To verify a scientific hypothesis, we conduct $textbf{experiments}$ to check if the predictions are supported by measurements.

In this problem, we are given $E = mc^{2}$, where $E$ stands for energy, $m$ stands for mass, and $c$ stands for the speed of light. We find the equation for mass $m$.

Using algebra, we have

$$
begin{align*}
E&= mc^{2} & mathrm{divide~by~c^{2}} \
frac{E}{c^{2}} &= m \
m &= frac{E}{c^{2}}
end{align*}
$$

$$
m = dfrac{E}{c^{2}}
$$

In this problem, we describe bias. A $textbf{bias}$ is a preference toward a particular point of view for personal and not scientific reasons. As scientists, we must aim to be aware and avoid our personal biases in order to conduct experiments in an objective manner.

In this problem, we discuss peer review. $textbf{Peer review}$ is sending a scientific paper to other experts to check for errors, biases, and possible oversights. This allows multi-perspective view on the research.

We know that $1~mathrm{kilometer} = 1000~mathrm{meter}$. We have

$$
begin{align*}
15~mathrm{kilometer} times frac{1000~mathrm{meter}}{1~mathrm{kilometer}} &= boxed{ 15000~mathrm{meters} }
end{align*}
$$

$$
15000~mathrm{meters}
$$

We know that $1~mathrm{kilometer} = 1000~mathrm{meter}$. We have

$$
begin{align*}
12000~mathrm{meter} times frac{1~mathrm{kilometer}}{1000~mathrm{meter}} &= boxed{ 12~mathrm{kilometers} }
end{align*}
$$

$$
12~mathrm{kilometers}
$$

In this problem, we find the length of an $textit{E. coli}$ bacteria in km. The bacteria has length $5 times 10^{-6}~mathrm{m}$.

From the table $1~mathrm{km} = 1 times 10^{3}~mathrm{m}$. We have

$$
begin{align*}
5 times 10^{-6}~mathrm{m} times frac{1~mathrm{km}}{1 times 10^{3}~mathrm{m}} &= boxed{ 5 times 10^{-9}~mathrm{km} }
end{align*}
$$

$$
5 times 10^{-9}~mathrm{km}
$$

Part A.

We know that $1~mathrm{kilo} = 1 times 10^{3}$. We have

$$
begin{align*}
33200~mathrm{dollars} times frac{1~mathrm{kilodollars}}{1 times 10^{3}~mathrm{dollars}} &= boxed{ 33.2 ~mathrm{kilodollars} }
end{align*}
$$

Part B.

We know that $1~mathrm{mega} = 1 times 10^{6}$. We have

$$
begin{align*}
33200~mathrm{dollars} times frac{1~mathrm{megadollars}}{1 times 10^{6}~mathrm{dollars}} &= boxed{ 0.0332 ~mathrm{megadollars} }
end{align*}
$$

begin{enumerate}
item [a)] $33.2 ~mathrm{kilodollars}$
item [b)] $0.0332 ~mathrm{megadollars}$
end{enumerate}

First, we find the time per flap, which must be the reciprocal for the flaps per second. We know that $1~mathrm{ms} = 10^{-3}~mathrm{s}$. We have

$$
begin{align*}
T &= frac{1~mathrm{s}}{200} \
T &= 5 times 10^{-3}~mathrm{s} \
&= left( 5 times 10^{-3}~mathrm{s} right) times left( frac{1~mathrm{ms}}{10^{-3}~mathrm{s}} right) \
T &= boxed{ 5~mathrm{ms} }
end{align*}
$$

$$
5~mathrm{ms}
$$

In this problem, a warehouse of the shape of a rectangular prism, has length $L = 0.012$ km, width $W = 10.5$ m, and height $H = 4.95$m. We find its volume.

First, we convert the length to meters. We know that $1~mathrm{km} = 1000~mathrm{m}$.

$$
begin{align*}
L &= left( 0.012~mathrm{km} right) left( frac{1000~mathrm{m}}{1~mathrm{km}} right) \
L &= 12.0~mathrm{m}
end{align*}
$$

We now get the volume

$$
begin{align*}
V &= LWH \
&= left( 12.0~mathrm{m} right) left( 10.5~mathrm{m} right) left( 4.95~mathrm{m} right) \
V &= boxed{ 623.7~mathrm{m^{3}} }
end{align*}
$$

$$
V = 623.7~mathrm{m^{3}}
$$

In this problem, we find the number of milliliters (mL) in $1.2$ L. We know that $1~mathrm{mL} = 0.001~mathrm{L}$.

We have

$$
begin{align*}
left( 1.2~mathrm{L} right) times left( frac{1~mathrm{mL}}{0.001~mathrm{L}} right) &= boxed{ 1200~mathrm{mL} }
end{align*}
$$

$$
1200~mathrm{mL}
$$

In this problem, we find the mass, in pounds, of the Star of Africa. It has mass $520.2~mathrm{carats}$. One carat is $0.20~mathrm{g}$ and $1~mathrm{kg} = 1000~mathrm{g}$.

First, we convert carat to gram.

$$
begin{align*}
520.2~mathrm{carats} times frac{0.20~mathrm{g}}{1~mathrm{carat}} &= 106.04~mathrm{g}
end{align*}
$$

Now, gram to kilogram,

$$
begin{align*}
106.04~mathrm{g} times frac{1~mathrm{kg}}{1000~mathrm{g}} &= 0.10604~mathrm{kg}
end{align*}
$$

Finally, kilogram to pounds

$$
begin{align*}
0.10604~mathrm{kg} times frac{2.21~mathrm{lb}}{1~mathrm{kg}} &= boxed{ 0.2343~mathrm{lb} }
end{align*}
$$

$$
0.2333~mathrm{lb}
$$

For mass, the base unit is $textbf{kilogram}$, for length it is $textbf{meter}$, and $textbf{second}$ for time.

In this problem, we describe the prefix $textit{kilo-}$ and the length $1450$ m in term of km.

The prefix $textit{kilo-}$ means 1000 of the attached unit.

The length in meters must be divided by 1000 to find the length in km, so we have $1450~mathrm{m} = 1.45~mathrm{km}$.

In this problem, we explain why terms in an equation must have the same units. One good analogy is the egg. Adding one egg to another egg is different from adding one $textit{dozen}$ of eggs to another dozen. When the units of the terms are different, we can not simply add the numerical coefficients, making the equation hard to solve.

In this problem, we differentiate between unit and dimension. A $textbf{dimension}$ is a measurement of physical quantity, and a $textbf{unit}$ is a way of reporting such measurement using numbers. For example, the $textbf{temperature}$ of an object is the measurement of its thermal energy (a dimension), and temperature can be reported in units of Celsius ($^{circ}mathrm{C}$), Fahrenheit ($^{circ}mathrm{F}$), or Kelvin (K).

In this problem, we convert Gm/s to m/s. It is given that the speed of light is $0.3$ Gm/s.

We know that $1~mathrm{Gm} = 1 times 10^{9}~mathrm{m}$, so we have

$$
begin{align*}
left( 0.3~mathrm{Gm/s} right) times left( frac{1 times 10^{9}~mathrm{m}}{1~mathrm{Gm}} right) &= boxed{ 3.0 times 10^{8}~mathrm{m/s} }
end{align*}
$$

$$
3.0 times 10^{8}~mathrm{m/s}
$$

In this problem, we are given that the speed limit in many US highways is $65$ mi/h. We compare this with $65$km/h, and convert into km/h.

Part A.

From the conversion factors, $1~mathrm{mi} = 1.61~mathrm{km}$. This means that one mile is longer than one kilometer, and thus $65$ mi/h must be traveling $textbf{faster}$ than $65$ km/h.

Part B.

