6. Dimensional Analysis and Using Units
Sign In
An error ocurred, try again later!
Chapter 1
6.
Dimensional Analysis and Using Units
| Student Learning Objectives: |
|---|
|
Why
This lesson delves into the concept of dimensional analysis, a technique used to convert units and solve complex problems. It explains how conversion factors act as a bridge between different units of measurement, making it easier to perform calculations. Derived units are also discussed, which are new units created by combining basic units like meters and seconds. These derived units are particularly useful in calculating speed and distance. For example, if you know the speed of a train and the time it takes to travel between two stations, you can easily find the distance between those stations using derived units. The lesson is a valuable lesson for anyone looking to understand these mathematical tools and how they apply to everyday situations like travel, sports, and science experiments.
Lesson Settings & Tools
| | 13 Theory slides |
| | 12 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Dimensional Analysis and Using Units
Slide of 13
Some of the most common base units are meters, kilograms, and seconds. Apart from these, are there more units that are used in daily life? The answer is yes! Area, volume, and speed are some examples of quantities that can be found by combining base units.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Share
Report error
Translate
Explore
Combining Base Units
The combination of base units creates new units of measure. Some of these combinations have their own names. Using the applet below, units like m/s^2 can be created following these two steps.
- Click on s and hold the click.
- Drag it to the denominator of the fraction and release the click.
To have a clean start, press the Reset
button. Any of the units inside the boxes can be drag either to the numerator or denominator of the fraction at the right. Build new units and find out if they have a particular name.
Share
Report error
Translate
Explore
Distance to the Lightning
One rainy morning, Jordan saw lightning from her bedroom window. After a few seconds of time had passed, she heard the sound of thunder, which shook the photos on her desk. Could Jordan know the distance from her house to the lightning struck? What information does she need?
Share
Report error
Translate
Challenge
Voyager 1 Arrival Time
Voyager 1 is a space probe launched by NASA in 1977. Currently, it is the most distant human-made object from Earth. It takes more than 21 hours for a radio signal to get from Earth to Voyager 1 — traveling at the speed of light! The nearest star to Earth (apart from the Sun) is Proxima Centauri, which is about 4.24 light years, or 4.014* 10^(13) kilometers away.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"years","answer":{"text":["75118"]}}
Share
Report error
Translate
Discussion
Combining Length and Time Units
In the sentence Voyager 1 travels at 61 000 kilometers per hour,
both a unit of time and a unit of length are involved. In this case, the unit of length is distance traveled, and the unit of time is the amount of time it takes for the item to travel the given distance. Combining these units leads to a new type of unit that describes how fast an object is moving.
Concept
Speed
The rate or speed of a moving object, denoted by r, measures how fast the object is moving — that is, the rate of motion of the object. It is calculated by dividing the distance traveled by the amount of time spent traveling.
Speed = Distance traveled/Time interval
From the above, three formulas can be derived relating speed, distance, and time. The triangular diagram below helps to set each formula.
Share
Report error
Translate
Example
Distance Between Train Stations
A train from New York departed at 10:30 A.M. and arrived in Boston at 2:30 P.M.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"miles","answer":{"text":["190"]}}
Hint
Multiply the speed of the train by the time it took to go from one city to the other.
Solution
Since the train left at 10:30 A.M. and arrived at 2:30 P.M., it took 4 hours to travel from one city to other.
t = 4 h
The speed of the train is 47.5 miles per hour. When the combined unit of miles per hour is multiplied by the base unit hours, the result will be in terms of miles. Therefore, to find the distance between the cities, multiply the speed by the time.
d = r* t
▼
Substitute values and simplify
SubstituteII
r= 47.5 mi/h, t= 4 h
d = 47.5 mi/h * 4 h
Rewrite
Rewrite 47.5 mi/h as 47.5 mih
d=47.5 mi/h* 4 h
CrossCommonFac
Cross out common factors
d = 47.5 mi/h* 4 h
CancelCommonFac
Cancel out common factors
d = 47.5 mi/1* 4
DivByOne
a/1=a
d = 47.5 mi* 4
Multiply
Multiply
d = 190 mi
In conclusion, the cities are 190 miles apart.
