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Moving Clocks Run Slower – Time Dilation

Special relativity theory introduced an interesting notion about time. Time does not pass at the same rate for moving frames of reference. Moving clocks run slower than clocks in a stationary frame of reference. This effect is known as time dilation. To calculate this time difference, a Lorentz transformation is used.

Time Dilation Lorentz Formula
TM is the time duration measured in the moving frame of reference
TS is the time duration measured from the stationary frame of reference
v is the velocity of the moving frame of reference
c is the speed of light

Time Dilation Example Problem

One way this effect was experimentally proven was measuring the lifetime of high energy muons. Muons (symbol μ) are unstable elementary particles that exist for an average of 2.2 μsec before decaying into an electron and two neutrinos. Muons are formed naturally when cosmic ray radiation interacts with the atmosphere. They can be produced as a by-product of particle collider experiments where their time of existence can measured accurately.

A muon is created in a laboratory and observed to exist for 8.8 μsec. How fast was the muon moving?


Time Dilation - Relativity Example Problem

A muon forms at t=0 moving at a velocity v. After 2.2 microseconds, the muon decays. A stationary observer measured the lifetime to be 8.8 microseconds. What was the velocity of the muon?

From the muon’s frame of reference, it exists for 2.2 μsec. This is the TM value in our equation.
TS is the time measured from the static frame of reference (the laboratory) at 8.8 μsec, or four times as long as it should exist: TS = 4 TM.

We want to solve for velocity, Let’s simplify the equation a little bit. First, divide both sides by TM.

Time Dilation Example Step 2

Flip the equation over

Time Dilation Step 3

Square both sides to get rid of the radical.

Time Dilation Step 4

This form is easier to work with. Use the TS = 4 TM relationship to get

time dilation step 5
Time Dilation step 6

Cancel out the TM2 to leave

Time Dilation Step 7

Subtract 1 from both sides

Time Dilation Example Step 8
Time Dilation Example step 9
Time Dilation Example Step 10

Multiply both sides by c2

Time Dilation Example Step 11

Take the square root of both sides to get v

Time Dilation Example step 12
v = 0.968c 


The muon was moving at 96.8% the speed of light.

One important note about these types of problems is the velocities must be within a few orders of magnitude of the speed of light to make a measurable and noticeable difference.

How To Find the Limiting Reactant – Limiting Reactant Example

Ammonia Ball and Stick Model

3D ball and stick model of the ammonia molecule. Todd Helmenstine

Many chemical reactions take place until one of the reactants run out. This reactant is known as the limiting reactant. Often it is straightforward to determine which reactant will be the limiting reactant, but sometimes it takes a few extra steps.

For example, burning propane in a grill. The propane and oxygen in the air combust to create heat and carbon dioxide. You are obviously more likely to run out of propane long before you run out of oxygen in the air. This makes the propane the limiting reactant. Other reactions aren’t quite as easy.

This example problem will show how to use the stoichiometric ratios between the reactants given in the balanced chemical equation to determine the limiting reactant.

Find the Limiting Reactant Example

Question: Ammonia (NH3) is produced when nitrogen gas (N2) is combined with hydrogen gas (H2) by the reaction

N2 + 3 H2 → 2 NH3

50 grams of nitrogen gas and 10 grams of hydrogen gas are reacted together to form ammonia. Which of the two gasses will run out first? (Which gas is the limiting reactant?)

Answer:  The reaction shows us for every mole of N2 consumed, 3 moles of H2 is also consumed. We need 3 moles of hydrogen gas for every mole of nitrogen gas. The first thing we need to find out is the number of moles of each gas is on hand.

N2 Gas: How many moles of nitrogen gas is 50 grams? One mole of nitrogen is 14.007 grams, so one mole of N2 will weigh 28.014 grams.

