examples

Exponential Functions and their Derivatives (including Examples)

Exponential functions have the general form:

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with two constants a and b (called base). It’s quite common to use Euler’s number e = 2.7182… as the base and the exponential function expressed as such:

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with two constants a and c. Converting from one form to the other is not that difficult, just use ec = b or c = ln(b). Here’s how it works:

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As for the plot, you should keep two special cases in mind. For b > 1 (which corresponds to c > 0 in case of base e), the function goes through the point P(0,a) and goes to infinity as x goes to infinity.

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(Exponential function with b > 1 or c > 0. For example: f(x) = 8·3x)

This is exponential growth. When 0 < b < 1 (or c < 0) this turns into exponential decline. The function again goes through the point P(0,a), but approaches zero as x goes to infinity.

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(Exponential function with 0 < b < 1 or c < 0. For example: f(x) = 0.5x)

Here’s how the differentiation of exponential functions works. Given the function:

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 The first derivative is:

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For the case of base e:

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We get:

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You should remember both formulas. Note that the exponential functions have the unique property that their first derivative (slope) is proportional to the function value (height above x-axis). So the higher the curve, the sharper it rises. This is why exponential growth is so explosive.

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Example 1:

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Example 2:

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Example 3:

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Example 4:

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If exponential functions are combined with power or polynomial functions, just use the sum rule.

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Example 5:

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Example 6:

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(This was an excerpt from the FREE ebook “Math Shorts – Derivatives”)

Motion With Constant Acceleration (Examples, Exercises, Solutions)

An abstraction often used in physics is motion with constant acceleration. This is a good approximation for many different situations: free fall over small distances or in low-density atmospheres, full braking in car traffic, an object sliding down an inclined plane, etc … The mathematics behind this special case is relatively simple. Assume the object that is subject to the constant acceleration a (in m/s²) initially has a velocity v(0) (in m/s). Since the velocity is the integral of the acceleration function, the object’s velocity after time t (in s) is simply:

1) v(t) = v(0) + a · t

For example, if a car initially goes v(0) = 20 m/s and brakes with a constant a = -10 m/s², which is a realistic value for asphalt, its velocity after a time t is:

v(t) = 20 – 10 · t

After t = 1 second, the car’s speed has decreased to v(1) = 20 – 10 · 1 = 10 m/s and after t = 2 seconds the car has come to a halt: v(2) = 20 – 10 · 2 = 0 m/s. As you can see, it’s all pretty straight-forward. Note that the negative acceleration (also called deceleration) has led the velocity to decrease over time. In a similar manner, a positive acceleration will cause the speed to go up. You can read more on acceleration in this blog post.

What about the distance x (in m) the object covers? We have to integrate the velocity function to find the appropriate formula. The covered distance after time t is:

2) x(t) = v(0) · t + 0.5 · a · t²

While that looks a lot more complicated, it is really just as straight-forward. Let’s go back to the car that initially has a speed of v(0) = 20 m/s and brakes with a constant a = -10 m/s². In this case the above formula becomes:

x(t) = 20 · t – 0.5 · 10 · t²

After t = 1 second, the car has traveled x(1) = 20 · 1 – 0.5 · 10 · 1² = 15 meters. By the time it comes to a halt at t = 2 seconds, it moved x(2) = 20 · 2 – 0.5 · 10 · 2² = 20 meters. Note that we don’t have to use the time as a variable. There’s a way to eliminate it. We could solve equation 1) for t and insert the resulting expression into equation 2). This leads to a formula connecting the velocity v and distance x.

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Solved for x it looks like this:

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It’s a very useful formula that you should keep in mind. Suppose a tram accelerates at a constant a = 1.3 m/s², which is also a realistic value, from rest (v(0) = 0 m/s). What distance does it need to go to full speed v = 10 m/s? Using equation 3)’ we can easily calculate this:

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Here are a few exercises and solutions using the equations 1), 2) and 3).

