All Solutions
Page 24: Assessment
Once the hypothesis is formed, a series of **experiments** can then be performed. As an observation, the data from the experiments will now undergo **measurement**. When analyzing the measured data, it is important for you to now the **dependent variable** and the **independent variable**.
Once you are done with the measurement, a **mathematical model** can then be created that will define the relation between the dependent and independent variables in the experiment.
Finally, the mathematical model can be used in order to **graph** the data and show how the model relates to the actual data from the measurements relate.
textbf{underline{textit{Solution}}}
$$
Moreover, we couldn’t count without mathematics a merely simple process which enables us to do science, moreover we couldn’t integrate or differentiate both are mathematical tools which enables us to have a deeper insight with the variation of a variable with respect to another variable, a simple example differentiating the distance with respect yields velocity.
Also, there is a lot of other branches of physics beside elementary algebra and calculus that have an impact to a lot of different branches of science.
In short mathematics enables us to have a greater insight to the nature of things and to predict as well as describe different phenomena, also it give us a scope and a reach into the imaginable as well as invisible.
textbf{underline{textit{Solution}}}
$$
* Length in meters.
* Mass in kilograms.
* Time in seconds.
* Current in ampere.
* Amount of molecules in units of mole.
* Temperature in kelvin.
* Luminosity intensity in candela.
Also, some example of derived units “units which are described by another units” in SI system are as follows:
* ize
* Force in newton.
* Energy in joule
* Power in watt.
textbf{underline{textit{Solution}}}
$$
item As 100 have only 1 significant figure as the zeros after the integer does not count as significant figure, therefor the actual result could vary from 55 gm to 144.44 gm and as we don’t know the precision of the measuring device, we can’t tell what is the actual measurement.\
The precision of the device, that is it the smallest possible weight the device could measure, to make it more clear, let’s imagine that the smallest weight that can be measured is 1 gm, and assume he made only 1 measurement and it was 98 gm.\
Since the precision of the measurement device is 1 gm, therefor we round the number to 1 significant figure, there for the answer would be 100 gm, however if the precision of the device was 0.1 gm and the measurement was 99.6 gm, then we would approximate the answer to 1 significant digit after the decimal thus the final answer would be 100.0 gm.\
If the precision of the device was 0.01 gm, and the measurement was 99.95 gm then the final answer would be rounded to 2 significant figures after the decimal and the final answer would be 100.00 gm.\
In short as the precision of the measuring device the actual result would vary even tough the final answer is rounded to the same final result, but the actual result would varies.\
item What would make the measurement more significant is adding decimal and zeros after the decimal to indicate the precision of the measurement.\
for example 100.0 would simply mean that this measurement was rounded to the 1 significant figure after the decimal, and 100.00 would mean that this measurement was rounded to 2 significant figures after the decimal and so on.\
Where the trailing zeros “zeros to the right of the decimal” are significant thus 100.0 is a 4 significant figure digit and 100.00 is a 5 significant figure digits thus by adding zeros after the decimal we could estimate what is the actual value.
item The precision of the measurement is not given, and so we don’t know how significant is this measurement, i.e. we don’t know whether it was rounded to at least 1 gm or 0.1 gm and so on.
item Adding decimal and zeros after the decimal, indicating the precision of the measurement and would give a better estimation of what is the actual measurement.
dfrac{60 minutes}{1 hour}
$$
1.8 hours times dfrac{60 minutes}{1 hour}
$$
a) 349,000g
$$
b) 287,000 J/cm^3
$$
textbf{underline{textit{Solution}}}
$$
Replacing proportionality with equality we add a constant Thus the equation expressing this relation is
$$
y= ax^2
$$
And observing the given figure we find that at $x=0$, $y$ has some value $y$, thus we add a constant to the above equation
$$
y= ax^2 + text{const.}
$$
And the general formula for such relations, is given by
$$
y= ax^2 + bx + c
$$
y= ax^2 + bx + c
$$
F = dfrac{1}{R}
$$
F = M
$$
F = v^2
$$
$$F = frac{mv^2}{R}$$
We can analyze the relationship between two variables by holding the other variables constant.
To analyze the relationship between $F$ and $R$, we can let $m$ and $v$ constant. As a result, we have:
$$F = frac{k}{R}$$
This implies that $F propto frac{1}{R}$, which means that $F$ and $R$ are $boxed{text{inversely proportional}}$ to each other.
To analyze the relationship between $F$ and $R$, we can let $v$ and $R$ constant. As a result, we have:
$$F = mk$$
This implies that $F propto m$, which means that $F$ and $m$ are $boxed{text{directly proportional}}$ to each other.
