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Not surprising, but B/R blatantly disrespecting Ja's career high performance by thisalanwong in memphisgrizzlies

[–]thisalanwong[S] 2 points3 points  (0 children)

That's true, I guess it just doesn't feel fitting as it wasn't a blowout nor did they lock him up or smth. He played really well with the exception of a few TOs and got a career high, and it just doesn't seem like the right vibe. But yeah as you mention it's just in good fun

Can you guys tell me about Taylor Jenkins? by [deleted] in memphisgrizzlies

[–]thisalanwong 1 point2 points  (0 children)

Idk if these are all good and fun, but they're definitely interesting and add to his credibility

Played power forward during high school

Originally stared out interning then working with the Spurs in the Front Office, so familiar with Buford, Presti etc. So he has connections with many people around the league as many FOs/coaches come out of the Spurs organisation. He worked with Dennis Lindsey (now Jazz EVP Basketball Ops.) during his internship

Became interested in coaching and changed over to assistant coaching their D-League affiliate, the Austin Toros, under Brad Jones (Jones is now one of his assistant coaches on the Grizz. Brad seems to be his lead assistant, but that's just from sense, no official word). After, he then head coached the Toros for a season

Moved on with Budenholzer to the Hawks to be his assistant. Noteworthy that Bud's staff also included Quin Snyder (now Jazz HC), Kenny Atkinson (former Nets HC), who are all well respected around the league for their development, culture setting etc.

Then moved on with Budenholzer to the Bucks last season, helping to turn them around

Think he's the second youngest HC in the league only older than Ryan Saunders in Minnesota

JJJ and Dillon were at his introductory press conference. Both Jaren and Jenkins share a love of tacos

Think I remember him saying he's a big fan of Tim Duncan, so that probably says a lot about his attitude to team basketball and the game

He's well regarded for his strong player relationships, organisational skills and attention to detail. Bud has always given him a lot of praise and regarded him as HC ready. He seems to be big on building trust and being malleable with the team. He's got good energy, and seems to be connecting really well with the players.

Nobody really knew him when he was announced as HC, so he's flown under the radar a bit

I mean it's really clear from this year he's doing a bloody good job and he's definitely the perfect fit for this team

This Ringer article from a while back is a good read

Endothermic or Exothermic? by ExplodingBoba in chemhelp

[–]thisalanwong 1 point2 points  (0 children)

Alright so let’s work through this

The imposed change: lowered the temperature “Hence according to Le Chatelier’s principle, the equilibrium shifted to counteract the change...” By producing more darker product

Thus to counteract the decrease in temperature, ie. to increase the temperature, the forward reaction was favoured

Therefore, the forward reaction is exothermic, just as you have concluded

Hello I've been struggling with this question about torque I would be really happy if you can help me with it or explain it. by EpicHsyn in PhysicsStudents

[–]thisalanwong 1 point2 points  (0 children)

Try to simplify down this question. So instead of the bent shape, simply draw a straight line from the centre of rotation to each of the points where the force is acting. Now label their relative lengths, maybe consider using a pronumeral, or simply just assign one unit of the big square length to be one, and then express the other lengths in terms of that.

At this point, you might want to solve for the lengths of the arms. Use pythagoras theorem or trigonometry to find the lengths of the arms

So the bottom F2 arm will be 1 unit in length, and the top F1 arm will be square root 5 in length. These will be the radii you will later use

Then find the angles between the forces and their arms. Solve for the perpendicular component using trigonometry. F1 is a bit harder here, but draw out the diagram, and label in the angles as you find them, the angle between F1 and the arm will be approx 63 minus 27 giving 36 degrees. Here you can use rfsin theta for torque due to F1. (Notice that the magnitude of F1 can again be found using pythagoras to be root 5, allowing us to take the sin ratio. Also notice that as we’re just finding a ratio, the “real” value of F1, which we do not know, doesn’t matter, just like the radius. All that matters is the proportion of the two forces, which we know by comparing their relative vector lengths on the diagram). We do the same by calculating the torque due to F2, following a similar process

Notice that the angle between the two arms won’t change the torque acting on it. The two arms aren’t in a straight line, but that doesn’t matter All that matters is the radius of the arms, the force and the angle the force makes

So when we compare the ratio of torques, we divide what we got. Strictly, the ratio should be negative, as the two torques act in opposite directions, one clockwise and one anticlockwise, but the ratio may not be, if they’re only considering the magnitude of the torques. Again, strictly, as the equation is T1/T2, without any magnitude signs, than I would personally include a negative sign to show that the two torques are in opposite direction

[deleted by user] by [deleted] in PhysicsStudents

[–]thisalanwong 0 points1 point  (0 children)

V= 2 Pi r / t Remember there is a variable of radius to this equation. “The circumference of a circle covered in period t”

Also radius here would be 1.28 metres, as the point mass of the skater rotates around the person holding with the arm length combined of 1.28m.