Limiting Reactant Example Step 1

Limiting Reactant Example Step 2

x moles N2 = 1.78

H2 Gas: How many moles of hydrogen gas is 10 grams? One mole of hydrogen is 1.008 grams so one mole of H2 is 2.016 grams.

Limiting Reactant step 3
Limiting Reactant Example Step 4
x moles H2 = 4.96

Now we know the number moles of each reactant, we can use the ratio from the chemical equation to compare the amounts.  The ratio between hydrogen gas and nitrogen gas should be:

Limiting Reactant Example Step 5

If we divide our moles of H2 into moles of N2, our value will tell us which reactant will come up short. Any value greater than the above ratio means the top reactant is in excess to the lower number. A value less than the ratio means the top reactant is the limiting reactant. The key is to keep the same reactant on top as the step above.

Limiting Reactant Example 6

Since our value is less than the ideal ratio, the top reactant is the limiting reactant. In our case, the top reactant is the hydrogen.

Answer: Hydrogen gas is the limiting reactant.

It doesn’t matter which reactant you put on top when you do this type of problem as long as you keep it the same throughout the calculations. If we had put nitrogen gas on top instead of hydrogen the ratio would have worked out the same way. The ideal ratio would have been 13 and the calculated ratio would have been 0.358 ( 1.78/4.96 ). The value would have been greater than the ideal ratio so the bottom reactant in the ratio would be the limiting reactant. In this case, it is the hydrogen gas.


SOHCAHTOA Example Problem – Trigonometry Help

SOHCAHTOA is the mnemonic used to remember which sides of a right triangle are used to find the ratios needed to determine the sine, cosine or tangent of an angle. Here are a pair of SOHCAHTOA example problems to help show how to use these relationships. If you have no idea what SOHCAHTOA means, check out this introduction to SOHCAHTOA.

SOHCAHTOA Example Problem 1

A mathematically inclined squirrel sits atop a 14-foot tall tree. He spies a nut on the ground some distance away. After careful measurements, he determines the nut is 74º from the base of his tree.
How far away is the nut from:
A) The base of the tree?
B) The math squirrel?

Here is a layout of the problem.

Squirrel and Nut SOHCAHTOA Example Problem

Part A: How far is the nut from the base of the tree?

Looking at our triangle, we see we know the angle and the length of the adjacent side. We want to know the length of the opposite side of the triangle. The part of SOHCAHTOA with these three parts is TOA, or tan θ = opposite / adjacent.

tan θ = opposite / adjacent
tan ( 74º ) = opposite / 14 ft

solve for ‘opposite’

opposite = 14 ft ⋅ tan ( 74º )
opposite = 14 ft ⋅ 3.487
opposite = 48.824 ft

The nut is 48.824 ft from the base of the tree.

Part B: How far is the nut from the math squirrel?

This time, the needed distance is the hypotenuse of the triangle. The part of SOHCAHTOA with  both adjacent and hypotenuse is CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos ( 74º ) = 14 ft / hypotenuse
hypotenuse = 14 ft / cos ( 74º )
hypotenuse = 14 ft / 0.276
hypotenuse = 50.791 ft

We can use the Pythagorean Theorem to check our work.

a2 + b2 = c2
(14 ft)2 + (48.824 ft)2 = c2
196 ft2+ 2383.783 ft2= c2
2579.783 ft2 = c2
50.792 ft = c

The two numbers are close enough to account for rounding errors.

SOHCAHTOA Example Problem 2

A prince has arrived to rescue a princess standing in a tower balcony 15 meters up across a moat 12 meters wide. He knew the princess was high up and there was a moat, so he brought the longest ladder in the kingdom, a 20-meter monster.

A) Is the ladder long enough?
B) If the top of the 20 m ladder touches the balcony, how far away from the tower wall is the bottom of the ladder?
C) What is the angle between the ground and the ladder?

Here is an illustration of our situation.