To analyze the relationship between $F$ and $v$, we can let $m$ and $R$ constant. As a result, we have:
$$F = kv^2$$
This implies that $F propto v^2$, which means that $F$ is $boxed{text{directly proportional}}$ to the square of $v$.
b). $F$ and $m$ are directly proportional
c). $F$ is directly proportional to the square of $v$
a) centimeters
$$
b) millimeters
$$
hline
Event & Time interval \
hline
Between hartbeat & 0.85 s\ hline
Time taken for earth rotation & 1 day\ hline
Time taken for earth revolution & 1 year\ hline
Presidintal election & 4 year\ hline
Average lifetime of human & 70 years\ hline
Age of earth & 4.5 Billion years\ hline
Age of universe & 14 Billion years\ hline
end{tabular}
The student who obtained a more precise measurement is **the first student** with $left( 3.001 pm 0.001right) times 10^8 text{ m/s}$. This is because of the uncertainty in its measurement which is only $0.001 times 10^8 text{ m/s}$ compared to that of the second student which is $0.006 times 10^8 text{ m/s}$.
The student who obtained a more accurate measurement is **the second student** with $left( 2.999 pm 0.006right) times 10^8 text{ m/s}$. This is because the measurement is closer to the actual speed of light which is about $2.998 times 10^8 text{ m/s}$.
b). The second student
$$132+83+76=291 text{ cm}$$
Now since all of the addends have no decimal places, the sum must retain the significance level of the addends.
On the other hand, the product of the dimensions is:
$$132 times 83 times 76 = 832656 text{ cm}^3$$
Since the least significant digits in the multiplicand is 2, then the product must also have 2 significant digits only. Hence, the product must be written as:
$$830000 text{ cm}^3 text{ or } 8.3 times 10^5 text{ cm}^3$$
zero and since y does not depend on x, our answer is zero.
d = av^{2} + bv + c
$$
m = a(m/s)^{2} + b(m/s) + c
$$
83 mm pm 0.5 mm
$$
Block A has a mass of 8.45 g while Block B has a mass of 45.87 g. Note that there are no leading/trailing zeroes, which means that all the digits in the measurements are significant. Block A has $boxed{text{3 significant digits}}$ while Block B has $boxed{text{4 significant digits}}$.
When adding two numbers, it is important for the sum to follow the least decimal places in the the addends. Since both addends have 2 decimal places, then the sum must also have two decimal places as well. Therefore, we have:
$$8.45+45.87= boxed{54.32 text{ g}}$$
From the previous step, we solved that the total mass is 54.32 g. Since there are no leading/trailing zeroes, it means that all of its digits are significant. Therefore, the total mass has $boxed{text{4 significant digits}}$.
The difference in the significant digits of the total mass from the individual masses arise from the difference in the number of the significant digits in the addends. In this problem, since the two addends have 3 and 4 significant digits, we should expect that the number of significant digits of the sum can only be the same to only one of the addends.
b). $54.32 text{g}$
c). $4$ significant digits
d). See the explanation.
textbf{underline{textit{Solution}}}
$$
item ~\
According the given statement in the problem, the relation between the speed of the falling object and the medium through which it falls, is inversely proportional.\
Which means that the as the density of the medium increases, the object falls more slowly, thus as it is given that the density of air is much smaller than the density of the water
[ rho_{air} ll rho_{water}]
Thus, the speed of the falling object in air would be much greater than that when falling through water
[ v_{air} gg v_{water}]
item~\
As the relation between the density of the medium and the speed of the object falling through it is given by – from Aristotle point of view – by the following relation
[ v propto dfrac{1}{rho}]
Where we can replace the proportionality sign with equal sign by adding a constant in the equation
[ v= dfrac{text{const.}}{rho} tag{1}]
Where, $rho$ is the density of the medium, and since the vacuum is the absence of any matter – mass – in a given volume, thus the density of the vacuum would be
begin{align*}
rho &= dfrac{0}{V}\
&= 0
intertext{And from equation (1), the velocity of the falling object through a vacuum would be infinite as}
v&=dfrac{text{const.}}{0}\
&= infty
end{align*}
Therefor Aristotle could have thought that as a very thing has a speed limit, thus rejecting the possibility of existence of something that would have a zero density.
* The speed of the falling object through air would be greater than the speed of the falling object through water.
* It speed will be infinite as, the density of the vacuum is zero thus Aristotle rejected the possibility of existence the vacuum.
of energy;
or
Newton’s laws of motion;
or
Law of conservation of
charge.
textit{color{#c34632} $ See $ $Explanation $}
$$
pm 0.5mL
$$
Problem A gives you Centimeters, so the multiplier on Page 6 says that you need to multiply that value by 0.01 (Or $10^{-2}$). Doing so converts the 42.3 cm into 0.423 m. It might also be required by your teacher to write in scientific notation.