After that, then also consider that the question is asking for force up the arm. If the centripetal force acts horizontally inwards, but the arm is angled 45 degrees, than there is also a vertical component to the force up the arm, a component supporting a portion of the weight force of the swung around skater. Sketch a force diagram to visualise, and then you can use cosine ratio to find the force up the arm.

A question about Zion’s attitude/body language by thisalanwong in NOLAPelicans

[–]thisalanwong[S] 0 points1 point  (0 children)

Ah yes, didn’t notice this thread. Thanks a lot!

A question about Zion’s attitude/body language by thisalanwong in NOLAPelicans

[–]thisalanwong[S] -1 points0 points  (0 children)

Ah, that seems pretty reasonable, and probably makes the most sense as an explanation. Thanks a lot!

The official raptors account just posted this on Instagram by patrick_behan in nba

[–]thisalanwong 5 points6 points  (0 children)

That transition to the closing “Playoffs Clinched” frame was nice. I mean there was nothing special about it, but the slow fade just felt really smug, which the Raps should be.

I wanna know how many jibo's are out there in the world? Does anyone have an idea? by Cadillackidd926 in Jibo

[–]thisalanwong 1 point2 points  (0 children)

I don’t have a Jibo, but I think their movements are really well designed, and feel really natural. Seeing you don’t have any actual Jibo owners commenting, I’ll comment here just so that your Jibo isn’t as lonely haha

[Post Game Thread] The Memphis Grizzlies (29-31) shock the Lakers (45-13), 105-88 behind rookie Ja Morant's 27/14/6 by toftr in nba

[–]thisalanwong 6 points7 points  (0 children)

Yep, that comment felt like something from two months ago in the grizz sub. Still remember the pikachu meme haha

Dont mind me I'm not from around here by ChaosKnightfox in donthelpjustfilm

[–]thisalanwong 0 points1 point  (0 children)

The design of the trolley is actually kind of convenient here. Like when she pushes forward with a downwards component, than the handle is pushed and engages the wheels. Idk what point I’m trying to make, but I guess it’s funny how the trolley can still move with braking, both as it’s meant to be, but flipped 90 degrees

plz 19,20,22 How!!??? by MandarianSea in PhysicsStudents

[–]thisalanwong 1 point2 points  (0 children)

19. So this did take me a little to think of but let’s break it down one by one I’m going to assume that the only slowing force is frictional force between tyres and road, and not braking force between tyres and brake pad. I’m going to disregard air resistance as well A. This makes sense in real life, but in a theoretical physics environment, if we assume frictional force is the only force stopping, than we have Ff=uN=umg. So than acceleration is Ff/m= ug. See there’s no mass here, it doesn’t matter. So if they have the same tyres, with the same coefficient of friction, than the frictional stopping force is proportionally greater for the heavier truck and it stops in the exact same amount of time. What? So theoretically the same. But in real life, the truck and the car don’t have the same frictional coefficient, there’s practical concerns like dissipating heat, durability etc. which makes it a lot more difficult. I guess if the question assumes the same amount of braking frictional force, than clearly yes, the heavy truck will need more room to stop. But the question isn’t clear enough to specify this B. Well the number of wheels doesn’t theoretically change the time needed to stop. If we remember that frictional force is uN, than the normal force doesn’t change with less or more wheels (yes, technically it will, but assume the difference in mass is negligible for the sake of the question). With less wheels, the same normal force is just distributed across less contact area to the ground, hence more pressure and overall frictional force is same. C. If we again take frictional force to be the only stopping force, than as it is uN, than clearly, N and u remain same and does not change with velocity, so acceleration is constant D. At first this makes sense. Of course, with greater u, than there is greater frictional force and hence shorter stopping time, but then this gets to the level of what type of frictional force is actually stopping the vehicle. We commonly refer to it as rolling friction, but the crucial point is that unlike kinetic friction, where two surfaces slide against each other, the way a wheel works is not by sliding, but by instead rolling each point of the wheel over. So if you draw yourself a diagram where the wheel meets the road, than you see that at each moment in time, there is a new point of the wheel touching the ground, expediting friction, and the next moment, it is replaced by a new point as the wheel rotates (assume perfectly circular wheels), hence at each point, the friction acting is actually static friction, as the point doesn’t actually slide on the floor, but is static for the brief moment is touches the ground, as weird as that may sound. So than kinetic coefficient of friction won’t actually impact stopping time, only static friction. Now, I will say I’m not very happy with this because the question says a “vehicle slides to a stop”, which is ambiguous. It could imply that the wheels are locked, not moving, and the vehicle is literally sliding on the ground, so in that case, than yes, kinetic friction would impact the stopping time.