Princess rescue trigonometry examplePart A: Is the ladder long enough?  To know if the ladder is long enough, we need to know how long it needs to be. At the very least, the prince’s ladder must reach the 15-meter balcony from the edge of the 12-meter moat. Use the Pythagorean theorem to find out how far up the tower wall a 20-meter ladder will reach.

a2 + b2 = c2
height2 + moat2 = ladder2
height2 = ladder2 – moat2
height2 = (20 m)2 – (12 m)2
height2 = 400 m2 – 144 m2
height2 = 256 m2
height = 16 m

A 20-meter ladder can reach 16 meters up the side of the tower. This is a meter longer than the 15 meters needed to reach the princess balcony.

Part B: When the ladder touches the edge of the balcony, how far away from the tower wall is the bottom of the ladder?

The first part measured the height the 20-meter ladder reached when the ladder was placed at the edge of the moat. We found we had more ladder than needed. If the ladder touches the balcony, we know the height it reaches is 15 meters. The ladder is still 20 meters long. We just need to find out how far from the tower to stick the bottom of the ladder. Use the Pythagorean theorem again.

a2 + b2 = c2
(balcony height)2 + (ground distance)2 = ladder2
(ground distance)2 = ladder2 – (balcony height)2

(ground distance)2 = (20 m)2 – (15 m)2
(ground distance)2 = 400 m2 – 225 m2
(ground distance)2 = 175 m2
ground distance = 13.23 m

The ladder’s base is 13.23 m away from the tower.

Part C: What is the angle between the ground and the ladder?

We were given the height of the balcony, which in this case is the ‘opposite’ side of the triangle from the needed angle. We also know the length of the ladder that forms the hypotenuse of the triangle. The part of SOHCAHTOA that has both of these parts is SOH, or sin θ = opposite/hypotenuse. Use this to solve for the angle.

sin θ = opposite/hypotenuse
sin θ = 15 m / 20 m
sin θ = 0.75
θ = sin-1(0.75)
θ = 48.59°

You can check your work in Part B if you use the answer we got as the ‘adjacent’ side of the triangle.

If you use the ladder as the hypotenuse, use CAH, or cos θ = adjacent / hypotenuse.

cos θ = adjacent / hypotenuse
cos θ = 13.23 m / 20 m
cos θ = 0.662
θ = cos-1(0.662)
θ = 48.59°

The angle matches the value above.

If you choose the balcony height as the opposite side, use TOA, or tan θ = opposite / adjacent

tan θ = opposite / adjacent
tan θ = 15 m / 13.23 m
tan θ = 1.134
θ = tan-1(1.134)
θ = 48.59°

Pretty neat how it doesn’t really matter which sides of the right triangle you use, as long as you use the correct part of SOHCAHTOA.


Right Triangle Trigonometry and SOHCAHTOA

Sohcahtoa isn't actually an Egyptian god, but if it helps to remember him that way, you'll have an easier time recalling right angle trig relationships.

Sohcahtoa isn’t actually an Egyptian god, but if it helps to remember him that way, you’ll have an easier time recalling right angle trig relationships.

Right triangles are extremely common in science homework. Even though they are common, they can be confusing to new students. That is why we have the Egyptian god SOHCAHTOA.

SOHCAHTOA is a handy mnemonic trigonometry students learn to remember which sides of a triangle are used for the three main trig functions: sine, cosine, and tangent.

These functions are defined by the ratios of various lengths of the sides of a right triangle. Let’s look at this right triangle.

Right TriangleThis triangle is made up of three sides of lengths a, b and c. Note the angle marked θ. This angle is formed by the intersection of b and c. The hypotenuse is always the longest of the three sides and opposite of the right angle. The side b is ‘adjacent’ to the angle, so this side is known as the adjacent side. It follows the side ‘opposite’ of the angle is known as the opposite side. Now that we have all our sides labelled, we can use SOHCAHTOA.


