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October 6

Fish in ocean different from fish we eat

Most fish that are caught by fishermen have silver scales and few white. We mostly eat fish with silver scales. Why we don't eat colourful fish that we see in aquarium?

And I have seen videos of fishermen catching fish, even there also mostly silver scale fishes get caught. Sundial687v (talk) 07:29, 6 October 2022 (UTC)[reply]

The fish we put in aquariums are colourful because we selected them for that property.

There's a difference in habitat. Colourful fish tend to live in places with lots of waterplants or coral reefs. There are places to hide there and some colours may even be good camouflage. Fish living in the open sea have nowhere to hide and their only way to camouflage is having some silvery colour. Fishermen prefer to fish in the open sea, because that's where they can use efficient nets. There are also some commonly eaten fish that are more brownish, at least on one side. They live on the bottom of shallow seas. PiusImpavidus (talk) 10:09, 6 October 2022 (UTC)[reply]

It's also worth noting that there are some brightly colored fish that are not uncommon to eat. In supermarkets in the United States, it's not uncommon to see red snapper. I've eaten schoolmaster snapper in Central America. Parrotfish, which are often brightly colored, are also caught for eating. I'm sure there are many others. As previously noted, the habitat of these brightly colored fish is often places like coral reefs, rather than open sea. In many cases, these colored fish have ecologically vital roles in these ecosystems, and overfishing them is causing big problems. Parrotfish, for example, are vital to preventing algal overgrowth on coral reefs that otherwise chokes them out of blocks sunlight. As a result, many of these fish have legal protections that would result in not seeing them as much on the menus of developed countries. In less developed countries, illegal practices (often out of desperation) like blast fishing are used for catching fish in coral reefs, which destroys the reefs in the process. These fish are then much more likely to be eaten locally, even by just the families of the fishermen, and not then sold on the open market. Beyond this, many brightly colored fish are too small to be useful for eating, or just don't have much accessibly meat on them. Most damselfish or clownfish aren't large enough to be useful for eating, for example, certainly not enough to justify the difficulty in catching them. They are, however, often illegally caught for the home aquarium trade. --OuroborosCobra (talk) 18:16, 6 October 2022 (UTC)[reply]

Not sure what these are, or this, but they eat them in Oman. Not the sort of thing you find in European waters though. Alansplodge (talk) 20:05, 6 October 2022 (UTC)[reply]

Lion fish have not invaded the Mediterranean, but other brightly coloured members of the Scorpaeniformes family such as the Scorpaena scrofa are widely consumed in dishes such as Bouillabaisse 2A01:E34:EF5E:4640:4D18:274A:1A7:CFF2 (talk) 13:07, 7 October 2022 (UTC)[reply]

To obfuscate things even more, the culinary name for a seafood meal served on your plate is "frequently unrelated" to the scientific name for the species of animal from which it was produced. And this is additionally "frequently unrelated" to the animal's name as known to recreational anglers, and commercial fishermen, and seafood wholesalers, and other-regions-of-the-world, and so on.... through the supply chain. When we include worldwide communities, and non-English-language names, things get very confusing and overlapping.

Here are a few tools to de-mangle the local, culinary, and scientific names:

Fish Names from NOAA
Seafood Species Directory from NOAA

Nimur (talk) 16:32, 7 October 2022 (UTC)[reply]

Simple baseball questions.

If 2 baseballs do a head-on collision, say 1 travelling 10 mph, and the other 40 mph. Will both balls rebound back at the same distance? 67.165.185.178 (talk) 11:13, 6 October 2022 (UTC).[reply]

Generally, I would expect some of the speed to transfer. There's a lot of information, starting with the Collision article. One thing to consider is that a baseball, unlike a billiard ball for example, is a rather irregular surface, due to the stitching and being covered by leather instead of a harder substance. ←Baseball Bugs ^{What's up, Doc?} carrots→ 11:27, 6 October 2022 (UTC)[reply]

Let's for simplicity assume the process takes place along a horizontal straight line in Newtonian spacetime. We can shift the reference frame of the observer. Let

x

denote the offset in the original reference frame, and

s

in that of the observer. A space unit of

1

will correspond to a mile, and a time unit of

1

to an hour. In the original frame, the positions of the balls before the collosion takes place at time

t = 0

are given by

x = 10 t

and

x = -40 t

. We introduce a moving observer, whose position in the original frame is given by

x = -15 t

. In their own frame, the observer is stationary, at the position

s = 0

. The transformation from

x

to

s

is then

s = x + 15 t

; the reverse transformation is

x = s - 15 t

.