1. During free fall (air resistance neglected) an object accelerates with about a = 10 m/s. Suppose the object is dropped, that is, it is initially at rest (v(0) = 0 m/s).

a) What is its speed after t = 3 seconds?
b) What distance has it traveled after t = 3 seconds?
c) Suppose we drop the object from a tower that is x = 20 meters tall. At what speed will it impact the ground?
d) How long does the drop take?

Hint: in exercise d) solve equation 1) for t and insert the result from c)

2. During the reentry of space crafts accelerations can be as high as a = -70 m/s². Suppose the space craft initially moves with v(0) = 6000 m/s.

a) What’s the speed and covered distance after t = 10 seconds?
b) How long will it take the space craft to half its initial velocity?
c) What distance will it travel during this time?

3. An investigator arrives at the scene of a car crash. From the skid marks he deduces that it took the car a distance x = 55 meters to come to a halt. Assume full braking (a = -10 m/s²). Was the car initially above the speed limit of 30 m/s?

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Solutions to the exercises:

Exercise 1

a) 30 m/s
b) 45 m
c) 20 m/s
d) 2 s

Exercise 2

a) 5,300 m/s and 56,500 m
b) 42.9 s (rounded)
c) 192,860 m (rounded)

Exercise 3

Yes (he was initially going 33.2 m/s)

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To learn the basic math you need to succeed in physics, check out the e-book “Algebra – The Very Basics”. For an informal introduction to physics, check out the e-book “Physics! In Quantities and Examples”. Both are available at low prices and exclusively for Kindle.

How To Calculate Maximum Car Speed + Examples (Mercedes C-180, Bugatti Veyron)

How do you determine the maximum possible speed your car can go? Well, one rather straight-forward option is to just get into your car, go on the Autobahn and push down the pedal until the needle stops moving. The problem with this option is that there’s not always an Autobahn nearby. So we need to find another way.

Luckily, physics can help us out here. You probably know that whenever a body is moving at constant speed, there must be a balance of forces in play. The force that is aiming to accelerate the object is exactly balanced by the force that wants to decelerate it. Our first job is to find out what forces we are dealing with.

Obvious candidates for the retarding forces are ground friction and air resistance. However, in our case looking at the latter is sufficient since at high speeds, air resistance becomes the dominating factor. This makes things considerably easier for us. So how can we calculate air resistance?

To compute air resistance we need to know several inputs. One of these is the air density D (in kg/m³), which at sea level has the value D = 1.25 kg/m³. We also need to know the projected area A (in m²) of the car, which is just the product of width times height. Of course there’s also the dependence on the velocity v (in m/s) relative to the air. The formula for the drag force is:

F = 0.5 · c · D · A · v²

with c (dimensionless) being the drag coefficient. This is the one quantity in this formula that is tough to determine. You probably don’t know this value for your car and there’s a good chance you will never find it out even if you try. In general, you want to have this value as low as possible.

On ecomodder.com you can find a table of drag coefficients for many common modern car models. Excluding prototype models, the drag coefficient in this list ranges between c = 0.25 for the Honda Insight to c = 0.58 for the Jeep Wrangler TJ Soft Top. The average value is c = 0.33. In first approximation you can estimate your car’s drag coefficient by placing it in this range depending on how streamlined it looks compared to the average car.

With the equation: power equals force times speed, we can use the above formula to find out how much power (in W) we need to provide to counter the air resistance at a certain speed:

P = F · v = 0.5 · c · D · A · v³

Of course we can also reverse this equation. Given that our car is able to provide a certain amount of power P, this is the maximum speed v we can achieve:

v = ( 2 · P / (c · D · A) )1/3

From the formula we can see that the top speed grows with the third root of the car’s power, meaning that when we increase the power eightfold, the maximum speed doubles. So even a slight increase in top speed has to be bought with a significant increase in energy output.

Note the we have to input the power in the standard physical unit watt rather than the often used unit horsepower. Luckily the conversion is very easy, just multiply horsepower with 746 to get to watt.

Let’s put the formula to the test.

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I drive a ten year old Mercedes C180 Compressor. According the Mercedes-Benz homepage, its drag coefficient is c = 0.29 and its power P = 143 HP ≈ 106,680 W. Its width and height is w = 1.77 m and h = 1.45 m respectively. What is the maximum possible speed?

