thought was that as the raindrop freezes its density decreases (H2O liquid --> H2O solid), therefore a decrease in density is an increase in vt. Book says B, I'm confused. If anyone can help in the same book while answering OP I'd appreciate it.

Where is this question and passage? I've looked in every chapter of my BR physics book and that's not question 20 in any of them.

To determine the terminal velocity, they must give you an equation with Area in it, because cross-sectional area impacts the drag force, which in turn impacts the terminal velocity. If the droplet expands upon freezing, then it will exhibit more drag. It also becomes rigid and can no long alter its shape to maximize its aerodynamics.