$$
tt{An area dimension is equal to $Length*Length=Length^2=L^2$}
$$

Dimensional consistency means that each term of an equation must have the same dimensions. Therefore any terms expressing an area should have the dimension of $L^2$:

$2pi r =L Rightarrow$ This does not conform to an Area.

$4pi r^2=L^2 Rightarrow$ This is the dimension of an Area.

The surface area of a sphere is $4pi r^2$

$$
4pi r^2
$$

$tt{Given Data:}$

$x$ Distance $Rightarrow$ Dimension: $L$

$t$ Time$Rightarrow$ Dimension: $T$

$v$ Speed $Rightarrow$ Dimension $frac{L}{T}$

$a$ Acceleration $Rightarrow$ Dimension $frac{L}{T^2}$

(a) $frac{x}{t^2}Rightarrow frac{L}{T^2} Rightarrow$ Acceleration.

(b) $vtRightarrow frac{L}{T}T=L Rightarrow$ Distance.

(c) $(2ax)^{frac{1}{2}}Rightarrow (frac{L}{T^2}L)^{frac{1}{2}}=(frac{L^2}{T^2})^{frac{1}{2}}=frac{L}{T} Rightarrow$ Speed.

(d)$frac{1}{2}at^2 Rightarrow frac{L}{T^2}T^2=L Rightarrow$Distance

$$
tt{(a)Acceleration, (b)Distance, (c)Speed, (d)Distance}
$$

Volume of a sphere: $frac{4}{3}pi r^3$

for the given sphere radius 0.735 m:

$$
v=frac{4}{3}pi(0.735)^3=1.66text{ m}^3
$$

$$
v=1.66text{ m}^3
$$

$tt{ The distance $x$ covered in the time $t$ by the object is:}$

$$
begin{align*}
x&=frac{1}{2}at^p \
L&=frac{L}{T^2}T^p \
L&=L*T^{P-2}
end{align*}
$$

In order for an equation to be dimenstionnaly consistant, all terms of the equation need to have the same dimesnsions ,this implies the following:

$$
Rightarrow T^{P-2}=T^{0}
$$

$$
Rightarrow P-2=0
$$

$$
P=boxed{2}
$$

$$
tt{$P=2$}
$$

$tt{It is estimated that human nail grow at a rate of 3mm per month:}$

We first convert the quantities in question to the desired units $frac{m}{s}$:

$$
3mm=3mmfrac{1m}{10^3mm}=3*10^{-3}m
$$

$$
1month=1monthfrac{1s}{frac{1}{2592000}1month}=2592000s
$$

Then we determine the order of magnitude of each quantity:

$$
3*10^{-3}mRightarrow 10^{-3}
$$

$$
2592000sRightarrow 10^{6}
$$

We can now determine the magnitude of nails growth:

$$
v=frac{d}{t}=frac{10^{-3}}{10^{6}}=10^{-9}frac{m}{s}
$$

$$
v=10^{-9}frac{m}{s}
$$

$$
tt{Helium-neon lasers produce a red beam of light with a wavelength of 632.8 nm, According to {color{#4257b2}{Table 1.4}}:}
$$

$$
1nm=10^{-9}m
$$

$$
1km=10^3m
$$

Therfore the length of laser beam in kilometers is :

$$
632.8 nm=632.8 nmfrac{1m}{10^{9}nm}=632.8*10^{-9}m
$$

$$
632.8*10^{-9}m=632.8*10^{-9}mfrac{1km}{10^3m}=boxed{632.8*10^{-12}km}
$$

$$
632.8*10^{-12}km
$$

$tt{The original speed of the Rover is expressed in inches per seconds.}$
(a) We are required to express the speed as meters per seconds, We proceed as follows:

$$
1m=39.37 text{inch}
$$

$$
Rightarrow 1.5 frac{text{inch}}{s}=1.5 frac{text{inch}}{s}frac{1m}{39.37text{inch}}=0.0381 frac{m}{s}
$$

(a) We are required to express the speed as Kilometers per hour, We proceed as follows:

$$
1km=10^3m
$$

$$
1s=frac{1}{3600}h
$$

$$
Rightarrow 0.0381frac{m}{s}=0.0381frac{m}{s}frac{km}{10^3m}=0.0381*10^{-3}frac{Km}{s}
$$

$$
0.0381*10^{-3}frac{Km}{s}=0.0381*10^{-3}frac{Km}{s}frac{s}{frac{1}{3600}h}=0.137frac{km}{h}
$$

$$
tt{(a) $0.0381frac{m}{s}$,(b) $0.137frac{km}{h}$}
$$

$tt{The average Diameter $d$ of a base ball=70mm}$

We then proceed to calculate the volume of a single ball, we will approximate the volume of the ball to the volume of a perfect sphere:

$$
v=frac{4}{3}pi r^3=frac{4}{3} pi(frac{d}{2})^3=1.8 *10^{5}mm^3=1.8 *10^{-4}m^3
$$

An average classroom volume is $V_{class}=200m^3$

The order of magnitude of a ball volume is:$10^{-4}m^3$

The order of magnitude of a classroom volume is:$10^{2}m^3$

Therefore the order of magnitude estimate of the number of balls $N$ that can fill a room is:

$$
N=frac{V_{class}}{v}=frac{10^{2}}{10^{-4}}=10^6
$$

$$
N=10^6
$$