First we need the projected area of the car:

A = 1.77 m · 1.45 m ≈ 2.57 m²

Now we can use the formula:

v = ( 2 · 106,680 / (0.29 · 1.25 · 2.57) )1/3

v ≈ 61.2 m/s ≈ 220.3 km/h ≈ 136.6 mph

From my experience on the Autobahn, this seems to be very realistic. You can reach 200 Km/h quite well, but the acceleration is already noticeably lower at this point.

If you ever get the chance to visit Germany, make sure to rent a ridiculously fast sports car (you can rent a Porsche 911 Carrera for as little as 200 $ per day) and find a nice section on the Autobahn with unlimited speed. But remember: unless you’re overtaking, always use the right lane. The left lanes are reserved for overtaking. Never overtake on the right side, nobody will expect you there. And make sure to check the rear-view mirror often. You might think you’re going fast, but there’s always someone going even faster. Let them pass. Last but not least, stay focused and keep your eyes on the road. Traffic jams can appear out of nowhere and you don’t want to end up in the back of a truck at these speeds.

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The fastest production car at the present time is the Bugatti Veyron Super Sport. Is has a drag coefficient of c = 0.35, width w = 2 m, height h = 1.19 m and power P = 1200 HP = 895,200 W. Let’s calculate its maximum possible speed:

v = ( 2 · 895,200 / (0.35 · 1.25 · 2 · 1.19) )1/3

v ≈ 119.8 m/s ≈ 431.3 km/h ≈ 267.4 mph

Does this seem unreasonably high? It does. But the car has actually been recorded going 431 Km/h, so we are right on target. If you’d like to purchase this car, make sure you have 4,000,000 $ in your bank account.

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This was an excerpt from the ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Sources:

http://ecomodder.com/wiki/index.php/Vehicle_Coefficient_of_Drag_List

http://www.mercedes-benz.de/content/germany/mpc/mpc_germany_website/de/home_mpc/passengercars/home/_used_cars/technical_data.0006.html

http://www.carfolio.com/specifications/models/car/?car=218999

What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

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(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.

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We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm

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If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.

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This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.

Intensity: How Much Power Will Burst Your Eardrums?

Under ideal circumstances, sound or light waves emitted from a point source propagate in a spherical fashion from the source. As the distance to the source grows, the energy of the waves is spread over a larger area and thus the perceived intensity decreases. We’ll take a look at the formula that allows us to compute the intensity at any distance from a source.

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First of all, what do we mean by intensity? The intensity I tells us how much energy we receive from the source per second and per square meter. Accordingly, it is measured in the unit J per s and m² or simply W/m². To calculate it properly we need the power of the source P (in W) and the distance r (in m) to it.

I = P / (4 · π · r²)

This is one of these formulas that can quickly get you hooked on physics. It’s simple and extremely useful. In a later section you will meet the denominator again. It is the expression for the surface area of a sphere with radius r.

Before we go to the examples, let’s take a look at a special intensity scale that is often used in acoustics. Instead of expressing the sound intensity in the common physical unit W/m², we convert it to its decibel value dB using this formula:

dB ≈ 120 + 4.34 · ln(I)

with ln being the natural logarithm. For example, a sound intensity of I = 0.00001 W/m² (busy traffic) translates into 70 dB. This conversion is done to avoid dealing with very small or large numbers. Here are some typical values to keep in mind:

0 dB → Threshold of Hearing
20 dB → Whispering
60 dB → Normal Conversation
80 dB → Vacuum Cleaner
110 dB → Front Row at Rock Concert
130 dB → Threshold of Pain
160 dB → Bursting Eardrums

No onto the examples.

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We just bought a P = 300 W speaker and want to try it out at maximal power. To get the full dose, we sit at a distance of only r = 1 m. Is that a bad idea? To find out, let’s calculate the intensity at this distance and the matching decibel value.