Share
Report error
Translate
Discussion
Converting Units of Measure
Sometimes, two quantities can be given using different units of measure. However, to operate with the quantities, they need to have the same unit. Consider the diagram below, where d_1 is given in meters, d_2 is given in miles, and the total distance d_1+d_2 is required to be calculated.
In such cases, one of the quantities needs to be converted into an equivalent quantity using the other quantity's unit of measure. For example, d_2 can be converted from miles to meters. In such instances, the conversion factor plays an important role.
Concept
Conversion Factor
A conversion factor is a fraction where the numerator and denominator represent the same quantity with different units. Example Conversion Factor [0.5em] 60 minutes/1 hour Recall that 1 hour and 60 minutes represent the same quantity. Multiplying a quantity by a conversion factor changes the quantity to an equivalent quantity in different units. Examine how to convert 2 hours to minutes using the above conversion factor.
| Given Quantity | Conversion | Result |
|---|---|---|
| 2 hours | 2 hours * 60 minutes/1 hour | 120 minutes |
Although the final result is in minutes, both quantities represent the same amount of time. Note that the opposite conversion, from minutes to hours, has a conversion factor of 1 hour60 minutes. If the task was to convert 120 minutes to hours, 120 minutes would be multiplied by this conversion factor.
| Given Quantity | Conversion | Result |
|---|---|---|
| 120 minutes | 120 minutes * 1 hour/60 minutes | 2 hours |
As shown in the examples above, the process of including units of measurement as factors is called dimensional analysis. Dimensional analysis can also be used when deciding which conversion factor will produce the desired units. In the table, some common conversion factors are used to convert the given measures.
| Given Quantity | Conversion | Result |
|---|---|---|
| 3 pounds | 3 pounds * 16 ounces/1 pound | 48 ounces |
| 160 ounces | 160 ounces * 1 pound/16 ounces | 10 pounds |
| 1 mile | 1 mile * 1760 yards/1 mile | 1760 yards |
Some common conversions involve distance, mass, area, volume, time, and temperature.
Why
The Reason That the Quantities Are EquivalentThe numerator and denominator of the conversion factor represent the same quantity. That means their quotient equals 1. Then, by the Identity Property of Multiplication, the amount of the given quantity does not change when multiplied by the conversion factor.
When converting from one unit to another, the desired unit needs to be in the numerator of the conversion factor while the given unit needs to be in the denominator. That way when the quantity is multiplied by the conversion factor, the given unit will cancel out and the desired unit will remain.
Keep in mind that, despite the given quantity and the new quantity have different values, they represent the same amount.
To convert d_2 from miles to meters, use the fact that 1 mile is the same as 1609.34 meters. 0.25 mi * 1609.34 m/1 mi = 402.335 m It has been shown that 0.25 miles is the same as 402.335 meters. Now that both d_1 and d_2 have the same unit of measure, d_1+d_2 can be found.
d_1+ d_2 &= 300 m + 402.335 m &= 702.335 m
Share
Report error
Translate
Example
Fastball Time to Plate
Tearrik is studying about the theory of baseball during study hall. He reads that on a baseball field, the distance from the pitcher's mound to home plate is 14.8 meters.
At his fastest, Tearrik can throw a fastball at 91 miles per hour. How many seconds does it take for the ball to reach home plate? Round the answer to two decimal places.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"seconds","answer":{"text":["0.36"]}}
Hint
Use the fact that 1 mile is 1609.34 meters and 1 hour is 3600 seconds.
Solution
From the given information, speed and distance are given and time is required.
Note that the speed of the ball is given in miles per hour, but the distance from the pitcher's mound to home plate is in meters. This suggests that a conversion from miles to meters is needed. Miles to Meters Conversion Factor = 1609.34 m/1 mi Multiplying the given speed by the conversion factor will convert the speed from miles per hour to meters per hour.
91 mi/h
Mult
Multiply by 1609.34 m/mi
91 mi/h * 1609.34 m/mi
▼
Simplify
WriteAsFrac
Write as a fraction
91 mi/1 h * 1609.34 m/1 mi
CrossCommonFac
Cross out common factors
91 mi/1 h * 1609.34 m/1 mi
CancelCommonFac
Cancel out common factors
91/1 h * 1609.34 m/1
Multiply
Multiply
146 449.94 m/h
Because the answer is required to be in seconds, a conversion from hours to seconds has to be made. To do so, the conversion factor needs to have a quantity in hours in the numerator and the equivalent quantity in seconds in the denominator. meters/hour * hour/seconds = meters/second Using the fact that that 1 hour is equivalent to 3600 seconds, the conversion factor can be set. Hours to Seconds Conversion Factor = 1 h/3600 s Having the conversion factor set, the conversion can be performed.