0.0000000000062 m or 6.2 $times 10^{-12}$ m
$$
5.80times10^9text{ s}+3.20times10^8text{ s}
$$
$$
=58.0times10^8text{ s}+3.20times10^8text{ s}
$$
$$
=left(58.0+3.20 right)times10^8text{ s}
$$
$$
=61.2times10^8text{ s}
$$
$$
4.87times10^{-6}text{ m}-1.93times10^{-6}text{ m}
$$
$$
=left(4.87-1.93 right)times10^{-6}text{ m}
$$
$$
=2.94 times10^{-6}text{ m}
$$
$$
3.14times10^{-5}text{ kg}+9.36times10^{-5}text{ kg}
$$
$$
=left( 3.14+9.36right)times10^{-5}text{ kg}
$$
$$
=12.50times10^{-5}text{ kg}
$$
$$
=1.250times10^{-4}text{ kg}
$$
$$
8.12times10^{7}text{ g}-6.20times10^{6}text{ g}
$$
$$
=81.2times10^{6}text{ g}-6.20times10^{6}text{ g}
$$
$$
=left( 81.2-6.20right)times10^{6}text{ g}
$$
$$
=75.0times10^{6}text{ g}
$$
$$
=7.50times10^{7}text{ g}
$$
$$
begin{align*}
1021 mathrm{ mu g} qty(dfrac{1 mathrm{mg}}{1000 mathrm{mu g}})&=1,021 mathrm{mg}\
0.000006 mathrm{kg} qty( dfrac{1 000 000 mathrm{mg}}{1 mathrm{kg}})&=6 mathrm{mg}
end{align*}
$$
0.31 mathrm{mg}<1.021 mathrm{mg}<6 mathrm{mg}<11.6 mathrm{mg}
$$
11.6text{ mg} = 11.6times10^{-3}text{ g}=1.16times10^{-2}text{ g}
$$
$$
1021text{ }mutext{g} = 1021times10^{-6}text{ g}=1.021times10^{-3}text{ g}
$$
$$
0.000006text{ kg} = 6times10^{-6}text{ kg}=6times10^{-3}text{ g}
$$
$$
0.31text{ mg} = 3.1times10^{-1}text{ mg}=3.1times10^{-4}text{ g}
$$
$$
0.31text{ mg}<1021text{}mutext{g}<0.000006text{ kg}<11.6text{ mg}
$$
text{color{#4257b2}Ccolor{#4257b2}lcolor{#4257b2}icolor{#4257b2}ccolor{#4257b2}k color{#4257b2}tcolor{#4257b2}o color{#4257b2}scolor{#4257b2}ecolor{#4257b2}e}
$$
$$
16.2text{ m}+5.008text{ m}+13.48text{ m}
$$
$$
=left( 16.2+5.008+13.48right)text{ m}
$$
$$
=34.688text{ m}
$$
Round it to the tenth place
$$
=34.7text{ m}
$$
textbf{color{#c34632}(b)}
$$
$$
5.006text{ m}+12.0077text{ m}+8.0077text{ m}
$$
$$
= left( 5.006+12.0077+8.0077right)text{ m}
$$
$$
= 25.0214text{ m}
$$
Round it upto 3 decimal places
$$
= 25.021text{ m}
$$
textbf{color{#c34632}(c)}
$$
$$
78.05text{ cm}^2-32.046text{ cm}^2
$$
$$
=left( 78.05-32.046right)text{ cm}^2
$$
$$
=46.004text{ cm}^2
$$
Round it up to 2 decimal places
$$
=46.00text{ cm}^2
$$
textbf{color{#c34632}(d)}
$$
$$
15.07text{ kg}-12.0text{ kg}
$$
$$
=left( 15.07-12.0right)text{ kg}
$$
$$
=3.07 text{ kg}
$$
Round it upto one decimal place
$$
=3.1 text{ kg}
$$
$$
left(6.2times10^{18}text{ m} right)left(4.7times10^{-10}text{ m} right)
$$
$$
=left(6.2times4.7right)times10^{18-10}text{ m}^2
$$
$$
=29.14times10^{8}text{ m}^2
$$
The result should have only two significant digits
$$
=29times10^{8}text{ m}^2
$$
$$
left(5.6times10^{-7}text{ m} right)/left(2.8times10^{-12}text{ s} right)
$$
$$
=dfrac{5.6}{2.8}times10^{-7+12}text{ m/s}
$$
$$
=2.0times10^{5}text{ m/s}
$$
$$
left(8.1times10^{-4}text{ km} right)left(1.6times10^{-3}text{ km} right)
$$
$$
=left(8.1times1.6 right)times10^{-4-3}text{ km}^2
$$
$$
=12.96times10^{-7}text{ km}^2
$$
The result should have only two significant digits
$$
=13times10^{-7}text{ km}^2
$$
$$
left(6.5times10^{5}text{ kg} right)/left(3.4times10^{3}text{ m}^3 right)
$$
$$
=dfrac{6.5}{3.4}times10^{5-3}text{ kg/m}^3
$$
$$
=1.911dotstimes10^{2}text{ kg/m}^3
$$
The result should have only significant digits
$$
=1.9 times10^{2}text{ kg/m}^3
$$
F =mg
$$
a) F = (41.63kg)(9.80m/s^2)
$$
632N = m (9.80m/s^2)
$$
By direct substitution to the formula, we get:
$$begin{aligned}
F &= mg \
&= left( 41.63right)left(9.80 right)\
&=407.974 text{ N}
end{aligned}$$
Notice that from the multiplicands, the value of the $g$ has the least significant digits with three. This means that our final answer must only have three significant digits as well. Rounding it to the nearest ones, we have $boxed{F = 408 text{ N}}$.