I = 300 W / (4 · π · (1 m)²) ≈ 23.9 W/m²

dB ≈ 120 + 4.34 · ln(23.9) ≈ 134 dB

This is already past the threshold of pain, so yes, it is a bad idea. But on the bright side, there’s no danger of the eardrums bursting. So it shouldn’t be dangerous to your health as long as you’re not exposed to this intensity for a longer period of time.

As a side note: the speaker is of course no point source, so all these values are just estimates founded on the idea that as long as you’re not too close to a source, it can be regarded as a point source in good approximation. The more the source resembles a point source and the farther you’re from it, the better the estimates computed using the formula will be.

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Let’s reverse the situation from the previous example. Again we assume a distance of r = 1 m from the speaker. At what power P would our eardrums burst? Have a guess before reading on.

As we can see from the table, this happens at 160 dB. To be able to use the intensity formula, we need to know the corresponding intensity in the common physical quantity W/m². We can find that out using this equation:

160 ≈ 120 + 4.34 · ln(I)

We’ll subtract 120 from both sides and divide by 4.34:

40 ≈ 4.34 · ln(I)   

9.22 ≈ ln(I)

The inverse of the natural logarithm ln is Euler’s number e. In other words: e to the power of ln(I) is just I. So in order to get rid of the natural logarithm in this equation, we’ll just use Euler’s number as the basis on both sides:

e^9.22 ≈ e^ln(I)

10,100 ≈ I

Thus, 160 dB correspond to I = 10,100 W/m². At this intensity eardrums will burst. Now we can answer the question of which amount of power P will do that, given that we are only r = 1 m from the sound source. We insert the values into the intensity formula and solve for P:

10,100 = P / (4 · π · 1²)

10,100 = 0.08 · P

P ≈ 126,000 W

So don’t worry about ever bursting your eardrums with a speaker or a set of speakers. Not even the powerful sound systems at rock concerts could accomplish this.

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This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00.

Increase Views per Visit by Linking Within your Blog

One of the most basic and useful performance indicator for blogs is the average number of views per visit. If it is high, that means visitors stick around to explore the blog after reading a post. They value the blog for being well-written and informative. But in the fast paced, content saturated online world, achieving a lot of views per visit is not easy.

You can help out a little by making exploring your blog easier for readers. A good way to do this is to link within your blog, that is, to provide internal links. Keep in mind though that random links won’t help much. If you link one of your blog post to another, they should be connected in a meaningful way, for example by covering the same topic or giving relevant additional information to what a visitor just read.

Being mathematically curious, I wanted to find a way to judge what impact such internal links have on the overall views per visit. Assume you start with no internal links and observe a current number views per visitor of x. Now you add n internal links in your blog, which has in total a number of m entries. Given that the probability for a visitor to make use of an internal link is p, what will the overall number of views per visit change to? Yesterday night I derived a formula for that:

x’ = x + (n / m) · (1 / (1-p) – 1)

For example, my blog (which has as of now very few internal links) has an average of x = 2.3 views per visit and m = 42 entries. If I were to add n = 30 internal links and assuming a reader makes use of an internal link with the probability p = 20 % = 0.2, this should theoretically change into:

x’ = 2.3 + (30 / 42) · (1 / 0.8 – 1) = 2.5 views per visit

A solid 9 % increase in views per visit and this just by providing visitors a simple way to explore. So make sure to go over your blog and connect articles that are relevant to each other. The higher the relevancy of the links, the higher the probability that readers will end up using them. For example, if I only added n = 10 internal links instead of thirty, but had them at such a level of relevancy that the probability of them being used increases to p = 40 % = 0.4, I would end up with the same overall views per visit:

x’ = 2.3 + (10 / 42) · (1 / 0.6 – 1) = 2.5 views per visit

So it’s about relevancy as much as it is about amount. And in the spirit of not spamming, I’d prefer adding a few high-relevancy internal links that a lot low-relevancy ones.

If you’d like to know more on how to optimize your blog, check out: Setting the Order for your WordPress Blog Posts and Keywords: How To Use Them Properly On a Website or Blog.