146 449.94 m/h
Mult
Multiply by 1 h/3600 s
146 449.94 m/h * 1 h/3600 s
▼
Simplify
WriteAsFrac
Write as a fraction
2 449.94 m/1 h* 1 h/3600 s
CrossCommonFac
Cross out common factors
2 449.94 m/1 h* 1 h/3600 s
CancelCommonFac
Cancel out common factors
146 449.94 m/3600 s
CalcQuot
Calculate quotient
40.680538... m/s
RoundDec
Round to 2 decimal place(s)
40.68 m/s
When units of distance are divided by units of speed, the units of distance cancel out, giving a result in units of time. meters/meterssecond = second Therefore, to compute the time it takes the ball to reach the home plate, the distance must be divided by the speed.
t = d/r
▼
Substitute values and simplify
SubstituteII
d= 14.8 m, r= 40.68 m/s
t = 14.8 m/40.68 m/s
DivByFracD
a/b/c= a * c/b
t = 14.8 m* s/40.68 m
CrossCommonFac
Cross out common factors
t = 14.8 m* s/40.68 m
CalcQuot
Calculate quotient
0.363815... s
RoundDec
Round to 2 decimal place(s)
t = 0.36 s
The ball reaches home plate in about 0.36 seconds, around one-third of a second. This is the same amount of time it takes to blink!
Share
Report error
Translate
Discussion
Defining New Units
As these examples show, speed combines two different base units, distance and time, in a quotient. Similarly, different combinations of base units can lead to new units, which are called derived units.
Concept
Derived Units
Derived units are units of measurement formed by multiplying or dividing base units in different combinations.
| Quantity (Symbol) | Expression | Base Units | Unit Symbol (Name) |
|---|---|---|---|
| Area (A) | length * width | m^2 | m^2 |
| Volume (V) | area * height | m^3 | m^3 |
| Speed (r) | distance/time | .m /s. | .m /s. |
| Acceleration (a) | speed/time | .m /s^2. | .m /s^2. |
| Density ( ) | mass/length^3 | .kg /m^3. | .kg /m^3. |
| Force (F) | mass * acceleration | .kg* m /s^2. | N (newton) |
| Pressure (P) | force/area | .N /m^2. | Pa (pascal) |
| Energy (E) | force * distance | N*m | J (joule) |
| Power (P) | force* distance/time | .J /s. | W (watt) |
| Frequency (F) | 1/time | .1 /s. | Hz (hertz) |
One common example of a derived unit is the acceleration of an object, found by dividing the change in speed by the change in time. The change in speed is measured as the difference between the final and initial speeds.
Acceleration = Final speed-Initial speed/Time taken
Share
Report error
Translate
Example
Acceleration of a Formula 1 Car
When crossing the starting line, the speed of a Formula 1 car is 50 meters per second. After 6 seconds, its speed is 86 meters per second. Find the acceleration rate, in meters per second squared, of the car in that time interval.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"a=","formTextAfter":"m\/s^2","answer":{"text":["6"]}}
Hint
The acceleration rate equals the change in speed divided by the time taken. The change in speed is the final speed minus the initial speed.
Solution
It is given that the initial speed of the Formula 1 car is 50 meters per second and the speed after 6 seconds is 86 meters per second. Since the final speed is greater, it means the car is accelerating.
Change in speed = Final speed-Initial speed
SubstituteII
Final speed= 86 m/s, Initial speed= 50 m/s
Change in speed = 86 m/s - 50 m/s
SubTerms
Subtract terms
Change in speed = 36 m/s
Now that the change in speed is known, the acceleration rate of the Formula 1 car can be computed.
a = Change in speed/Time taken
SubstituteII
Change in speed= 36 m/s, Time taken= 6 s
a = 36 m/s/6 s
WriteAsFrac
Write as a fraction
a =36 m1 s/6 s
DivFracD
a/c/b= a/b* c
a = 36 m/6 s * s
Multiply
Multiply
a = 36 m/6 s^2
CalcQuot
Calculate quotient
a = 6 m/s^2
In conclusion, the acceleration rate of the car during those 6 seconds was 6 meters per second squared.
