Category: AQA Unit 2 Forces/ Motion/ Waves

New Scientist News: Beer brewing could help make better bricks

Beer brewing could help make better bricks

THERE’S more to brewing than beer. A by-product of the process could be about to give an upgrade to a workhorse building material – red clay bricks. By blending in the grains left over from making beer, the bricks can be more environmentally friendly and better insulators.

Bricks are often impregnated with polystyrene as a way to enhance their heat-trapping abilities. This is appealing, because the bricks remain strong, and they can be built into energy-efficient buildings, says Eduardo Ferraz of the Polytechnic Institute of Tomar in Portugal. However, EU restrictions on carbon emissions have made it expensive to incorporate polystyrene and other synthetic materials into bricks.

Ferraz and his colleagues have now shown that brewery grains can be mixed into clay bricks to enhance their ability to trap heat, without compromising strength.

Spent grain for the process should be easily available, because commercial breweries produce huge quantities of it as a pulpy mixture that is usually used in animal feed or ends up in landfill.

With a clay paste containing 5 per cent spent grains, the team was able to create bricks just as strong as the conventional type, while reducing the amount of heat they lost by 28 per cent (Journal of Materials in Civil Engineering, doi.org/p6k). The reason for this, the team says, is that the grains make the bricks more porous, and so they trap more air, which increases heat retention.

One thing could stand in the way of using this process, though: the smell. Bill Daidone of the Acme Brick Company, one of the largest brick manufacturers in the US, says his lab abandoned experiments because the stench of the moist grains was overpowering. “We opened up the bucket and it was terrible,” he says. This problem vanishes once the bricks are fired, though, says Ferraz.

Bricks that provide insulation without sacrificing strength could be a big boost to the brick industry, says John Sanders, a researcher at the National Brick Research Center at Clemson University in South Carolina.

“With the current concern for energy codes, I think the industry is open to change,” Sanders says.

This article appeared in print under the headline “Brewing benefits? Alcohol, hangovers and better bricks”

http://google.com/producer/s/CBIw2Kfm0gM

Permanent link to this article: https://animatedscience.co.uk/2013/new-scientist-news-beer-brewing-could-help-make-better-bricks

Secret of Bolt’s speed unveiled

Secret of Bolt’s speed unveiled http://www.bbc.co.uk/news/science-environment-23462815

 

Secret of Usain Bolt’s speed unveiled

By Melissa HogenboomScience reporter, BBC News

Usain Bolt wins the 100mBolt’s 2012 Olympic record of 9.63 seconds in the 100m final was not his fastest 100m sprint

Scientists say they can explain Usain Bolt’s extraordinary speed with a mathematical model.

His 100m time of 9.58 seconds during the 2009 World Championships in Berlin is the current world record.

They say their model explains the power and energy he had to expend to overcome drag caused by air resistance, made stronger by his frame of 6ft 5in.

Writing in the European Journal of Physics, the team hope to discover what makes extraordinary athletes so fast.

According to the mathematical model proposed, Bolt’s time of 9.58 seconds in Berlin was achieved by reaching a speed of 12.2 metres per second, equivalent to about 27mph.

The team calculated that Bolt’s maximum power occurred when he was less than one second into the race and was only at half his maximum speed. This demonstrates the near immediate effect of drag, which is where air resistance slows moving objects.

They also discovered less than 8% of the energy his muscles produced was used for motion, with the rest absorbed by drag.

Jamaica's Usain Bolt celebrates after winning the men's 100m final The 2012 gold medallist became a worldwide sensation

When comparing Bolt’s body mass, the altitude of the track and the air temperature, they found out that his drag coefficient – which is a measure of the drag per unit area of mass – was actually less aerodynamic than that of the average man.

Jorge Hernandez of the the National Autonomous University of Mexico said: “Our calculated drag coefficient highlights the outstanding ability of Bolt. He has been able to break several records despite not being as aerodynamic as a human can be.

“The enormous amount of work that Bolt developed in 2009, and the amount that was absorbed by drag, is truly extraordinary.

“It is so hard to break records nowadays, even by hundredths of a second, as the runners must act very powerfully against a tremendous force which increases massively with each bit of additional speed they are able to develop.

“This is all because of the ‘physical barrier’ imposed by the conditions on Earth. Of course, if Bolt were to run on a planet with a much less dense atmosphere, he could achieve records of fantastic proportions.

“The accurate recording of Bolt’s position and speed during the race provided a splendid opportunity for us to study the effects of drag on a sprinter.

“If more data become available in the future, it would be interesting to see what distinguishes one athlete from another,” added Mr Hernandez.

Bolt 2012Bolt (L) is known to be a

Bolt’s time in Berlin was the biggest increase in the record since electronic timing was introduced in 1968.

John Barrow at Cambridge University who has previously analysed how Bolt could become even faster, explained that his speed came in part due his “extraordinary large stride length”, despite having such an initial slow reaction time to the starting gun.

“He has lots of fast twitch muscle fibres that can respond quickly, coupled with his fast stride is what gives him such an extraordinary fast time.”

He said Bolt has lots of scope to break his record if he responded faster at the start, ran with a slightly stronger tail-wind and at a higher altitude, where there was less drag.