Mathematics of Explosions

When a strong explosion takes place, a shock wave forms that propagates in a spherical manner away from the source of the explosion. The shock front separates the air mass that is heated and compressed due to the explosion from the undisturbed air. In the picture below you can see the shock sphere that resulted from the explosion of Trinity, the first atomic bomb ever detonated.

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Using the concept of similarity solutions, the physicists Taylor and Sedov derived a simple formula that describes how the radius r (in m) of such a shock sphere grows with time t (in s). To apply it, we need to know two additional quantities: the energy of the explosion E (in J) and the density of the surrounding air D (in kg/m3). Here’s the formula:

r = 0.93 · (E / D)0.2 · t0.4

Let’s apply this formula for the Trinity blast.

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In the explosion of the Trinity the amount of energy that was released was about 20 kilotons of TNT or:

E = 84 TJ = 84,000,000,000,000 J

Just to put that into perspective: in 2007 all of the households in Canada combined used about 1.4 TJ in energy. If you were able to convert the energy released in the Trinity explosion one-to-one into useable energy, you could power Canada for 60 years.

But back to the formula. The density of air at sea-level and lower heights is about D = 1.25 kg/m3. So the radius of the sphere approximately followed this law:

r = 542 · t0.4

After one second (t = 1), the shock front traveled 542 m. So the initial velocity was 542 m/s ≈ 1950 km/h ≈ 1210 mph. After ten seconds (t = 10), the shock front already covered a distance of about 1360 m ≈ 0.85 miles.

How long did it take the shock front to reach people two miles from the detonation? Two miles are approximately 3200 m. So we can set up this equation:

3200 = 542 · t0.4

We divide by 542:

5.90 t0.4

Then take both sides to the power of 2.5:

t 85 s ≈ 1 and 1/2 minutes

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Let’s look at how the different parameters in the formula impact the radius of the shock sphere:

  • If you increase the time sixfold, the radius of the sphere doubles. So if it reached 0.85 miles after ten seconds, it will have reached 1.7 miles after 60 seconds. Note that this means that the speed of the shock front continuously decreases.

For the other two parameters, it will be more informative to look at the initial speed v (in m/s) rather the radius of the sphere at a certain time. As you noticed in the example, we get the initial speed by setting t = 1, leading to this formula:

v = 0.93 · (E / D)0.2

  • If you increase the energy of the detonation 35-fold, the initial speed of the shock front doubles. So for an atomic blast of 20 kt · 35 = 700 kt, the initial speed would be approximately 542 m /s · 2 = 1084 m/s.

  • The density behaves in the exact opposite way. If you increase it 35-fold, the initial speed halves. So if the test were conducted at an altitude of about 20 miles (where the density is only one thirty-fifth of its value on the ground), the shock wave would propagate at 1084 m/s

Another field in which the Taylor-Sedov formula is commonly applied is astrophysics, where it is used to model Supernova explosions. Since the energy released in such explosions dwarfs all atomic blasts and the surrounding density in space is very low, the initial expansion rate is extremely high.

This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00. You can take another quick look at the physics of shock waves here: Mach Cone.

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.

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The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes

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We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.

Image
———————-

I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).

Physics: Free Fall and Terminal Velocity

After a while of free fall, any object will reach and maintain a terminal velocity. To calculate it, we need a lot of inputs.

The necessary quantities are: the mass of the object (in kg), the gravitational acceleration (in m/s²), the density of air D (in kg/m³), the projected area of the object A (in m²) and the drag coefficient c (dimensionless). The latter two quantities need some explaining.

The projected area is the largest cross-section in the direction of fall. You can think of it as the shadow of the object on the ground when the sun’s rays hit the ground at a ninety degree angle. For example, if the falling object is a sphere, the projected area will be a circle with the same radius.

The drag coefficient is a dimensionless number that depends in a very complex way on the geometry of the object. There’s no simple way to compute it, usually it is determined in a wind tunnel. However, you can find the drag coefficients for common shapes in the picture below.

Now that we know all the inputs, let’s look at the formula for the terminal velocity v (in m/s). It will be valid for objects dropped from such a great heights that they manage to reach this limiting value, which is basically a result of the air resistance canceling out gravity.

v = sq root (2 * m * g / (c * D * A) )

Let’s do an example.