We use the conversion factor $1~mathrm{mi} = 1.61~mathrm{km}$.

$$
begin{align*}
left( 65~mathrm{mi/h} right) times left( frac{1.61~mathrm{km}}{1~mathrm{mi}} right) &= boxed{ 104.65~mathrm{km/h} }
end{align*}
$$

In this problem, we verify that the equation $v_text{f} = v_text{i} + at$ is dimensionally consistent. The quantities $v_text{f}$ and $v_text{i}$ are velocities, $a$ is an acceleration, and $t$ is in time.

Let $L$ stand for dimension of length, and $T$ for time. The dimension of velocity is $L/T$, for acceleration it is $L/T^{2}$, and for time, the dimension is $T$. We have

$$
begin{align*}
v_text{f} &= v_text{i} + at \
frac{L}{T} &= frac{L}{T} + left( frac{L}{T^{2}} right)left( T right) \
frac{L}{T} &= frac{L}{T} + frac{L}{T} \
frac{L}{T} &stackrel{checkmark}{=} frac{L}{T}\
end{align*}
$$

We see that the equation is indeed dimensionally consistent.

In this problem, we are given that a tortoise has speed $v = 2.51~mathrm{cm/s}$. We calculate the time $t$ it takes for the tortoise to reach $d = 17~mathrm{cm}$.

We know the equation relating speed, time, and distance.

$$
begin{align*}
vt &= d & text{divide both sides by $v$} \
frac{vt}{v} &= frac{d}{v} \
implies t &= frac{d}{v} \
&= frac{17~mathrm{color{#c34632}cm}}{left( 2.5~mathrm{{color{#c34632}cm}/s} right)} \
t &= 6.7729~mathrm{s} \
t &= boxed{ 6.8~mathrm{s} }
end{align*}
$$

We report the answer with the same number of significant digits has the given with the least number of significant digits. In this case, the distance $d = 17$ cm only has 2 significant digits.

$$
t = 6.8~mathrm{s}
$$

In this problem, we find the area of a circle with radius $r = 12.77~mathrm{m}$. The area $A$ of a circle with radius $r$ is $A = pi r^{2}$.

Using the formula, we have

$$
begin{align*}
A &= pi r^{2} \
&= pi left( 12.77~mathrm{m} right)^{2} \
&= 512.3086~mathrm{m^{2}} \
A &= boxed{ 512.3~mathrm{m^{2}} }
end{align*}
$$

We report the answer with the same number of significant digits has the given with the least number of significant digits. In this case, the radius $r = 12.55$ m only has 4 significant digits.

$$
A = 512.3~mathrm{m^{2}}
$$

In this problem, we find the area of a triangular sail. It has height $h = 4.1~mathrm{m}$ and base $b = 6.15~mathrm{m}$. It is given that the area $A$ of a triangle is $A = frac{1}{2}bh$.

Using the given formula, we have

$$
begin{align*}
A &= frac{1}{2}bh \
&= frac{1}{2} left( 4.1~mathrm{m} right) left( 6.15~mathrm{m} right) \
&= 12.6075~mathrm{m^{2}} \
A &= boxed{ 13~mathrm{m^{2}} }
end{align*}
$$

We report the answer with the same number of significant digits has the given with the least number of significant digits. In this case, the height $h = 4.1$ m only has 2 significant digits.

$$
A = 13~mathrm{m^{2}}
$$

In this problem, we are given that the masses of the caught bass, rock cod, and salmon are $m_{1} = 1.07~mathrm{kg}$, $m_{2} = 6.0~mathrm{kg}$, and $m_{3} = 6.05~mathrm{kg}$. We find the total mass of the catch.

To find the total mass, we simply add the given masses

$$
begin{align*}
M &= m_{1} + m_{2} + m_{3} \
&= left( 1.07~mathrm{kg} right) + left( 6.0~mathrm{kg} right) + left( 6.05~mathrm{kg} right) \
&= 13.12~mathrm{kg} \
M &= boxed{ 13~mathrm{kg} }
end{align*}
$$

We report the answer with the same number of significant digits has the given with the least number of significant digits. In this case, the mass of the rock cod $m_{2} = 6.0$ kg only has 2 significant digits.

$$
M = 13~mathrm{kg}
$$

In this problem, we find the perimeter of a rectangular paper. It has length $l = 25.2~mathrm{cm}$ and width $w = 18.1~mathrm{cm}$.

The perimeter $P$ of a rectangle is $P = 2left( l + w right)$. Using this formula,

$$
begin{align*}
P &= 2left( l + w right) \
&= 2left( left[ 25.2~mathrm{cm} right] + left[ 18.1~mathrm{cm} right] right) \
&= 2left( 43.3~mathrm{cm} right)\
P &= boxed{86.6~mathrm{cm}}
end{align*}
$$

We report the answer with the same number of significant digits has the given with the least number of significant digits. In this case, both the given quantities have 3 significant digits.

$$
P = 86.6~mathrm{cm}
$$

Part A.

For this part, the given number is $0.000054$. Notice that the leading zeroes do not contribute to the number of significant figures, so the number must be $textbf{2}$.

Part B.

For this part, the given number is $3.001 times 10^{5}$. We only count the significant figures in the number $3.001$. Since the zeroes are between nonzero numbers, they are accounted in the number of significant figures. Therefore, there are $textbf{4}$ significant figures in the given number.

begin{enumerate}
item [a)] $2$
item [b)] $4$
end{enumerate}

In this problem, we are given that the square root of 2, up to6 significant figures, is $1.41421$. We find the number in 4 significant figures.

All the digits are nonzero, so they account for the number of significant figures. When written in 4 significant digits, we round off at the $4$th number 1.41$text{color{#c34632}4}$21, which gives us
$$
boxed{ 1.414 }
$$

$$
1.414
$$

The product must have the same significant figure as the least significant figure in the given quantities. The width has the least number of significant figures, and the number is $textbf{2}$.

One possible way is using scientific notation. We can add some trailing zeroes to increase the number of significant figures. For this quantity, we may have

$$
begin{align*}
100~mathrm{m/s} &= boxed{ 1.00 times 10^{2}~mathrm{m/s} }
end{align*}
$$

In this problem, a poster is $0.95$ m high and $1.0$ m wide. In calculating the perimeter, we find the number of digits after the decimal point.

In calculating the perimeter, the operation involved is addition. Hence, the sum must have the same number of digits following the decimal point as the given quantity with the least number of digits following the decimal point. The corresponding given is the width $1.0$m, which has one. Hence, the perimeter must have $textbf{one}$ digit following the decimal point.

In this problem, we are given the speed of light $c = 2.9979 times10^{8}~mathrm{m/s}$. We write this quantity in 3 significant digits.

We round off the quantity in the $3$rd significant digit, which i $2.9{color{#c34632}9}79$, which gives us
$$
boxed{ c = 3.00 times 10^{8}~mathrm{m/s} }
$$

$$
3.00 times 10^{8}~mathrm{m/s}
$$

In this problem, we are given the initial speed and change in speed of a bus. We find the final speed if the initial speed is $v_text{i} = 2.2~mathrm{m/s}$, the change in speed is $Delta v = 5.225~mathrm{m/s}$, and the time it accelerates is $t = 20~mathrm{s}$.

We use the following equation

$$
begin{align*}
v_text{f} &= v_text{i} + Delta v\
&= left( 2.2~mathrm{m/s} right) + left( 5.225~mathrm{m/s^{2}} right) \
&= 7.425~mathrm{m/s} \
v_text{f} &= boxed{ 7.4~mathrm{m/s} }
end{align*}
$$

$$
v_text{f} = 7.4~mathrm{m/s}
$$

In this problem, we are given a computer monitor of height $h = 31.25~mathrm{cm}$ and width $w = 47~mathrm{cm}$. We find the number of significant figures of the area, and the area in the proper number of significant digits.

Part A.

The given quantity with the least number significant figures is $w = 47~mathrm{cm}$. Since the area is a product, its number of significant figures is the same as the least number of significant figures in the given. Hence, the number is $textbf{2}$.

Part B.