Share
Report error
Translate
Example
Impact Force
During baseball practice, a ball hit by Tearrik traveled 4 seconds before hitting the center field wall at a speed of 49 meters per second. Tearrik wondered about the force with which the ball impacted the wall. Since force is mass times acceleration, to help him, the coach told Tearrik that a baseball has a mass of 145 grams.
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":"F\u2248","formTextAfter":"N","answer":{"text":["1.78"]}}
Hint
From the derived units table, the base units for newtons are kilograms multiplied by meters per second squared. Therefore, the mass of the ball needs to be converted from grams to kilograms. Acceleration equals the change in speed divided by the change in time.
Solution
According to the derived units table, force is obtained by multiplying the mass of the object, in kilograms, by its acceleration in meters per second squared.
F = mass_(kg)* acceleration_(m/s^2)
First, convert the mass of the baseball from grams to kilograms using the fact that 1 kilogram is 1000 grams.
145 g
Mult
Multiply by 1 kg/1000 g
145 g * 1 kg/1000 g
▼
Simplify
CrossCommonFac
Cross out common factors
145 g * 1 kg/1000 g
CancelCommonFac
Cancel out common factors
145 * 1 kg/1000
Multiply
Multiply
145 kg/1000
CalcQuot
Calculate quotient
0.145 kg
It is given that the ball hit the center field wall at a speed of 49 meters per second. This is the final speed of the ball. Also, it is said that the speed of the ball at the moment it was hit — the initial speed — was at 0 meters per second. Change in speed = 49 m/s - 0 m/s ⇓ Change in speed = 49 m/s Using this speed and the fact that the ball traveled 4 seconds before hitting the wall, its acceleration can be found.
a = r/t
▼
Substitute values and evaluate
SubstituteII
r= 49 m/s, t= 4 s
a = 49 m/s/4 s
WriteAsFrac
Write as a fraction
a = 49 m1 s/4 s
DivFracD
a/c/b= a/b* c
a = 49 m/4 s*s
Multiply
Multiply
a = 49 m/4 s^2
CalcQuot
Calculate quotient
a = 12.25 m/s^2
Finally, the force with which the ball impacted the center field wall can be calculated.
F = m * a
▼
Substitute values and simplify
SubstituteII
m= 0.145 kg, a= 12.25 m/s^2
F = 0.145 kg * 12.25 m/s^2
Multiply
Multiply
F = 1.77625 kg*m/s^2
RoundDec
Round to 2 decimal place(s)
F ≈ 1.78 kg*m/s^2
Substitute
kg*m/s^2= N
F ≈ 1.78 N
The ball impacted the wall with about 1.78 netwons of force.
Share
Report error
Translate
Example
Time vs. Speed Graph
Saturday morning before baseball practice, Tearrik went out for a jog beginning at 10:00 A.M. He was 1500 meters away from his starting location at 10:10 A.M. Over the next 5 minutes he jogged another 500 meters. Tearrik then jogged another 1120 meters before finally stopping at 10:22 A.M. Make a graph representing the time and Tearrik's speed during his jog.
Answer
Hint
Place time on the horizontal axis and speed on the vertical axis. Find the speed in each time interval by dividing the distance covered by the time interval.
Solution
Start by organizing the given information in a table. Note that information for three time intervals is given.
| Start Time | End Time | Time Interval (minutes) | Distance Covered (meters) |
|---|---|---|---|
| 10:00 A.M. | 10:10 A.M. | 10 | 1500 |
| 10:10 A.M. | 10:15 A.M. | 5 | 500 |
| 10:15 A.M. | 10:22 A.M. | 7 | 1120 |
To find the speed in each interval, divide the distance by the time interval.
| Time Interval (minutes) | Distance Covered (meters) | Speed (meters per minute) |
|---|---|---|
| 10 | 1500 | 1500/10 = 150 |
| 5 | 500 | 500/5 = 100 |
| 7 | 1120 | 1120/7 = 160 |
Next, place time on the horizontal axis and speed on the vertical axis. Each number to the right of the origin corresponds to the minutes after 10:00 A.M. Based on the table, it is convenient to divide the horizontal axis into 5-minute intervals and the vertical axis into intervals of 50 meters per minute.