Bolt’s Berlin record was won with a tail wind of only 0.9m per second, which didn’t give him “the advantage of helpful wind assistance”, he added.

“You’re allowed to have a wind no greater than 2m per second to count for record purposes, so without becoming any faster he has huge scope to improve,” Prof Barrow told BBC News.

Permanent link to this article: https://animatedscience.co.uk/2013/secret-of-bolts-speed-unveiled

Material halts liquid flow on demand

Material halts liquid flow on demand http://www.bbc.co.uk/news/science-environment-22079600

Permanent link to this article: https://animatedscience.co.uk/2013/material-halts-liquid-flow-on-demand

Cross Winds Calculations..

Try out this game which is all about defeating a cross wind…

Permanent link to this article: https://animatedscience.co.uk/2011/cross-winds-calculations

Bowman Game

This one is all about the angles and the Kinematics, looks simple but think about projectile motion…

https://www.animatedscience.co.uk/fun/bowman.swf

Permanent link to this article: https://animatedscience.co.uk/2011/bowman-game

8 Equations of Motion…

A common problem for teaching Physics when you are not a Physics teacher is that you make many mistakes with complex ideas which appear simple on the surface.

I drop a ball to the earth and want to work out the distance fallen in a certain time. I then do a calculation of..

distance x speed = time

Oh dear but when I do the experiment it does not work like that but it works like this…

[latex size=”3″]s =\frac{1}{2}at^{2}[/latex]

Now we can reason this out. If I allow a ball to fall to earth it must accelerate due to the field of gravity around the earth 9.81N/kg. Or 9.81 m/s/s.

But where does the formulae  come from that we general use?

[latex size=”3″]s = ut+\frac{1}{2}at^{2}[/latex]

This post is simply to give some advice about this formulae. To start with define everything we use…

  1. s = the distance between initial and final positions (displacement) (sometimes denoted as x)
  2. u = the initial velocity (speed in a given direction)
  3. v = the final velocity
  4. a = the constant acceleration
  5. t = the time taken to move from the initial state to the final state

So the question is what is it all about.

Well we need to think way back to the idea of a simple idea of how to work out the distance travelled by  a runner in a race.

Think of an athlete travelling 100m in 10s at a constant speed. His velocity or speed is…

[latex size=”3″] \frac{d}{t}= speed[/latex]  OR  [latex size=”3″] \frac{s}{t}= v[/latex]

Now this works fine if we are travelling at a constant speed but hey as you know this is not always the case and sometimes an object has a constant acceleration or deceleration. If you think about a graph of a person who got faster and faster then we would have a slope or triangular area on a speed-time or velocity-time graph. Now the area under the graph would be the distance travelled on the journey. But if our speed changed s=vt (for constant speed) becomes….

[latex size=”3″] s=\frac{vt}{2}[/latex]

Now you have an expression for the average speed but only from a standing start. Imagine now the same velocity time graph but this time the runner was already travelling at a velocity at the start of timing. Our area would become a triangle and rectangle…

[latex size=”3″] s=\frac{(v-u)t}{2}+ut[/latex]

Simplifies to

[latex size=”3″] s=\frac{(v+u)t}{2}[/latex]  – Eq 2

This is the formulae for average speed that takes care of all situations even when u or v is 0. Now then think graphically again if we are travelling at an initial velocity and then accelerate we are back to triangle and square again when looking at a vt graph. Hence…

[latex size=”3″] v=u+at[/latex]  – Eq 1

Now we can use this in a rearranged form..

[latex size=”3″] t=\frac{(v-u)}{a}[/latex]

Substitute into Eq 1 in new form into 1 to remove t so we now have an expression…

[latex size=”3″] s=\frac{(v+u)}{2}*\frac{(v-u)}{a} [/latex]

Which simplifies to..

[latex size=”3″] s=\frac{(v^{2}-u^{2})}{2a} [/latex]  – Equ 4

Now we have this we can work also work backwards to get Eq 3 …

[latex size=”3″] v=u+at[/latex]  – Eq 1

And *t gives..

[latex size=”3″]vt = ut+at^{2}[/latex]

Divide both sides by two…

[latex size=”3″]\frac{vt}{2} = \frac{ut}{2} +\frac{ at^{2}}{2} [/latex]

Add ut to both sides, multiply by 2 and tidy up..

[latex size=”3″]s = ut+\frac{ at^{2}}{2} [/latex] – Eq 3

So we now have the four equations of motion for an object which is travelling at a constant velocity or accelerating at a constant rate. As shown you can reason them all out with very simple ideas from first principals. You can also call them whatever number you want as some people label them differently.

So hopfully we have now understood where the formulae comes from and realise that if we drop a ball we must apply the formulae to work out the distance fallen in a more complex way.

Then it gets really interesting when you think about a tanker travelling at a constant velocity which then slows down. How far does it travel as it slows…

[latex size=”3″]s = ut-\frac{ at^{2}}{2} [/latex]

Of course, it will be ut for the whole time and then you take away a little bit of “s” from the other term as you slow.

Permanent link to this article: https://animatedscience.co.uk/2011/equations-of-motion

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