The area is the product of the height and width. We have

$$
begin{align*}
A &= hw \
&= left( 31.25~mathrm{cm} right) left( 47~mathrm{cm} right) \
&= 1468.75~mathrm{cm^{2}} \
A &= boxed{1500~mathrm{cm^{2}}}
end{align*}
$$

begin{enumerate}
item [a)] $2$
item [b)] $1500~mathrm{cm^{2}}$
end{enumerate}

In this problem, we are given the length $l = 144.3~mathrm{m}$ and width $w = 47.66~mathrm{m}$. We find the perimeter and area.

Part A.

The perimeter $P$ of a rectangle is given by $P = 2left( l + w right)$. Using this formula,

$$
begin{align*}
P &= 2left( l + 2 right)\
&= 2left( left[ 144.3~mathrm{m} right] + left[ 47.66~mathrm{m} right] right)\
&= 2left( 191.96~mathrm{m} right) \
&= 383.92~mathrm{m} \
P &= boxed{ 383.9~mathrm{m} }
end{align*}
$$

Part B.

The area $A$ is given by $A = lw$. We have

$$
begin{align*}
A &= lw \
&= left( 144.3~mathrm{m} right) left( 47.66~mathrm{m} right) \
&= 6877.338~mathrm{m^{2}} \
A&= boxed{ 6877~mathrm{m^{2}} }
end{align*}
$$

begin{enumerate}
item [a)] $P = 383.9~mathrm{m}$
item [b)] $A = 6877~mathrm{m^{2}}$
end{enumerate}

$textbf{An inference is a conclusion based on facts, evidence, and logic,}$ or, in other words, a logical interpretation of one’s observations.

$textbf{A hypothesis is a proposed explanation for an observed phenomenon that can be verified or rejected by conducting careful experiments.}$ It can also be thought of as a combination of one’s observations and inferences.

In physics, we say that $textbf{a quantity is conserved when it is constant in time,}$ i.e., it has the same value before and after the chosen point of reference.

$textbf{A force causes a change in motion,}$ motion itself doesn’t require a force.

$textbf{The advantage of the metric system is that it’s simple to convert from one unit to another, as it is based on powers of ten.}$ This means that you essentially just have to shift the decimal point to a new place value.

$textbf{Peer review doesn’t guarantee that the conclusions of a scientific report are correct;}$ it just guarantees that the report meets the minimum requirements of a scientific journal the author sent their report to.

The area of a circle has a dimension of length squared, whether it be $text{m}^{2}$, $text{cm}^{2}$ or any other unit for length. Let’s see which of the two expressions gives $L^{2}$:

$$
begin{align*}
pi r^{2}&=pi[L^{2}]={[L^2]}\
\
\
\
2pi r&=2pi[L]={[L]}\
end{align*}
$$

Therefore, the first expression gives the area of a circle.

The dimension of $pi r^{2}$ is $[L^{2}]$, and the dimension of $2pi r$ is $[L]$. Thus, the first expression gives the area of a circle.

$textbf{The quantity $T+d$ does not make sense physically.}$ The dimension of a physical quantity refers to the type of quantity in question. In this case, we have two distinct quantities, and adding them makes no sense. Moreover, $textbf{any valid physical equation has to be dimensionally consistent, meaning that each term has to have the same dimensions}$ and that’s not the case when adding time and distance.

$textbf{The quantity $frac{d}{T}$ does make sense physically.}$ Dividing distance by time, we get a new quantity, which is velocity. Velocity tells us what distance did an object traverse in a certain amount of time. $textbf{Dividing two different quantities often results in a new quantity that makes sense physically,}$ and, in this case, we did just that.

The quantity $T+d$ makes no sense physically because we would be adding two different quantities with different dimensions. On the other hand, the quantity $frac{d}{T}$ does make sense because dividing distance by time we get a new quantity, which is velocity.

$textbf{(a)}$     $textbf{No, it’s not possible for two quantities to have the same units but different dimensions. If two quantities have the same units, then they are of the same type and have the same dimensions.}$

After all, it doesn’t make sense to measure both time and distance in centimeters, as they are two fundamentally different quantities with different dimensions; one being time, and the other being length. As mentioned before, the dimension of a physical quantity refers to the type of quantity in question. Therefore, $textbf{if two quantities have different dimensions, then they must be of a different type, and are measured in different units that can’t be converted to one another.}$

$textbf{(b)}$     $textbf{Yes, two quantities can have the same dimensions but different units.}$ For example, you can measure distance in centimeters or inches; the units are clearly different, but they both have the same dimension, and refer to the same type of quantity.

The important thing to notice here is that you can convert one unit to another, which is an indicator that the quantities are of the same type and dimension.

$textbf{(a)}$      No, two quantities can’t have the same units but different dimensions. If two quantities have the same units, then they must be of the same type and have the same dimensions.

$textbf{(b)}$      Yes, two quantities can have the same dimensions but different units. As long as you can convert units to one another, you are dealing with the same dimension.

Distance has a dimension of length, which we will denote by [L]. Case by case, we have the following dimensions:

$$
begin{align*}
& textbf{A.} quad vt=[dfrac{L}{T}][T]=[L]\
\
\
\
& textbf{B.} quad dfrac{1}{2}at^{2}=dfrac{L}{T^{2}}[T^{2}]=[L]\
\
\
\
& textbf{C.} quad 2at=dfrac{[L]}{[T^{2}]}[T]=dfrac{[L]}{[T]}\
\
\
\
& textbf{D.} quad dfrac{v^{2}}{a}=dfrac{[L^{2}]}{[T^{2}]}dfrac{[T^{2}]}{[L]}=[L]\
end{align*}
$$

Therefore, quantities $textbf{A., B.,}$ and $textbf{D.}$ have the same dimension as distance.

$textbf{A.}$, $textbf{B.}$, and $textbf{D.}$

Speed has a dimension of length over time, which we will denote by $frac{[L]}{[T]}$. Case by case, we have the following dimensions:

$$
begin{align*}
& textbf{A.} quad frac{1}{2}at^{2}=frac{L}{T^{2}}[T^{2}]=[L]\
\
\
\
& textbf{B.} quad at=frac{[L]}{[T^{2}]}[T]=frac{[L]}{[T]}\
\
\
\
& textbf{C.} quad sqrt{frac{2x}{a}}=sqrt{[L]frac{[T^{2}]}{[L]}}=[T]\
\
\
\
& textbf{D.} quad sqrt{2ax}=sqrt{frac{L}{[T^{2}]}[L]}=sqrt{frac{[L]}{[T^{2}]}}=frac{[L]}{[T]}\
end{align*}
$$

Therefore, quantities $textbf{B.}$ and $textbf{D.}$ have the same dimension as distance.

$textbf{B.}$ and $textbf{D.}$

$textbf{(a)}$ One gigadollar is $10^{9}$ dollars, so

$$
begin{align*}
114 000 000 text{dollars}&=114 000 000cdot10^{-9} text{gigadollars}\
&=quadboxed{0.114 text{gigadollars}}\
end{align*}
$$

$textbf{(b)}$ One teradollar is $10^{12}$ dollars, so

$$
begin{align*}
114 000 000 text{dollars}&=114 000 000cdot 10^{-12} text{teradollars}\
&=quadboxed{1.14cdot 10^{-4} text{teradollars}}
end{align*}
$$

$$
begin{align*}
textbf{(a)}quadquad &boxed{0.114 text{gigadollars}}\
quad\
textbf{(b)}quadquad &boxed{1.14cdot 10^{-4} text{teradollars}}\
end{align*}
$$

One kilometer is 1000 meters, and there are 3600 seconds in one hour which means that

$$
begin{align*}
1 dfrac{text{m}}{text{s}}&=left(1 dfrac{text{m}}{text{s}}right)left(dfrac{1 text{km}}{1000 text{m}}right)left(dfrac{3600 text{s}}{1 text{h}}right)\
&=3.6 frac{text{km}}{text{h}}\
end{align*}
$$