S – Sine
O – Opposite
H – Hypotenuse

C – Cosine
A – Adjacent
H – Hypotenuse

T – Tangent
O – Opposite
A – Adjacent

SOH = sin θ = opposite over hypotenuse = ac
CAH = cos θ = adjacent over hypotenuse = bc
TOA = tan θ = opposite over adjacent = ab

Easy to remember. Now let’s see how easy it is to apply.

Example Problem

Consider this triangle.

trig example for SOHCAHTOAThe hypotenuse has a length of 10 and one angle of the triangle is 40º. Find the lengths of the other two sides.

Let’s start with the side with length a. This side is opposite the angle and we know the length of the hypotenuse. The part of SOHCAHTOA with both hypotenuse and opposite is SOH or sine.

sin 40º = opposite / hypotenuse
sin 40º = a / 10

solve for a by multiplying both sides by 10.

10 sin 40º = a

Punch 40 into your calculator and hit the sin key to find the sine of 40º.

sin 40º = 0.643

a = 10 sin 40º
a = 10 (0.643)
a = 6.43

Now let’s do side b. This side is adjacent to the angle, so we should use CAH or cosine.

cos 40º = adjacent / hypotenuse
cos 40º = b / 10

solve for b

b = 10 cos 40º

Enter 40 and hit the cos button on your calculator to find:

cos 40º = 0.766

b = 10 cos 40º
b = 10 (0.766)
b = 7.66

The sides of our triangle are 6.43 and 7.66. We can use the Pythagorean equation to check our answer.

a2 + b2 = c2
(6.43)2 + (7.66)2 = c2
41.35 + 58.68 = c2
100.03 = c2
10.00 = c

10 is the length of the triangle’s hypotenuse and matches our calculation above.

As you can see, our friend SOHCAHTOA can help us calculate the angles and lengths of the sides of right triangles with very little information. Make him your friend too.


Conservation of Momentum Example Problem

Momentum is a measurement of inertia in motion. When a mass has velocity, it has momentum. Momentum is calculated by the equation

momentum = mass x velocity
momentum = mv

This conservation of momentum example problem illustrates the principle of conservation of momentum after a collision between two objects.


Consider a 42,000 kg train car travelling at 10 m/s toward another train car. After the two cars collide, they couple together and move along at 6 m/s. What is the mass of the second train car?

Momentum Example Problem 1


In a closed system, momentum is conserved. This means the total momentum of the system remains unchanged before and after the collision.

Before the collision, the total momentum was the sum of the momentums of both train cars.

The first car’s (blue freight car) momentum is

momentumBlue = mv
momentumBlue = (42,000 kg)(10 m/s)
momentumBlue = 420,000 kg·m/s

The second car’s (tanker car) momentum is

momentumtanker = mv
momentumtanker = m(0 m/s)
momentumtanker = 0 kg·m/s

Add these two together to get the total momentum of the system prior to collision.

Total momentum = momentumBlue + momentumtanker
Total momentum = 420,000 kg·m/s + 0 kg·m/s
Total momentum = 420,000 kg·m/s

After the collision, the two cars couple together into one mass. This new mass is

massBlue + masstanker
40,000 kg + masstanker

The velocity the new pair of cars is travelling is 6 m/s. Because momentum is conserved, we know the total momentum of the cars after the collision is equal to the momentum prior to the collision.

Total Momentum = 420,000 kg·m/s
Total Momentum = mv
Total momentum = (42,000 kg + masstanker)·(6 m/s)

420,000 kg·m/s = (42,000 kg + masstanker)·(6 m/s)

Divide both sides by 6 m/s

70,000 kg = 42,000 kg + masstanker

Subtract 40,000 kg from both sides

70,000 kg – 40,000 kg = masstanker
30,000 kg = masstanker


The mass of the second car is equal to 30,000 kg.

Remember, the momentum of a system is conserved. The momentum of the individual masses may change, but the net momentum of the system does not change.