Skydivers are in free fall after leaving the plane, but soon reach the terminal velocity. We will set the mass to m = 75 kg, g = 9.81 (as usual) and D = 1.2 kg/m³. In a head-first position the skydiver has a drag coefficient of c = 0.8 and a projected area A = 0.3 m². What is the terminal velocity of the skydiver?

v = sq root (2 * 75 * 9.81 / (0.8 * 1.2 * 0.3) )

v ≈ 70 m/s ≈ 260 km/h ≈ 160 mph

Let’s take a look how changing the inputs varies the terminal velocity. Two bullet points will be sufficient here:

  • If you quadruple the mass (or the gravitational acceleration), the terminal velocity doubles. So a very heavy skydiver or a regular skydiver on a massive planet would fall much faster.
  • If you quadruple the drag coefficient (or the density or the projected area), the terminal velocity halves. This is why parachutes work. They have a higher drag coefficient and larger area, thus effectively reducing the terminal velocity.

This was an excerpt from the Kindle ebook: Great Formulas Explained – Physics. Mathematics, Economics. Check out my BEST OF for more interesting physics articles.

My Fair Game – How To Use the Expected Value

You meet a nice man on the street offering you a game of dice. For a wager of just 2 $, you can win 8 $ when the dice shows a six. Sounds good? Let’s say you join in and play 30 rounds. What will be your expected balance after that?

You roll a six with the probability p = 1/6. So of the 30 rounds, you can expect to win 1/6 · 30 = 5, resulting in a pay-out of 40 $. But winning 5 rounds of course also means that you lost the remaining 25 rounds, resulting in a loss of 50 $. Your expected balance after 30 rounds is thus -10 $. Or in other words: for the player this game results in a loss of 1/3 $ per round.

 Let’s make a general formula for just this case. We are offered a game which we win with a probability of p. The pay-out in case of victory is P, the wager is W. We play this game for a number of n rounds.

The expected number of wins is p·n, so the total pay-out will be: p·n·P. The expected number of losses is (1-p)·n, so we will most likely lose this amount of money: (1-p)·n·W.

 Now we can set up the formula for the balance. We simply subtract the losses from the pay-out. But while we’re at it, let’s divide both sides by n to get the balance per round. It already includes all the information we need and requires one less variable.

B = p · P – (1-p) · W

This is what we can expect to win (or lose) per round. Let’s check it by using the above example. We had the winning chance p = 1/6, the pay-out P = 8 $ and the wager W = 2 $. So from the formula we get this balance per round:

B = 1/6 · 8 $ – 5/6 · 2 $ = – 1/3 $ per round

Just as we expected. Let’s try another example. I’ll offer you a dice game. If you roll two six in a row, you get P = 175 $. The wager is W = 5 $. Quite the deal, isn’t it? Let’s see. Rolling two six in a row occurs with a probability of p = 1/36. So the expected balance per round is:

B = 1/36 · 175 $ – 35/36 · 5 $ = 0 $ per round

I offered you a truly fair game. No one can be expected to lose in the long run. Of course if we only play a few rounds, somebody will win and somebody will lose.

It’s helpful to understand this balance as being sound for a large number of rounds but rather fragile in case of playing only a few rounds. Casinos are host to thousands of rounds per day and thus can predict their gains quite accurately from the balance per round. After a lot of rounds, all the random streaks and significant one-time events hardly impact the total balance anymore. The real balance will converge to the theoretical balance more and more as the number of rounds grows. This is mathematically proven by the Law of Large Numbers. Assuming finite variance, the proof can be done elegantly using Chebyshev’s Inequality.

The convergence can be easily demonstrated using a computer simulation. We will let the computer, equipped with random numbers, run our dice game for 2000 rounds. After each round the computer calculates the balance per round so far. The below picture shows the difference between the simulated balance per round and our theoretical result of – 1/3 $ per round.

Image

(Liked the excerpt? Get the book “Statistical Snacks” by Metin Bektas here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2)