Therefore, in kilometers per hour, the speed of 23 meters per second is

$$
begin{align*}
23 frac{text{m}}{text{s}}&=left(23 frac{text{m}}{text{s}}right)left(dfrac{3.6 frac{text{km}}{text{h}}}{1 frac{text{m}}{text{s}}}right)\
&=82.8 frac{text{km}}{text{h}}\
&=quadboxed{83 frac{text{km}}{text{h}}}
end{align*}
$$

$boxed{text{speed}=83 frac{text{km}}{text{h}}}$

We know that $1 text{m}=100 text{cm}$, which means that

$$
begin{align*}
(1 text{m})^{2}=(100 text{cm})^{2}=1cdot 10^{4} text{cm}^{2}\
(1 text{m})^{3}=(100 text{cm})^{3}=1cdot 10^{6} text{cm}^{2}
end{align*}
$$

Multiplying the dimensions, the volume in cubic meters is

$$
begin{align*}
V&=Ltimes Wtimes H\
&=631 text{m}times 646 text{m}times34 text{m}\
&=13 859 248 text{m}^{3}\
&=1.4 cdot 10^{7} text{m}^{3}\
end{align*}
$$

Therefore, the volume in cubic centimeters is

$$
begin{align*}
V&=1.39cdot 10^{7} text{m}^{3}=1.39cdot 10^{7} text{m}^{3}timesleft(dfrac{10^{6} text{cm}^{3}}{1 text{m}^{3}}right)\
&=quadboxed{1.4 cdot 10^{13} text{cm}^{3}}
end{align*}
$$

$$
begin{align*}
boxed{V=1.4cdot 10^{13} text{cm}^{3}}\
end{align*}
$$

Since the atom completes 9 192 631 770 cycles per second, we can conclude right away that it will take less than a second for the same atom to complete 1.5 million cycles.

First, we write the following equation:

$$
begin{align*}
dfrac{9 192 631 770 text{cycles}}{1 text{s}}&=dfrac{1.5cdot 10^{6} text{cycles}}{t}\
end{align*}
$$

Solving the above equation for $t$, and using $textbf{the rule for multiplication and division for the number of significant figures in the result,}$ we get

$$
begin{align*}
t&=dfrac{1.5cdot 10^{6} text{cycles}}{9 192 631 770 text{cycles}}times 1 text{s}\
&=1.631741636cdot 10^{-4} text{s}\
&=quadboxed{1.6cdot 10^{-4} text{s}}\
end{align*}
$$

$$
begin{align*}
boxed{t=1.6cdot 10^{-4} text{s}}\
end{align*}
$$

One meter is 1000 millimeters. Therefore, the length of a $33 text{m}$ long blue whale in millimeters is

$$
begin{align*}
33 text{m}&=(33 text{m})left(dfrac{1000 text{mm}}{1 text{m}}right)\
&=quadboxed{33 000 text{mm}}\
end{align*}
$$

$$
begin{align*}
boxed{33 000 text{mm}}\
end{align*}
$$

$textbf{(a)}$      One meter is $10^{6} mutext{m}$. Therefore, the thickness in meters is

$$
begin{align*}
70 mutext{m}&=70 mutext{m}left(dfrac{1 text{m}}{10^{6} mutext{m}}right)\
&=quadboxed{70cdot 10^{-6} text{m}}\
end{align*}
$$

$textbf{(b)}$      Using the result from $textbf{(a)}$, we get that the thickness in kilometers is

$$
begin{align*}
70 mutext{m}&=70cdot 10^{-6} text{m}\
&=(70cdot 10^{-6} text{m})left(dfrac{1 text{km}}{10^{3} text{m}}right)\
&=quadboxed{70cdot 10^{-9} text{km}}\
end{align*}
$$

$textbf{(a)}$     $boxed{70cdot 10^{-6} text{m}}$

$textbf{(b)}$     $boxed{70cdot 10^{-9} text{km}}$

136.8 teracalculations is $136.8cdot 10^{12} text{calculations}$, and one second is $10^{6} text{microseconds}$. Denoting the number of calculations by $n$, we write the following equation:

$$
begin{align*}
dfrac{136.8cdot 10^{12} text{calculations}}{1 text{s}}=dfrac{n}{1 mutext{s}}
end{align*}
$$

Converting microseconds to seconds and solving the equation for $n$, we get

$$
begin{align*}
n&=dfrac{136.8cdot 10^{12} text{calculations}}{1 text{s}}times 1cdot 10^{-6} text{s}\
&=136.8cdot 10^{6} text{calculations}\
end{align*}
$$

Since $136.8cdot 10^{12} text{calculations}$ has 4 significant figures, and $1 text{s}$ has only one significant figure, $textbf{using the rule for multiplication and division for significant figures we write the final result as}$

$$
begin{align*}
boxed{n=1cdot 10^{8} text{calculations}}\
end{align*}
$$

$$
begin{align*}
boxed{1cdot 10^{8} text{calculations}}\
end{align*}
$$

To find the speed of light in $frac{text{cm}}{text{s}}$, we just have to multiply the expression given in $frac{text{m}}{text{s}}$ by $left(frac{100 text{cm}}{1 text{m}}right)$, since one meter is 100 centimeters. So,

$$
begin{align*}
3.00cdot 10^{8} frac{text{m}}{text{s}}=3.00cdot 10^{10} frac{text{cm}}{text{s}}\
end{align*}
$$

$textbf{(a)}$      1 jiffy is defined as the time it takes light to travel 1 centimeter. Using this definition, as well as the formula for speed, we obtain that 1 jiffy is

$$
begin{align*}
text{speed}&=dfrac{text{distance}}{text{time}}\
rightarrow text{time}&=dfrac{text{distance}}{text{speed}}\
\
\
\
1 text{jiffy}&=dfrac{1 text{cm}}{3.00cdot 10^{10} frac{text{cm}}{text{s}}}\
&=3.33cdot 10^{-11} text{s}\
&=quadboxed{3cdot 10^{-11} text{s}}\
end{align*}
$$

* We used the rule for multiplication and division for significant figures, as per usual.

$textbf{(b)}$      The conversion factor is

$$
begin{align*}
1 text{jiffy}=3cdot 10^{-11} text{s}\
\
\
\
dfrac{1 text{jiffy}}{3cdot 10^{-11} text{s}}=1\
end{align*}
$$

Therefore, the number of jiffys in one minute is

$$
begin{align*}
1 text{min}&=(60 text{s})left(dfrac{1 text{jiffy}}{3cdot 10^{-11} text{s}}right)\
\
\
\
&=quadboxed{2cdot 10^{12} text{jiffys}}\
end{align*}
$$

$$
begin{align*}
textbf{(a)} quad &boxed{1 text{jiffy}=3cdot 10^{-11} text{s}}\
\
\
\
textbf{(b)} quad &boxed{1 text{min}=2cdot 10^{12} text{jiffys}}\
end{align*}
$$

The volume of a cube with a side $a$ is given by $V=a^{3}$, or, as we shall denote it

$$
begin{align*}
V=atimes A\
end{align*}
$$

In our case, $A$ is the area of the resulting slick, and $a$ its thickness.

Keep in mind that the total volume of the oil remains unchanged, the molecules just rearrange themselves one next to another so that the slick is one molecule thick.