Periodic Table Scavenger Hunt Worksheet

Grab your periodic table and search for the answers to this Periodic Table Scavenger Hunt!

Periodic Table Scavenger Hunt

How To Use the Scavenger Hunt Worksheets

Fun Periodic Table Scavenger Hunt

Fun Periodic Table Scavenger Hunt

Use a periodic table to hunt up the answers to these 20 periodic table questions. The reward for your hard work is an expanded knowledge of how to use a periodic table!

Download the PDF of this worksheet for easier printing. If you need the answers, simply take a look at the answer key. If you need to print the answer key, use this PDF.

Happy Hunting!

How to Calculate Standard Deviation

Statistics Bar GraphStandard deviation is a measurement of how spread out the numbers are of a set of data values. The closer the standard deviation is to zero, the closer the data points are to the mean. Large values of standard deviation is an indication the data is spread out away from the mean. This will show how to calculate the standard deviation of a set of data.

Standard deviation, represented by the lower case Greek letter, σ is calculated from the variance from the mean of each data point. Variance is simply the average of the squared difference of each data point from the mean.

There are three steps to calculating variance:

  1. Find the mean of the data.
  2. For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
  3. Find the mean of the values found in step 2.

Example: Let’s take a set of test scores from a math class of nine students. The scores were:

65, 95, 73, 88, 83, 92, 74, 83, and 94

Step 1 is find the mean. To find the mean, add all these scores together.

65 + 95 + 73 + 88 + 83 + 92 + 74 + 83 + 94 = 747

Divide this value by the total number of tests (9 scores)

747 ÷ 9 = 83

The mean score on the test was a score of 83.

For step 2, we need to subtract the mean from each test score and square each result.

(65 – 83)² = (-18)² = 324
(95 – 83)² = (12)² =144
(73 – 83)² = (-10)² = 100
(88 – 83)² = (5)² = 25
(83 – 83)² = (0)² = 0
(92 – 83)² = (9)² = 81
(74 – 83)² = (-9)² = 81
(83 – 83)² = (0)² = 0
(94 – 83)² = (11)² = 121

Step 3 is find the mean of these values. Add them all together:

324 + 144 + 100 + 25 + 0 + 81 + 81 + 0 + 121 = 876

Divide this value by the total number of scores (9 scores)

876 ÷ 9 = 97 (rounded to the nearest whole score)

The variance of the test scores is 97.

The standard deviation is simply the square root of the variance.

σ = √97 = 9.8 (round to nearest whole test score = 10)

This means scores within one standard deviation, or 10 points of the average score could be all considered ‘average scores’ of the class. The two scores 65 and 73 would be considered ‘below average’ and the 94 would be ‘above average’.

This calculation of standard deviation is for population measurements. This is when you can account for all the data in the population of the set. This example had a class of nine students. We know all the scores of all the students in the class. What if these nine scores were randomly taken from a larger set of scores, say the entire 8th Grade. The set of nine test scores is considered a sample set from the population.

Sample standard deviations are calculated slightly different. The first two steps are identical. In step 3, instead of dividing by the total number of tests, you divide by one less than the total number.

In our example above, the total from step 2 added together was 876 for 9 test scores. To find the sample variance, divide this number by one less than 9, or 8

876 ÷ 8 = 109.5

The sample variance is 109.5. Take the square root of this value to get the sample standard deviation:

sample standard deviation =  √109.5 = 10.5


To find the population standard deviation:

  1. Find the mean of the data.
  2. For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
  3. Find the mean of the values found in step 2.
  4. Divide the value of step 3 by the total number of values.
  5. Take the square root of the result of step 4.

To find the sample standard deviation:

  1. Find the mean of the data.
  2. For each number in the data set, subtract the mean found in step 1 from each value and then square each value.
  3. Find the mean of the values found in step 2.
  4. Divide the value of step 3 by the total number of values minus 1.
  5. Take the square root of the result of step 4.