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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 22: Lesson Check

Exercise 16

Step 1

1 of 2

The SI base units for mass, length, and time are, respectively, $textbf{kilogram (kg), meter (m), second (s).}$

Result

2 of 2

Kilogram (kg), meter (m), second (s).

Exercise 17

Solution 1

Solution 2

Step 1

1 of 3

The prefix $textit{kilo}$ means $textbf{one thousand}$.

Step 2

2 of 3

With this prefix, we can express $1450 text{m}$ as $textbf{1.45 text{km}.}$

Result

3 of 3

The prefix $textit{kilo}$ means one thousand, and we can use it to express

$1450 text{m}$ as $1.45 text{km}$.

Step 1

1 of 2

Kilo(k) = 1,000
1450 meters = ?k meters

1,450/1,000 = 1.45

Result

2 of 2

1450 m = 1.45 km

Exercise 18

Step 1

1 of 2

$textbf{All terms in a physics equation must have the same dimensions because only then the equation is correct.}$ You can’t equate length and mass, or add time to acceleration for example, because it’s simply wrong and doesn’t make sense. This is why dimensional analysis is very useful when working on problems; it’s a good indication if your solution might be correct or incorrect.

Result

2 of 2

The reason why all terms in a physics equation must be consistent with one another is because only then the equation is correct.

Exercise 19

Solution 1

Solution 2

Step 1

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$textbf{Dimension is more fundamental than unit, as it refers to the type of a physical quantity, whereas units are used when measuring dimensions of that quantity.}$ Moreover, you can change the units of measurements, but dimensions stay the same.

Step 2

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For example, both accelerations $17 frac{text{mm}}{text{s}^{2}}$ and $5 frac{text{km}}{text{h}^{2}}$ have dimensions of length over time squared, but each of them has a different unit.

Result

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Dimension refers to the type of a physical quantity, whereas units are used for measuring quantities. Using different units doesn’t change the dimension of a given quantity.

Accelerations $17 frac{text{mm}}{text{s}^{2}}$ and $5 frac{text{km}}{text{h}^{2}}$ are measured in different units, but they both have the same dimension, which is length over time squared.

Step 1

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The dimension represents a fundamental quantity, whereas the unit is a measurement of that dimension.

For example, time is a dimension and second or minute are the unit to measure the dimension.

Exercise 20

Solution 1

Solution 2

Step 1

1 of 2

We know that $1 text{Gm}=10^{9} text{m}$, or, conversely, $1 text{m}=10^{-9} text{Gm}$. The speed of light in meters per second is

$$
begin{align*}
0.3 frac{text{Gm}}{text{s}}&=left(0.3 frac{text{Gm}}{text{s}}right)left(dfrac{1 frac{text{m}}{text{s}}}{10^{-9} frac{text{Gm}}{text{s}}}right)\
&=quadboxed{3cdot 10^{8} frac{text{m}}{text{s}}}\
end{align*}
$$

Result

2 of 2

$$
begin{align*}
boxed{0.3 frac{text{Gm}}{text{s}}=3cdot 10^{8} frac{text{m}}{text{s}}}\
end{align*}
$$

Step 1

1 of 2

The speed of light is

$$
0.3 Gm/s=(0.3 Gm/s)left( frac{1 m/s}{10^{-9} Gm/s} right)=3times 10^{8} m/s
$$

Result

2 of 2

$3times 10^8$ m/s

Exercise 21

Solution 1

Solution 2

Step 1

1 of 3

$textbf{(a)}$ $65 frac{text{mi}}{text{h}}$ is greater than $65 frac{text{km}}{text{h}}$ because 1 mile is approximately equal to 1.61 kilometers.

Step 2

2 of 3

$textbf{(b)}$ The speed limit in kilometers per hour that corresponds to 65 miles per hour is

$$
begin{align*}
65 frac{text{mi}}{text{h}}&=
left(65 frac{text{mi}}{text{h}}right)left(dfrac{1.61 frac{text{km}}{text{h}}}{1 frac{text{mi}}{text{h}}}right)\
&=boxed{104.65 frac{text{km}}{text{h}}}\
end{align*}
$$

Result

3 of 3

$textbf{(a)}$ $65 frac{text{mi}}{text{h}}$ is greater than $65 frac{text{km}}{text{h}}$.

$textbf{(b)}$ $boxed{65 frac{text{mi}}{text{h}}=104.65 frac{text{km}}{text{h}}}$

Step 1

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65 km/h is less than 65 mi/h since 1 km is less than 1 mi.

Step 2

2 of 3

$$
65 mi/h=(65 mi/h)left( frac{1 km/h}{0.62 mi/h} right)=104.8 km/h
$$

so 104.8 km/h will correspond to 65 mi/h

Result

3 of 3

See answers.

Exercise 22

Step 1

1 of 2

Denoting the dimensions of $v_{f,i} , a$ and $t$ as $left[frac{text{L}}{text{T}}right], left[frac{text{L}}{text{T}^{2}}right]$ and $[text{T}]$, respectively, we have

$$
begin{align*}
v_{f}&=v_{i}+at\
\
\
\
left[dfrac{text{L}}{text{T}}right]&=left[dfrac{text{L}}{text{T}}right]+left[dfrac{text{L}}{text{T}^{2}}right][text{T}]\
\
\
\
rightarrowquadquad &boxed{left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}}right]+left[dfrac{text{L}}{text{T}}right]}\
end{align*}
$$

All terms have the same dimension, which means that the equation is dimensionally consistent.

Result

2 of 2

The equation is dimensionally consistent because all terms have the same dimension;

$$
begin{align*}
boxed{left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}}right]+left[dfrac{text{L}}{text{T}}right]}\
end{align*}
$$

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