Converting $mutext{m}$ to meters and solving for $A$, we obtain

$$
begin{align*}
1.0 text{m}^{3}&=0.50cdot 10^{-6} text{m}times A\
A&=dfrac{1.0 text{m}^{3}}{0.50cdot 10^{-6} text{m}}\
&=quadboxed{2.0cdot 10^{6} text{m}^{2}}\
end{align*}
$$

$boxed{A=2.0cdot 10^{6} text{m}^{2}}$

To find the acceleration due to gravity in $frac{text{cm}}{text{s}^{2}}$, we essentially just have to multiply the given value by $left(frac{100 text{cm}}{1 text{m}}right)$. A more detailed solution in which $frac{text{m}}{text{s}^{2}}$ cancel out is:

$$
begin{align*}
9.81 dfrac{text{m}}{text{s}^{2}}&=left(9.81 dfrac{text{m}}{text{s}^{2}}right)left(dfrac{100 frac{text{cm}}{text{s}^{2}}}{1 frac{text{m}}{text{s}^{2}}}right)\
&=9.81times 100 frac{text{cm}}{text{s}^{2}}\
&=quadboxed{981 frac{text{cm}}{text{s}^{2}}}\
end{align*}
$$

$boxed{quad 981 frac{text{cm}}{text{s}^{2}}quad}$

$textbf{An equation is dimensionally consistent when all terms have the same dimension.}$ To find the power $p$, we will use dimensional analysis.

$$
begin{align*}
v^{2}&=2ax^{p}\
\
\
\
left[dfrac{L^{2}}{T^{2}}right]&=left[dfrac{L}{T^{2}}right][L]^{p}\
[L]^{2}&=[L]^{1+p}\
\
\
\
2&=1+p\
& boxed{p=1}\
end{align*}
$$

$boxed{p=1}$

$textbf{An equation is dimensionally consistent when all terms have the same dimension.}$ To find the power $p$, we will use dimensional analysis.

$$
begin{align*}
a&=2xt^{p}\
\
\
\
left[dfrac{L}{T^{2}}right]&=[L][T]^{p}\
dfrac{1}{[T]^{2}}&=[T]^{p}\
[T]^{-2}&=[T]^{p}\
\
\
\
&boxed{p=-2}\
end{align*}
$$

$boxed{p=-2}$

Let’s start with this: one hour is 3600 seconds, which is on the order of $10^{3}$, and 24 hours is 86 400 seconds, which is on the order of $10^{4}$. Knowing this makes it easier to give other estimates.

$textbf{(a)}$      Since 84 000 is closer to 100 000, or $10^{5}$, then a good estimate would be $10^{7}$.

$textbf{(b)}$      A baseball game lasts about three hours, which is somewhere around 11 000 seconds, so an estimate would be $10^{4}$.

$textbf{(c)}$      Usually, a heart rate is 60 to 100 beats in a minute, or roughly one heartbeat per second, which is $10^{0}$.

$textbf{(d)}$      The Earth is about 4.6 billion, or $4.6cdot 10^{9}$, years old. In seconds, that would be approximately $10^{17}$.

$textbf{(e)}$      Assuming that you are in your teens or early twenties, the estimated order-of-magnitude of your age in seconds would be $10^{8}$.

To sum it all up, these are the order-of-magnitude estimates in seconds:

$textbf{(a)}$      a year: $10^{7} text{s}$

$textbf{(b)}$      a baseball game: $10^{4} text{s}$

$textbf{(c)}$      a heartbeat: $10^{0}=1 text{s}$

$textbf{(d)}$      the age of Earth: $10^{17} text{s}$

$textbf{(e)}$      your age: $10^{8} text{s}$

$textbf{(a)}$      $10^{7} text{s}$

$textbf{(b)}$      $10^{4} text{s}$

$textbf{(c)}$      $10^{0}=1 text{s}$

$textbf{(d)}$      $10^{17} text{s}$

$textbf{(e)}$      $10^{8} text{s}$

An order-of-magnitude calculation is defined as an estimate accurate to within the nearest power of 10. That being said, keep in mind that we have to give an estimate for the length in meters: an estimate of $10^{1}$ is 10 meters, and $10^{0}$ is one meter.

$textbf{(a)}$      An order-of-magnitude estimate for the length in meters of your height is $10^{0}$. Although this is equal to 1 meter, to assume that the correct answer is $10^{1}$ is wrong: no human being is 10 meters tall.

$textbf{(b)}$      A fly is usually smaller than 1 centimeter, so it’s safe to assume that its length is in milimeters. Therefore, a rough estimate for the length of a fly in meters is $10^{-2}$.

$textbf{(c)}$      A normal car is usually up to 3 meters long, so the answer is again $10^{0}$.

$textbf{(d)}$      An ordinary commercial jetliner is roughly 50 meters long, which is on the order of $10^{1}$.

$textbf{(e)}$      The longest interstate highway stretching from coast to coast is more than 4000 kilometers long, so an order-of-magnitude estimate in meters of such an interstate is $10^{6}$.

The order-of-magnitude estimates in meters:

$textbf{(a)}$      your height: $10^{0} text{m}$

$textbf{(b)}$      a fly: $10^{-2} text{m}$

$textbf{(c)}$      a car: $10^{0} text{m}$

$textbf{(d)}$      a jetliner: $10^{1} text{m}$

$textbf{(e)}$      an interstate highway stretching from coast to coast: $10^{6} text{m}$

$textbf{(a)}$      $10^{0} text{m}$

$textbf{(b)}$      $10^{-2} text{m}$

$textbf{(c)}$      $10^{0} text{m}$

$textbf{(d)}$      $10^{1} text{m}$

$textbf{(e)}$      $10^{6} text{m}$

$textbf{(a)}$      To three significant figures, $pi$ equals

                            $boxed{pi=3.14}$

$textbf{(b)}$      To five significant figures, $pi$ equals

$$
begin{align*}
boxed{pi=3.1416}\
end{align*}
$$

Note that it’s not 3.1415 because the first digit to be dropped (which is 9) is equal to or greater than 5, so it’s correct to increase the previous digit (5 in this case) by 1.

$$
begin{align*}
textbf{(a)} quad &boxed{pi=3.14}\
textbf{(b)} quad &boxed{pi=3.1416}\
end{align*}
$$

The area of a circle is given by the formula

$$
begin{align*}
A=r^{2}pi\
end{align*}
$$

So, the area of the circle in question is

$$
begin{align*}
A&=(24.87 text{m})^{2}pi\
&=1943.128149 text{m}^{2}\
&=quadboxed{1943 text{m}^{2}}\
end{align*}
$$

We dropped all the decimals because of the rule for significant figures for multiplication and division: the result can’t be more precise than the least accurately known input value. In this case, the radius is known to four significant figures; thus, the final result can’t have more than 4 significant figures.

$$
begin{align*}
boxed{A=1943 text{m}^{2}}\
end{align*}
$$

For starters, we can most definitely say that the number of seats in a typical Major League stadium is not in the hundreds; rather, it’s in thousands.

A couple thousand is also too little, so we can conclude that we’re dealing with tens of thousands. At this point, we can eliminate a hundred thousand because that really is $textit{a lot}$ of seats for a typical stadium; in fact, it’s way too many.

Our best guess would be that such stadiums can hold a capacity of 30 to 50 thousand people; anything above 60 000 is a stretch.

The best way to approach such problems is first to eliminate what’s too little and what’s too much, based on your intuition. From that point, maybe it’s a little hard to narrow down to something that makes sense, so it’s a good idea to come up with a reasonable order-of-magnitude, which, in this case, is $10^{4}$. Anything more than that is definitely a bonus: an estimate of $3cdot 10^{4}$ is already more precise, but not always entirely necessary.

Keep in mind that you’re only giving a rough estimate of something, not the precise value, and with practice, your intuition and reasoning will get better.

Roughly around $3cdot 10^{4}$ to $5cdot 10^{4}$ seats.

$textbf{(a)}$      Knowing that a year is 52 weeks, and taking the abovementioned assumptions into consideration, we can conclude the following:

The population of the US is on the order of $10^{8}$, and the order-of-magnitude of the number of weeks in a year is $10^{1}$. So, the roughest estimate would be that the number of gallon containers of milk purchased in a year is on the order of $10^{9}$.

However, since we’ve assumed that the population is $3cdot 10^{8}$, and the number of weeks is 52 (or $5.2cdot 10^{1}$), $textbf{a better estimate would be on the order of $10^{10}$,}$ because when multiplying 52 by 3, you get 156, which is on the order of $10^{2}$.