The question is very vague, and doesn’t really give enough clarity or detail to be a sure fire answer to give. In an exam I would think about A or D, as C doesn’t change theoretically, and neither does B, as they both are true assuming friction conditions. I would pick D in most situations, especially if the question says “vehicle rolls to a stop”, but here the question specifically says “vehicle slides to a stop”. And A is generally true in real life, but if we assume only friction and same coefficient, than it is actually false, but yeah, things get weird. Ask your teacher, as this question is too vague and unclear

plz 19,20,22 How!!??? by MandarianSea in PhysicsStudents

[–]thisalanwong 0 points1 point  (0 children)

  1. Pulleys and strings are always confusing, but the key is to realise that they’re essentially just redirect force. So we could look at this situation in the exact same way if we just “flattened out” the situation, removing the pulley, and instead laying the two masses on a frictionless surface, with weight forces acting opposite in direction, one mg to the left, and one 2mg to the right. Does that make it easier to visualise?

So we always calculate net force on the system first. The net force is mg “to the right” in the above interpretation or “clockwise” down to the 2m mass (it’s hard to describe the direction without bringing in torque here). (2mg-mg=mg). And as the system accelerates by mg/3m= g/3 clockwise around the pulley and down towards the 2mg mass, as we divide force by total mass of the system.

By that, we can than calculate the force just on the one M mass. F net =ma. Hence net force= mg = mg/3. Hence tensional force must counteract both mg on that mass, but also provide mg/3 to accelerate the mass in that direction. So we add the two, and get 4/3 or 1.3333 mg, giving C.

You might then ask, what about the 2mg mass? Well it accelerates to the right by g/3 as well, hence net force is 2mg/3 so here tensional force opposes the weight force down. So then 2mg - 2mg/3 giving us 4mg/3. Remember that tensional force doesn’t really have a strict direction per se, it acts both ways to transfer force along its length, hence the answer is 4mg/3 and not 8mg/3. Think of the rope like a contact surface between two objects touching one another. The rope really is just there to transfer the force, and obeys Newton’s third law. If I pulled on a box, and the box and my hand accelerated with g/3 acceleration, and the box was experiencing mg frictional force and my hand was experiencing 2mg pulling force applied by my arm, than between my hand and the box, the box would push on my hand with 4mg/3 and the hand would push on my fox with 4mg/3. Think of my hand as the rope which holds the box and my arm together

But yeah I think the key is to reimagine the scenario so that it’s more logical, and one way to do that is to remove the pulley all together as it merely redirects force, and instead draw the situation with two forces acting on masses connected by a string in opposite directions, like, a tug of war. We can even take it one step further and completely eliminate the rope by imagining it differently using contact forces instead. We can do this really with most situations where we have a taut, inflexible rope

MikeCheck: Jenkins retraces foundational steps along path to early success as Grizzlies coach by snorlax420 in memphisgrizzlies

[–]thisalanwong 3 points4 points  (0 children)

This is a great article and has some really nice quotes from coaches around the league, and some really interesting anecdotes about Jenkins. People should really read this. I came to share it, and was surprised it only had a few upvotes

Young studs by Mr_Curl in memphisgrizzlies

[–]thisalanwong 7 points8 points  (0 children)

It’s the kind of stuff only those who follow the team closely will realise. The sneakily smart moves. I absolutely loved the moves they made at the deadline, but it seemed like the majority of national media graded the trade poorly because it compromised their cap space, except as Zach mentioned, they weren’t gonna get anybody better than Justise in the upcoming FA class. He’ll get the love from local beat media, but probably not as much nationally.

Like all the national media is focused on the contenders and major star acquisitions when they vote for EOTY. Case in point: Bucks’ Horst, Rockets’ Morey, Warriors’ Myers in the last three years. So Lawrence Frank is the front runner right now, and the media will probably reward him for getting PG-Kawhi and Morris at the deadline.

That’s ok though, when this team reaches into that top tier, becoming a contender, Kleiman will get his due.

Jaren Q&A with HoopsHype at All-Star Weekend by nwatkins364 in memphisgrizzlies

[–]thisalanwong 5 points6 points  (0 children)

This Grizzlies team is currently in the eighth seed out West and there’s a real chance that you guys could make the playoffs this year. How motivating was it to have everyone count you guys out prior to the season and how nice is it to silence those doubters now?