Next, translate the information from the table to the graph.
- The first 10 minutes, Tearrik's speed was at 150 meters per minute. This information is represented by a horizontal line from (0,150) to (10,150).
- The next 5 minutes, Tearrik's speed was at 100 meters per minute. A line from (10,100) to (15,100) represents this information.
- The last 7 minutes, Tearrik's speed was 160 meters per minute. This is represented by a line from (15,160) to (22,160).
Consequently, the graph representing the time and speed during Tearrik's jog is represented as follows.
Share
Report error
Translate
Closure
Voyager 1 Arrival Year
With all this new information learned, the time it takes Voyager 1 to reach Proxima Centauri can now be calculated. Recall that Voyager 1 travels at a constant speed of 61 000 kilometers per hour and that the distance from Earth to Proxima Centauri is 4.24 light years, which is the same as 4.014* 10^(13) kilometers. Speed &= 61 000 km/h Distance &= 4.014* 10^(13) km To determine when Voyager 1 will reach Proxima Centauri, use the formula below.
Time= Distance/Speed
Next, substitute the values and simplify. Remember, r stands for the speed (rate).
t = d/r
▼
Substitute values and simplify
SubstituteII
d= 4.014* 10^(13) km, r= 61 000 km/h
t = 4.014* 10^(13) km/61 000 km/h
DivByFracD
a/b/c= a * c/b
t = 4.014* 10^(13) km* h/61 000 km
CrossCommonFac
Cross out common factors
t = 4.014* 10^(13) km* h/61 000 km
CancelCommonFac
Cancel out common factors
t = 4.014* 10^(13) h/61 000
4.014* 10^(13) h/61 000
Mult
Multiply by 1 day/24 h*1 year/365 days
4.014* 10^(13) h/61 000* 1 day/24 h*1 year/365 days
▼
Simplify
CrossCommonFac
Cross out common factors
4.014* 10^(13) h/61 000* 1 day/24 h*1 year/365 days
MultFrac
Multiply fractions
4.014* 10^(13) years/61 000* 24 * 365
Multiply
Multiply
4.014* 10^(13) years/534 360 000
CalcQuot
Calculate quotient
75 117.898046... years
RoundInt
Round to nearest integer
75 118 years
Share
Report error
Translate
Dimensional Analysis and Using Units
Exercise 2.1
Ali, the plumber, has discovered a leaky pipe in his house. He calculates that the pipe is leaking at a rate of 8 fluid ounces per minute. How fast is the pipe leaking in gallons per hour?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"gal\/h","answer":{"text":["3.75"]}}
security
report
code
security
report
code
We want to convert the current rate of leaking, 8 fluid ounces per minute, to gallons per hour. First let's express 8 fluid ounces per minute as a unit rate. 8 oz/1 min Next, we need to make two separate conversions, gallons to ounces and minutes to hours. Each conversion should have a conversion factor.
| Measure | Conversion Factor |
|---|---|
| 1gallon equals 128 ounces | 1gal/128 oz |
| 60 minutes equals 1 hour | 60 min/1 h |
We can convert 8 fluid ounces per minute into gallons per hour by multiplying the conversion factors by the current leaking rate.
The pipe leaks at a rate of 3.75 gallons per hour.
security
report
code
E
Hint & Answer
S
Solution
Without performing the conversion, will the new number of units be greater or less than the number of original units when converting from kilometers per hour to miles per hour?
{"type":"choice","form":{"alts":["Less","Greater"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":0}
security
report
code
security
report
code
Let's write the two units as ratios. kilometers/hour and miles/hour Since hours is the same in both ratios, the only unit we need to convert is kilometers to miles. We have the following equivalence. 1.609 km = 1 mi Since 1 mile represents the same distance as 1.609 kilometers, one mile is greater than one kilometer. 1 mile > 1 kilometer Therefore, when converting from kilometers to miles, the resulting number of miles must always be less than the number of kilometers. Consequently, the number of miles per hour is also a lesser number than the corresponding number of kilometers per hour.