$textbf{(b)}$      Nowadays, an empty one gallon milk jug weighs a little more that 60 grams, but we’ll round it to 60 to make calculations easier.

Multiplying only the powers of 10, we get that the approximate weight is on the order of $10^{11}$ grams. However, $textbf{a more precise number would be on the order of $10^{12}$,}$ because we’re actually multiplying 60 by $156cdot 10^{8}$, which is just under 10 000. Therefore, we can safely say that the approximate weight of empty one gallon plastic milk jugs is $10^{12}$ grams, or $10^{6}$ tons.

$textbf{(a)}$      $10^{10}$ containers of milk.

$textbf{(b)}$      $10^{6}$ tons of plastic.

$textbf{(a)}$      The rotational speed of the Earth’s surface can be estimated by simply dividing the distance between the two cities by their time difference. Using the well-known formula for average speed, we get:

$$
begin{align*}
v_{avg}&=dfrac{Delta x}{Delta t}\
&=dfrac{4800 text{km}}{3 text{h}}\
&=quadboxed{1600 frac{text{km}}{text{h}}}\
end{align*}
$$

$textbf{(b)}$      Since we don’t know the radius of Earth, we can’t calculate its circumference by using the usual formula $c=2rpi$. Instead, we will use the result from $textbf{(a)}$, and multiply it by 24 hours, because that’s how long it takes for Earth to make one rotation about its axis. Therefore, the Earth’s circumference is

$$
begin{align*}
c&=v_{avg}times 24 text{h}\
&=1600 frac{text{km}}{text{h}}times 24 text{h}\
&=quadboxed{38 400 text{km}}\
end{align*}
$$

$textbf{(c)}$      Using the result from $textbf{(b)}$, as well as the usual formula for circumference, and solving for $r$ yields

$$
begin{align*}
r&=dfrac{c}{2pi}\
&=dfrac{38 400 text{km}}{2pi}\
&=quadboxed{6112 text{km}}\
end{align*}
$$

* When you plug those values into your calculator, you’ll get a decimal number. However, since we only have to $textit{estimate}$ the radius, it’s more than okay to round up the result like above.

$$
begin{align*}
textbf{(a)} quad &boxed{v_{avg}=1600 frac{text{km}}{text{h}}}\
\
textbf{(b)} quad &boxed{c=38 400 text{km}}\
\
textbf{(c)} quad &boxed{r=6112 text{km}}\
end{align*}
$$

Our goal is to establish which of the following equations are dimensionally consistent using $textbf{dimensional analysis.}$

$textbf{A.}$     $v=at$

*      $left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}^{2}}right]cdotleft[text{T}right]$

*      $left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}}right]$

The equation $v=at$ is $textbf{dimensionally consistent.}$

$textbf{B.}$     $v=frac{1}{2}at^{2}$

*      $left[dfrac{text{L}}{text{T}}right]=left[dfrac{text{L}}{text{T}^{2}}right]cdotleft[text{T}^{2}right]$

*      $left[dfrac{text{L}}{text{T}}right]neqleft[text{L}right]$

The equation $v=frac{1}{2}at^{2}$ is $textbf{dimensionally inconsistent.}$

$textbf{C.}$     $t=frac{a}{v}$

*      $left[text{T}right]=left[dfrac{text{L}}{text{T}^{2}}right]cdotleft[dfrac{text{T}}{text{L}}right]$

*      $left[text{T}right]neqleft[dfrac{1}{text{T}}right]$

The equation $t=frac{a}{v}$ is $textbf{dimensionally inconsistent.}$

$textbf{D.}$     $v^{2}=2ax$

*      $left[dfrac{text{L}^{2}}{text{T}^{2}}right]=left[dfrac{text{L}}{text{T}^{2}}right]cdotleft[text{L}right]$

*      $left[dfrac{text{L}^{2}}{text{T}^{2}}right]=left[dfrac{text{L}^{2}}{text{T}^{2}}right]$

The equation $v^{2}=2ax$ is $textbf{dimensionally consistent.}$

Our goal is to establish which of the following quantities have the dimension of acceleration using $textbf{dimensional analysis.}$ The dimension of acceleration is length per time squared;

$$
begin{align*}
[a]=left[dfrac{text{L}}{text{T}^{2}}right]
end{align*}
$$

$textbf{A.}$     $xt^{2}$

*      $left[xt^{2}right]overset{?}{=}left[aright]$

*      $left[text{L}right]cdotleft[text{T}^{2}right]neqleft[dfrac{text{L}}{text{T}^{2}}right]$

Quantity $xt^{2}$ $textbf{doesn’t have the dimension of acceleration.}$

$textbf{B.}$     $frac{v^{2}}{x}$

*      $left[dfrac{v^{2}}{x}right]overset{?}{=}left[a right]$

*      $left[dfrac{text{L}^{2}}{text{T}^{2}}right]cdotleft[dfrac{1}{text{L}}right]=left[dfrac{text{L}}{text{T}^{2}}right]$

*      $left[dfrac{text{L}}{text{T}^{2}}right]=left[dfrac{text{L}}{text{T}^{2}}right]$

Quantity $frac{v^{2}}{x}$ $textbf{has the dimension of acceleration.}$

$textbf{C.}$     $frac{x}{t^{2}}$

*      $left[dfrac{x}{t^{2}}right]overset{?}{=}left[aright]$

*      $left[text{L}right]cdotleft[dfrac{1}{text{T}^{2}}right]=left[dfrac{text{L}}{text{T}^{2}}right]$

Quantity $frac{x}{t^{2}}$ $textbf{has the dimension of acceleration.}$

$textbf{D.}$     $frac{v}{t}$

*      $left[dfrac{v}{t}right]overset{?}{=}left[aright]$

*      $left[dfrac{text{L}}{text{T}}right]cdotleft[dfrac{1}{text{T}}right]=left[dfrac{text{L}}{text{T}^{2}}right]$

Quantity $frac{v}{t}$ $textbf{has the dimension of acceleration.}$

$textbf{(a)}$     Wavelength of $675 text{nm}$ expressed in mm:

$$
begin{align*}
675 text{nm}&=675 text{nm}timesleft(dfrac{1cdot 10^{-6} text{mm}}{1 text{nm}}right)=boxed{675cdot 10^{-6} text{mm}}\
end{align*}
$$

$textbf{(b)}$     Wavelength of $675 text{nm}$ expressed in m:

$$
begin{align*}
675 text{nm}&=675 text{nm}timesleft(dfrac{1cdot 10^{-9} text{m}}{1 text{nm}}right)=boxed{675cdot 10^{-9} text{m}}\
end{align*}
$$

$$
begin{align*}
textbf{(a)}quad &boxed{675 text{nm}=675cdot 10^{-6} text{mm}}\
textbf{(b)}quad &boxed{675 text{nm}=675cdot 10^{-9} text{m}}\
end{align*}
$$

$$
tt{The speed of the glacier is : $64frac{m}{day}=frac{64mfrac{10^{-3}km}{1m}}{1dayfrac{24h}{1day}}=boxed{2.7*10^{-3}frac{km}{h}} $}
$$

$$
2.7*10^{-3}frac{km}{h}
$$

$textbf{(a)}$      If the number of wingbeats per second is 605, then the number of wingbeats in 1 minute is

$$
begin{align*}
dfrac{605 text{wingbeats}}{1 text{s}}&=dfrac{n_{wb}}{60 text{s}}\
\
rightarrow n_{wb}&=dfrac{605 text{wingbeats}}{1 text{s}}times 60 text{s}\
\
&= quadboxed{36300 text{wingbeats}}\
end{align*}
$$