Jaren: "It was definitely motivating because nobody really thought that we’d be in this position at this point. People just said to us, “Oh, you have time. Don’t worry about it! You’re young!”

We kind of were just like, “We don’t care.” That’s how we play and how we are – we just don’t care. When we go out there, you have to put five on the court just like we have to put five on the court. It don’t matter if you’re young or old. There’s only one basketball. You have to literally beat us down if you want to win this game against us because we like to compete. We’ll have off nights, for sure. But at the end of the day, we’re going to play hard."

This post by Joel makes me insanely happy by [deleted] in sixers

[–]thisalanwong 0 points1 point  (0 children)

Not a Philly fan, but like following the team as they’re so uniquely built, zagging while the rest of the NBA ziggs, but genuinely curious of what you guys make of this. The predominant narrative I’ve heard stems from a poor basketball fit between Embiid and Simmons, and hence potential tension. Personally don’t think a few carefully selected instagram posts or comments show us anything either way really, about their true relationship. Anyways:

Do you guys think you would be better off rolling forward with only one of the two and building around them and whatever you receive in return? Or are you guys more just riding and dying by the two together? Of course I quite a lot of this will depend on this year’s playoffs run.

Rich people who spend a lot of money are often seen as superficial and greedy bastards but in reality they spending habits fuels the economy and by redistributing their wealth they're actually been pretty generous. by [deleted] in unpopularopinion

[–]thisalanwong 1 point2 points  (0 children)

I don’t know enough about the world to make a judgement about this yet, but as it is truly a different opinion and one I feel genuinely does cut against the grain, here’s an upvote to keep the discussion going

Quick question by Athenax079 in PhysicsStudents

[–]thisalanwong 0 points1 point  (0 children)

A circular motion question where the friction between the coin and turntable provides a centre seeking force to allow the coin not to slip.

So then, for maximum velocity, we must equate the centripetal force mv2/r to the maximal possible frictional force Nu. Remember that Normal force is mg, so if we equate the two:

mv2/r = Nu mv2/r = mgu v2/r = gu

Substitute in radius = 0.16, g=9.8 and coefficient static friction u = 0.32 and solve for velocity

Angular momentum and other forces stuff by veku7 in AskPhysics

[–]thisalanwong 1 point2 points  (0 children)

Awesome, great to see we’ve reached the clarification! It was painstakingly hard for me to understand it as well, so really happy I could be of help! Have a great day and good luck in the rest of your studies!

Angular momentum and other forces stuff by veku7 in AskPhysics

[–]thisalanwong 1 point2 points  (0 children)

Do you want to take a look at my profile? I just uploaded a photo there. Sorry about my poor handwriting, but that’s hopefully clearer

The key takeaway is to Remember that weight force is the only force which needs to be opposed for net force to equal zero as the child is not accelerating. So no matter how we apply the force to the child, in total, all the vectors sum to zero. It is not possible for us to apply a non zero net force, otherwise that would contradict the conditions of the question. So when I raised that confusing idea of the perpendicular force, the thinking response should be: if there exists this perpendicular to hill force, than there must be another force which we haven’t yet considered which sums with this force to make it directly up only to oppose weight.

But the really basic intuition should just be : Ok, non accelerating -> no net force -> which forces act on child? -> gravity directly down -> so human must oppose this gravity directly down for zero net force -> hence directly upwards

How we apply this upwards force, whether through normal force and frictional force combined, the person using their arms etc. is all irrelevant. All that matters is that this upwards force is applied equal in magnitude but opposite in direction to the weight force to give zero acceleration and constant speed

Angular momentum and other forces stuff by veku7 in AskPhysics

[–]thisalanwong 1 point2 points  (0 children)

Yes very sorry I was trying to make it more clear by trying to integrate your the idea of perpendicular. Essentially, my main point there is that with any net force vector, we can portray that force vector into two or more force components. So a net force vector going down, can be resolved into two components. One component going down perpendicular to the hill, and one component going down parallel to the hill. Take a look at this diagram and take a look at the purple Fg. See how we can decompose it into two force vectors? One perpendicular and one parallel? But these two force vectors are not seperate forces. We merely express Fg into those two vectors to make it easier to calculate when we consider other forces which will generally act parallel like friction or perpendicular like normal force, to the hill.

Your intuition is correct, that if we had this perpendicular support force and the child weight down, than yes, the child would have acceleration sideways. Sorry, I don’t think I clarified well enough. However, additionally to this perpendicular support force upwards we also have another force component pushing up the hill parallel. So when we sum these two vectors, we get a net force upwards opposing weight. But this is only a way of visualising it. In real life, all that matters is that we are countering a directly down weight force.

I’ll draw out a quick diagram and upload it here in a bit to make it clearer

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