Question

use dimensional analysis to convert each measurement .
24.L = _______ gal

Answer 1

i got this ok so

Answer 2

there are 3.78 liters in a gallon so u divide that by 24 and get what

Answer 3

about 6.34 gallons

Answer 4

ok so how u do that

Answer 5

oh ok well then listen to him or write 3 73/100 = 6 34/100

Answer 6

To use dimensional analysis for unit conversions, you must write each unit conversion as a fraction that equals 1

Answer 7

The conversion of liters to gallons is approximately:

1 gal = 3.78 gal (Eq. 1)

Dividing both sides of Eq. 1 by 1 gal, you get:

\(1 = \dfrac{3.78~L}{1~gal}\) (Eq. 2)

Dividing both sides by sides of Eq. 1 by 3.78 L, you get:

\(\dfrac{1~gal} {3.78~L} = 1\) (Eq. 3)

Since both Eq. 2 and Eq. 3 are equal to 1, all these conversion fractions are equal to each other and equal to 1.

\(\dfrac{3.78~L}{1~gal} =\dfrac{1~gal} {3.78~L} = 1\)

Since we know that multiplying a number by 1 does not change the number (the identity property of multiplication), we can multiply a measure in liters by the appropriate conversion factor without changing the amount of liters. The only thing that changes is the units.

The way we choose the conversion fraction is by figuring out which conversion fraction will cancel the units we don't want and will leave the units we want.

Answer 8

\(24~\cancel{L} \cdot \dfrac{1 ~gal}{3.78~\cancel{L}} = 6.3~ gal\)

Answer 9

Ohhhh cool good job i understand that :D @mathstudent55

Answer 10

ohhh

Answer 11

so is that the answer

Answer 12

whatever he told you yes and describe it like him as well :D

Answer 13

This is the idea behind dimensional analysis for the conversion of units.
If you do it correctly, there should be no errors in conversions because you always make sure you cancel out the unwanted units leaving the units you want.

Answer 14

Here is an example:

Convert 65.60 m/sec into mph.

We need these basic conversions:
1 m = 100 cm
2.54 cm = 1 in.
12 in. = 1 ft
5280 ft = 1 mi
60 sec = 1 min
60 min = 1 hr

\(65 \dfrac{m}{sec} \times \dfrac{100 ~cm}{1~m} \times \dfrac{1~in.}{2.54~cm} \times
\dfrac{1~ft}{12~in.} \times \dfrac{1~mi}{5280~ft} \times \dfrac{60~sec}{1~min} \times
\dfrac{60~min}{1~hr} =\)

\(= 146.7~ mph\)

Answer 15

See how the units cancel out leaving only the units we want:

\(65 \dfrac{\color{red}{\cancel{m}}}{\color{brown}{\cancel{sec}}} \times \dfrac{100 ~\color{blue}{\cancel{cm}}}{1~\color{red}{\cancel{m}}} \times \dfrac{1~\color{green}{\cancel{in.}}}{2.54~\color{blue}{\cancel{cm}}} \times
\dfrac{1~\color{orange}{\cancel{ft}}}{12~\color{green}{\cancel{in.}}} \times \dfrac{1~\bf{\Large{mi}}}{5280~\color{orange}{\cancel{ft}}} \times \dfrac{60~\color{brown}{\cancel{sec}}}{1~\color{purple}{\cancel{min}}} \times
\dfrac{60~\color{purple}{\cancel{min}}}{1~\bf{\Large{hr}}} =\)

Only units of mi/hr are left.

Answer 16

yes