$textbf{(b)}$      One mosquito wingbeat lasts

$$
begin{align*}
dfrac{605 text{wingbeats}}{1 text{s}}&=dfrac{1 text{wingbeat}}{t_{wb}}\
rightarrow t_{wb}&=dfrac{1 text{s}}{605}\
end{align*}
$$

and a cesium-133 atom completes 9 192 631 770 cycles per second. The number of cycles $n_{c}$ that this atom completes during one mosquito wingbeat is

$$
begin{align*}
dfrac{9 192 631 770 text{cycles}}{1 text{s}}&=dfrac{n_{c}}{t_{wb}}\
\
rightarrow n_{c}&=dfrac{9 192 631 770 text{cycles}}{1 text{s}}times t_{wb}\
&=dfrac{9 192 631 770 text{cycles}}{1 text{s}}times dfrac{1 text{s}}{605}\
&=15 194 432.68 text{cycles}\
&=quadboxed{15.2cdot 10^{6} text{cycles}}\
end{align*}
$$

Again, we used the rule for multiplication and division for significant figures when writing the final result.

$$
begin{align*}
textbf{(a)} quad &boxed{n_{wb}=36300 text{wingbeats}}\
\
textbf{(b)} quad &boxed{n_{c}=15.2cdot 10^{6} text{cycles}}\
end{align*}
$$

1 meter is approximately 3.28 feet; so, 10 feet in SI units is

$$
begin{align*}
10 text{feet}&=10 text{feet}timesleft(dfrac{1 text{m}}{3.28 text{feet}}right)\
&=quadboxed{3 text{m}}\
end{align*}
$$

A knot is one nautical mile per hour, so 10 knots is 10 nautical miles per hour. Furthermore, one nautical mile is 1.852 kilometers, which means that 10 nautical miles per hour is 18.52 kilometers per hour. Converting this value to meters per second, we get

$$
begin{align*}
18.52 frac{text{km}}{text{h}}&=left(18.52 frac{text{km}}{text{h}}right)left(dfrac{frac{1000 text{m}}{1 text{km}}}{frac{3600 text{s}}{1 text{h}}}right)\
&=18.52times dfrac{5}{18}frac{text{m}}{text{s}}\
&=5.14444 frac{text{m}}{text{s}}\
&=quadboxed{5 frac{text{m}}{text{s}}}\
end{align*}
$$

Notice that the least accurately known input value is, in fact, 10, which has only 1 significant figure.

“10 and 10” is in SI units “3 and 5”;

$rightarrow quadquad boxed{10 text{ft}=3 text{m}}$

$rightarrow quad boxed{10 text{knots}=5 frac{text{m}}{text{s}}}$

$textbf{(a)}$     The speed of the nerve impulses in kilometers per hour is:

$$
begin{align*}
140 dfrac{text{m}}{text{s}}&=140 dfrac{text{m}}{text{s}}timesleft(dfrac{10^{-3} text{km}}{1 text{m}}right)timesleft(dfrac{3600 text{s}}{1 text{h}}right)\
&=140timesdfrac{10^{-3} text{km}cdot3600}{1 text{h}}\
&=quadboxed{504 dfrac{text{km}}{text{h}}}\
end{align*}
$$

$textbf{(b)}$     The distance that the impulses cover in $5.0 text{ms}$ is:

$$
begin{align*}
d&=vt\
&=140 dfrac{text{m}}{text{s}}cdot 5.0 text{ms}timesleft(dfrac{10^{-3} text{s}}{1 text{ms}}right)\
&=140 dfrac{text{m}}{text{s}}cdot5.0cdot 10^{-3} text{s}\
&=quadboxed{0.70 text{m}}\
end{align*}
$$

*     In both cases we applied $textbf{the rule for multiplication and division for significant figures}$ when writing the final result.

$$
begin{align*}
textbf{(a)}quad &boxed{v=504 dfrac{text{km}}{text{h}}}\
textbf{(b)}quad &boxed{d=0.70 text{m}}\
end{align*}
$$

$textbf{(a)}$      In one day, the newborn’s brain increases in mass by

$$
begin{align*}
dfrac{1.6 text{mg}}{1 text{min}}&=dfrac{Delta m}{1 text{day}}\
Delta m&=dfrac{1.6 text{mg}}{1 text{min}}times 1 text{day}\
&=dfrac{1.6 text{mg}}{1 text{min}}times 1440 text{min}\
&=2304 text{mg}\
&=quadboxed{2.3cdot 10^{3} text{mg}}\
end{align*}
$$

$textbf{(b)}$      The time it takes for the brain’s mass to increase by $0.0075 text{kg}$ is

$$
begin{align*}
dfrac{1.6cdot 10^{-6} text{kg}}{1 text{min}}&=dfrac{0.0075 text{kg}}{Delta t}\
Delta t&=0.0075 text{kg}timesdfrac{1 text{min}}{1.6cdot 10^{-6} text{kg}}\
&=4687.5 text{min}\
&=78.125 text{h}\
&=quadboxed{78 text{h}}\
end{align*}
$$

$$
begin{align*}
textbf{(a)} quad &boxed{Delta m=2.3cdot 10^{3} text{mg}}\
\
textbf{(b)} quad &boxed{Delta t=78 text{h}}\
end{align*}
$$

$textbf{(a)}$      First, we will convert the speed from centimeters per second to meters per minute:

$$
begin{align*}
31 frac{text{cm}}{text{s}}&=left(31 frac{text{cm}}{text{s}}right)left(dfrac{frac{1 text{m}}{100 text{cm}}}{frac{1 text{min}}{60 text{s}}}right)\
&=31times 0.6 frac{text{m}}{text{min}}\
&=18.6 frac{text{m}}{text{min}}\
end{align*}
$$

Next, we’ll calculate the time it took to traverse 150 meters:

$$
begin{align*}
v&=dfrac{Delta x}{Delta t}\
Delta t&=dfrac{Delta x}{v}\
&=dfrac{150 text{m}}{18.6 frac{text{m}}{text{min}}}\
&=8.1 text{min}
end{align*}
$$

Finally, we know that the rate of rotation is 7 revolutions per minute. The number of completed revolutions $n$ after 150 meters is

$$
begin{align*}
dfrac{7 text{revolutions}}{1 text{min}}&=dfrac{n}{8.1 text{min}}\
n&=8.1 text{min}timesdfrac{7 text{revolutions}}{1 text{min}}\
&=56.7 text{revolutions}\
&=quadboxed{56 text{revolutions}}\
end{align*}
$$

In this case, we can’t just round up the number in the final result like we usually do; instead, we have to think about what is asked and what’s the correct answer. The probe $textit{didn’t complete}$ 57 revolutions by the time it was 150 meters away from the spacecraft; it completed 56 full revolutions and was about to make the 57th revolution. That’s the reason why we wrote the result like we did.

$textbf{(b)}$      Before we calculate how far the probe moves away from the spacecraft during each revolution, we have to determine how long it takes to complete one revolution:

$$
begin{align*}
dfrac{7 text{revolutions}}{60 text{s}}&=dfrac{1 text{revolution}}{t}\
t&=1 text{revolution}timesdfrac{60 text{s}}{7 text{revolutions}}\
&=8.6 text{s}\
end{align*}
$$

Using the formula for speed, solving for distance $d$, and applying the rule for multiplication and division for significant figures, we get the following result:

$$
begin{align*}
d&=vt\
&=31 frac{text{cm}}{text{s}}times 8.6 text{s}\
&=266.6 text{cm}\
&=quadboxed{2.7 text{m}}\
end{align*}
$$

$$
begin{align*}
textbf{(a)} quad &boxed{n=56 text{revolutions}}\
\
textbf{(b)} quad &boxed{d=2.7 text{m}}\
end{align*}
$$

$$
tt{to have a dimensionally consistent equation, all terms of the equation must have the same dimension: }
$$

$$
begin{align*}
a&=vt^p\
frac{L}{T^2}&=frac{L}{T}T^p\
T^{-2}&=T^{p-1}\
-2&=p-1 \
&Rightarrow boxed{p=-1}
end{align*}
$$

$$
p=-1
$$

$$
tt{to have a dimensionally consistent all the terms of the equation must have the same dimension: }
$$

$$
begin{align*}
T&=2pi L^pg^q\
T&=L^p(frac{L}{T^2})^q\
T&=frac{L^{p+q}}{T^{2q}}\
-2q=1& &p+q=0\
q=frac{-1}{2}& &p=-q=frac{1}{2}
end{align*}
$$

$$
q=frac{-1}{2}, p={1}{2}
$$

$textbf{The metric system is a system of measurement designed for transnational use.}$ A vital part of science are experiments, and the ability of scientists in different location to reproduce it and verify the consistency of reported results. This is where the metric system comes into play: scientists all over the world use it for collecting data and performing experiments. The metric system is like a universal language used to make communication and collaboration easier.