Let 5 kilometers per hour be the speed of a certain object. To convert it to miles per hour, multiply the speed by the corresponding conversion factor.
5 km/1 h* 1 mi/1.609 km ≈ 3.11 mi/h
As can be seen, 3.11, the new number of units, is less than 5, the number of original units.
security
report
code
E
Hint & Answer
S
Solution
Copy and complete the statement. Assume that 1 foot equals 0.305 meters. 60ft/s= km/h
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":"60ft\/s=","formTextAfter":"km\/h","answer":{"text":["65.88"]}}
security
report
code
security
report
code
The left-hand side of the equation can be written as a ratio. 60ft/s In this exercise, we need to change feet per second into kilometers per hour. To do that, we should first convert feet to meters and then meters to kilometers. Similarly, seconds should be converted to minutes and then minutes to hours. Since multiple conversion factors are involved, let's organize them in a table.
| Measure | Conversion Factor |
|---|---|
| 1 foot equals 0.305 meters | 0.305m/1 ft |
| 1000 meters equals 1 kilometers | 1 km/1000 m |
| 60 seconds equals 1 minute | 60 s/1 min |
| 60 minutes equals 1 hour | 60 min/1 h |
To perform this conversion, we need to multiply the given value by each conversion factor.
We can now complete the statement. 60ft/s=65.88km/h
security
report
code
E
Hint & Answer
S
Solution
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":null,"answer":{"text":["31536000"]}}
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"%","answer":{"text":["33"]}}
security
report
code
security
report
code
To convert the age of the Universe from seconds to years, we need to multiply 435 quadrillion by multiple conversion factors. cl seconds & ↓ & minutes & (60 s/min) ↓ & hours & (60 min/h) ↓ & days & (24 h/day) ↓ & years & (365 days/year) Let's construct the conversion factors so that we end up with years as the unit of measure.
| Measure | Conversion Factor |
|---|---|
| 60 second equals 1 minute | 1min/60 s |
| 60 minutes equals 1 hour | 1h/60 min |
| 24 hours equals 1 day | 1day/24 h |
| 1 year equals 365 days | 1 year/365 days |
To perform the conversion, each conversion factor should be multiplied by the age of the Universe. Notice that 435 quadrillion can be written as 435* 10^(15).
Consequently, to convert the age of the Universe from seconds to years, 435* 10^(15) must be divided by 31 536 000.
To determine the percentage of the age of the Universe that the Earth has existed, we divide the age of Earth by the age of the Universe. Here, for the age of the Universe, we will use the fraction obtained in the previous part.
The Earth has existed for approximately 33 % of the time that has passed since the Big Bang.
security
report
code
E
Hint & Answer
S
Solution
Ramsha is driving her car at 90 kilometers per hour. How many seconds does it take for Ramsha to travel 200 meters?
{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]},"rendered":false},"text":""},"formTextBefore":null,"formTextAfter":"seconds","answer":{"text":["8"]}}
security
report
code
security
report
code
To determine how many seconds it takes to travel 200 meters when Ramsha drives with a speed of 90 kilometers per hour, we must first convert the speed to meters per second. This requires us to use two conversion factors. Remember that we must write the conversion factors in a way that cancels out kilometers and hours.
| Measure | Conversion Factor |
|---|---|
| 1000 meters equals 1 kilometer | 1000m/1 km |
| 1 hour equals 3600 seconds | 1h/3600 s |
To convert the speed, we must multiply 90 kilometers per hour by both of these conversion factors.
Having converted the speed into meters per second, we can use the following formula to determine time. Time=Distance/Speed By substituting distance and speed, we can figure out the time it takes Ramsha to travel 200 meters.
It will take Ramsha 8 seconds to travel 200 meters.
security
report
code
E
Hint & Answer
S
Solution
Dimensional Analysis and Using Units
Recommended exercises
To get personalized recommended exercises, please login first.
Loading content
Edit Lesson
≥
>
■2
■▢
e▢
7
8
9
×
÷■1
=
=
√▢
▢√
∫▢▢
4
5
6
+
−
≤
<
log
ln
log▢
1
2
3
()
∣
sin
cos
tan
0
.
π
x
y
Loading content