In the observer's frame, the pre-collision equations for the moving balls are then

s = 25 t

and

s = -25 t

. For the observer, the situation is perfectly symmetric. If the balls have the same physical characteristics, they will bounce back symmetrically and when they get to rest at some time

t = t r > 0

in the original reference frame, it will be at equal distances

s = \pm r

from the observer. Transforming this back to the original reference frame gives us the positions

x = - 15 t r - r

and

x = - 15 t r + r

. These positions are at different positions from the point of collision

x = 0

. --Lambiam 17:42, 6 October 2022 (UTC)[reply]

Perhaps a simpler way to think of this is to think of the conservation of momentum. Since momentum is a vector (i.e. directional) quantity and is equal to mass times velocity, the faster baseball brings more momentum to the collision, and the total of the two momentums cannot be zero. But in order for them to bounce back equally, it would have to be zero. --174.95.81.219 (talk) 22:20, 6 October 2022 (UTC)[reply]

Okay, is conservation of momentum is Newton's 2nd or 3rd law? And I see we don't have a conservation of force? 67.165.185.178 (talk) 02:46, 8 October 2022 (UTC).[reply]

You mentioned "conservation of force" but that does not exist. We have Newton's second law, which relates acceleration to force (for constant mass). Conservation of Momentum is not one of Newton's laws, but is implied by it (as mentioned in the article on Momentum) Rmvandijk (talk) 14:20, 12 October 2022 (UTC)[reply]

Where did you get the original x = -15t from? 67.165.185.178 (talk) 00:24, 7 October 2022 (UTC).[reply]

One can freely introduce an observer moving in any uniform way in the original reference frame. The specific motion of this observer was chosen as the unique choice that makes the situation symmetric. --Lambiam 19:39, 7 October 2022 (UTC)[reply]

Okay, the slower ball just slows down the faster ball, so they both travel together in the same direction after collision. At half of their average speed. 67.165.185.178 (talk) 00:31, 10 October 2022 (UTC).[reply]

Okay, the centerpoint between 10 and 40 is 25 again. If 2 baseballs head-on collide both at 25 m/s, I would need to know if the collision is inelastic or not, if it's inelastic I would need to know the coefficient of restitution. The more elastic the more faster, but loses more energy. What if instead of 25 m/s, they were both 25 m/s^2? Then I would need to know a whole different set of factors? 67.165.185.178 (talk) 00:12, 7 October 2022 (UTC).[reply]

You've now given them an acceleration, but for the conservation of energy and momentum equation you need a velocity. So you need to define the initial distance between them, or the velocity just before collision. In the last case, not much changes compared to before, but with an initial distance you need to determine the speed before collision, adding a step. Rmvandijk (talk) 11:51, 7 October 2022 (UTC)[reply]

I think what you just did was just change the question to a velocity problem. That is, find the velocity, then continue on. But I've had conversations with people that say you don't need to know the velocity to solve the problem... Do people here disagree? 67.165.185.178 (talk) 22:24, 7 October 2022 (UTC).[reply]

You need to know something to solve the problem. Will two colliding objects bounce back at the same distance? As a yes/no question this is unanswerable. It depends. --Lambiam 08:07, 8 October 2022 (UTC)[reply]

No. My point is, what if my question was 2 balls collide at m/s, then couldn't someone say "let's find the acceleration 1st, find the m/s^2, then treat it as an acceleration/force problem." As I asked for m/s^2, someone made the "let's find the velocity at collision, the m/s, then continue it as a velocity-momentum question." 67.165.185.178 (talk) 13:52, 8 October 2022 (UTC).[reply]

To answer the original questions that you have asked, you need to have sufficient input data. If we have a collision, momentum is conserved. Momentum is mass times velocity, so to determine what happens after, we need the momentum going in. That means we need their mass and velocity, and with that we can calculate their mass and velocity afterwards. But you ask if they bounce back the same distance. To answer that question, you'd need to know the distance they had when they started, and what other forces will act on them, because in a friction-less world, the answer would be "No", because they'd never stop after the collision. Since we have a momentum-problem in collisions, we need to know the velocities involved (or directly know their momentum, but that is not a measurable quantity, unlike velocity). If you were to discuss a bouncing ball, you wouldn't need the velocity, you could use its mass, Coefficient of restitution, gravity constant and initial height to determine how high it bounces, because that is an energy problem. Note that this could also be calculated with velocity just prior to impact, in which case you would not need the initial height. Rmvandijk (talk) 14:20, 12 October 2022 (UTC)[reply]

If the initial situation is symmetric, it will remain symmetric. The answer I gave above does not depend on the elasticity of the balls or any other material property. --Lambiam 19:44, 7 October 2022 (UTC)[reply]

I tried assuming a perfectly elastic collision and projectile motion with the assumption that there could be a head on collision with one baseball on the upward part of the arc and the other on the downward. One ball could then spend a longer time in the air after the collision to land the same distance horizontally away from point of impact. There don't seem to be any solutions when the baseballs have opposite but unequal velocities at impact, only a solution with equal velocities with impact at the top of the arc. The only valid solution seems to be when both baseball are thrown perfectly straight in the air, one after another, with each landing directly below the point of impact. However, no one here should trust my algebra to any degree at all. fiveby(zero) 18:22, 8 October 2022 (UTC)[reply]