The SI units are units of measurement used in almost every country in the world. It has seven base units, and allows for an unlimited number of additional units that are derived from base units. The base units are $textbf{the second, meter, kilogram, ampere, kelvin, mole, and candela.}$ It’s important to note that the SI is an evolving system, meaning that unit definitions are modified as the technology and precision improve.

Probably the main advantage of the metric system is that $textbf{it’s a decimal system of units, which means that the units are related to one another by multiples of 10.}$ This makes it incredibly easy to convert from one unit to another: you just have to move the decimal to a new place value. These power-of-ten multiples of base units have standard prefixes that you can “glue” to any base unit. For example, the prefix kilo means one thousand, or $10^{3}$; one kilometer is one thousand meters, one kilogram is one thousand grams, and so on. Overall, the metric system is very intuitive and easy to use.

$textbf{One of the disadvantages when changing from one system of units to another is having to know the conversion factors from one unit to another.}$ This is often not as straightforward as converting one SI unit to another. For example, if you want to convert feet to meters, you have to keep in mind that $1 text{m}=3.281 text{ft}$. Moreover, $textbf{errors are more likely to occur when dealing with such calculations,}$ which could in various industries lead to potentially dangerous outcomes.

$textbf{Officially, only three countries in the world don’t use the metric system, but rather the imperial system: the U.S., Liberia and Myanmar.}$ In Liberia and Myanmar, both the metric and imperial measurements are used, while in the U.S. imperial units are the norm.

It’s worth mentioning that $textbf{some other countries use both systems as well.}$ The imperial system originated in the UK, where it’s still used in everyday life. In addition to the UK, the former Commonwealth countries like Canada, India, and Australia, also often use the imperial system.

Simply put, chemistry is a study of matter and the changes it undergoes during different processes. There is much overlap between physics and chemistry, as $textbf{chemistry uses various theories and branches of physics, such as the kinetic theory of gasses, nuclear and atomic physics, as well as quantum physics}$ just to name a few. In terms of measuring techniques, mass spectrometry is extensively used in chemistry to identify substances based on their mass-to-charge ratios.

Meteorology is a branch of physics, or more precisely, geophysics. It’s the study of the Earth’s atmosphere with a major emphasis on weather forecast.

In their line of work, $textbf{meteorologists have to be comfortable with the physics of fluids and waves to be able to predict the local weather as precisely as possible.}$ The applications of those branches include determining the atmospheric pressure over a certain area, tracking the motion of storm clouds (based on something called the Doppler shift), and many more.

Biology studies life and living organisms, their physical structure, chemical processes, molecular interactions and so forth. That being said, there are many applications of physics in biology. For example, $textbf{classical electrostatics is a huge part of modern molecular biology:}$ large molecules, such as proteins, are usually electrically charged. DNA itself is highly charged; the electrostatic force holds the molecule together and gives it structure. Moreover, physics helps describe the properties of cell walls and membranes Already mentioned $textbf{mass spectrometry and fluid dynamics are also used in biology, as well as wave optics,}$ which is a fundamental principle on which microscopes work.

$circquad$ Chemistry: kinetic theory of gasses, nuclear and quantum physics

$circquad$ Meteorology: physics of fluids and waves

$circquad$ Biology: electrostatics, wave optics

* Click for more detail.

Given the equation

$$
begin{align*}
N=T-5.0
end{align*}
$$

where $N$ is the number of chirps in $7.0 text{s}$, we are to $textbf{determine how many times does the cricket chirp in $21 text{s}$ at a temperature of $22^{circ} text{C}$.}$ Since the equation gives the number of chirps in $7.0 text{s}$, we’ll $textbf{multiply the result by 3}$ to get the correct answer.

The number of chirps in $7.0 text{s}$ is:

$$
begin{align*}
N_{7.0}&=22-5.0\
&=17\
end{align*}
$$

Multiplying the obtained result by 3 gives us the number of chirps in $21 text{s}$:

$$
begin{align*}
N&=3cdot N_{7.0}\
&=3cdot 17\
&=quadboxed{ 51 }\
end{align*}
$$

$$
begin{align*}
boxed{N=51}\
end{align*}
$$

$tt{The function relating chirping rate and temperature is \$N=T-5$}$

for $T=10 Rightarrow N=5$

From this information, we can deduce that (B) and (C) plots are two possible solutions, but we also know that$N=T-5$ is a linear function,therefore the correct answer is (C)

$$
tt{Plot (C)}
$$

The goal is to determine the time it takes for the cricket to chirp 80 times at $T=15 ^{circ}text{C}$, knowing that the equation describing the rate of chirping is given by:

$$
begin{align*}
N=T-5.0
end{align*}
$$

where $N$ is the number of chirps in $7.0 text{s}$.

Notice that $textbf{the rate of chirping is constant at a fixed temperature:}$ the cricket will chirp the same number of times in each $7.0 text{s}$ interval.

The number of chirps in $7.0 text{s}$ at $15 ^{circ}text{C}$ is:

$$
begin{align*}
N&=15-5.0\
&=10\
end{align*}
$$

which we will conveniently write as

$$
begin{align*}
boxed{N=dfrac{10 text{chirps}}{7.0 text{s}}}\
end{align*}
$$

Since the rate of chirping is constant, $textbf{the following expression must hold true:}$

$$
begin{align*}
dfrac{10 text{chirps}}{7.0 text{s}}&=dfrac{80 text{chirps}}{t}\
end{align*}
$$

which solving for $t$ gives:

$$
begin{align*}
t&=80 text{chirps}timesdfrac{7.0 text{s}}{10 text{chirps}}\
&=quadboxed{56 text{s}}\
end{align*}
$$

The correct answer is, therefore, $textbf{C.}$

$textbf{C.}$

The equation describing the rate of chirping is given by:

$$
begin{align*}
N=T-5.0
end{align*}
$$

where $N$ is defined as the number of chirps in $7.0 text{s}$.

The goal is to determine the temperature at which the cricket chirps 113 times per minute using the equation above, which requires calculating $N$ first.

At a fixed temperature, the cricket will chirp the same number of times in a given time interval. In other words, $textbf{the following ratio is constant:}$

$$
begin{align*}
boxed{dfrac{n}{t}=const.}\
end{align*}
$$

In this case, the cricket chirps 113 times in a minute,

$$
begin{align*}
dfrac{113 text{chirps}}{60 text{s}}=const.
end{align*}
$$

Since $N$ is the number of chirps in 7.0 seconds, which is also constant at a fixed temperature, we can write the following equation:

$$
begin{align*}
dfrac{n}{7.0 text{s}}=dfrac{113 text{chirps}}{60 text{s}}
end{align*}
$$

Solving for $n$ and applying $textbf{the rule for multiplication and division for significant figures}$ gives

$$
begin{align*}
n&=dfrac{113 text{chirps}}{60 text{s}}times 7.0 text{s}=13 text{chirps}\
end{align*}
$$

meaning that

$$
begin{align*}
Rightarrowquad &boxed{N=13}\
end{align*}
$$

Finally, the temperature in degrees Celsius is:

*          $13=T-5.0$

*          $boxed{T=18 ^{circ}text{C}}$

Therefore, the correct option is $textbf{C.}$

$textbf{C.}$