By manipulating the formula, we can directly substitute the given values:
$$begin{aligned}
m &= frac{F}{g} \
&= frac{632}{9.80} \
&=64.4898 text{ kg}
end{aligned}$$
Notice that from the given, the value of the $g$ and $F$ both have three significant digit. This means that our final answer must only have three significant digits as well. Rounding it to the nearest tenth, we have $boxed{m = 64.5 text{ kg}}$.
b). $64.5 text{ kg}$
51.8kg – 3.64kg
$$
48.2kg
$$
textbf{underline{textit{Solution}}}
$$
[ V= l times w times h tag{1}]
Where,\
$l$: Is the length of the room.\
$w$: Is the width of the room.\
$h$: Is the height of the room.\
Given that the room has a length of 16.40 m, and its width is 4.5 m and the height of the room is 3.26 m, we can find the volume the room encloses using equation (1), as follows
begin{align*}
V &= 16.40 times 4.5 times 3.26 \
&= 240.588 ~ rm{m}^3
intertext{The number of significant figures in the answer can’t exceed the least number of significant figures in the measuring lengths, thus the answer is rounded to 2 significant figures}
&= fbox{$241 ~ rm{m}^3$}
end{align*}
241 ~ rm{m}^3
$$
v = ltimes wtimes h
$$
v = 16.40m times 4.5m times 3.26 m
$$
$$V = l times w times h$$
Where $l$ is the length, $w$ is the width and $h$ is the height. By a direct substitution from the given data, we get:
$$begin{aligned}
V &=16.40 times 4.5 times 3.26 \
&=240.588 text{ m}^3
end{aligned}$$
Notice that from the multiplicands, the value of the $w$ has the least significant digits with two. This means that our final answer must only have two significant digits as well. Rounding it to the nearest tens, we have $boxed{V= 240 text{ m}^3}$.
132.68m + 48.3m + 132.736m + 48.37m
$$
362.1m
$$
$$begin{aligned}
P &= 132.68 + 48.3 + 132.736 + 48.37 \
&= 362.086 text{ m}
end{aligned}$$
Notice that from the addends, the least decimal place is from 48.3 m which has 1. Therefore, the perimeter must also have 1 decimal place in it. Rounding our answer to the nearest tenths, we get $boxed{P = 362.1 text{ m}}$
textbf{underline{textit{Solution}}}
$$
And as the pointer is much closer to the 18th more like 3/4 the distance between the two marks thus the value of the current is
$$
I = left( 17 times 0.2 right) + left( dfrac{3}{4} times 0.2 right)= 3.55~ rm{A}
$$
Where one division is 0.2 A thus the current is nearly 3.55 A.
The uncertainty of the measurements is half the smallest division which is 0.2 A, thus the uncertainty is $pm 0.1$A.
Thus the reading of the ammeter reads
$$
3.55 pm 0.1 ~ rm{A}
$$
3.55 pm 0.1 ~ rm{A}
$$
textbf{underline{textit{Solution}}}
$$
begin{center} textbf{underline{textit{Accuracy of estimation}}} end{center}The accuracy of our measurement is given by the following equation
begin{align*}
text{accuracy} &= dfrac{text{actual value} – text{estimated value} }{text{actual value}}\
intertext{Thus the accuracy is equal to }
&=dfrac{213.1 – 220}{213.1}\
&= -3.24 %
end{align*}
begin{center} textbf{underline{textit{Precision of estimation}}} end{center}Our estimation was given with precision of 0.1 m or 10 cm, therefor in meters that would be 220 cm with uncertainty of $pm 20 ~ rm{cm}$, thus the estimated door height would be
[ 220 pm 20 ~ rm{cm}]
textbf{underline{textit{note:}}} uncertainty here is chosen to be large enough so that the actual values of the door height lies within the range of estimation, in other words if the door is height is very well known to someone then he could state the height of the door in centimeters with a precision to the nearest mm.\
begin{center} textbf{underline{textit{Precision of measurement}}}end{center}Measuring the door is height with a meter stick, would provide the actual value of the door is height with a precision to the nearest mm.\
But when we measured the height of the door using a meter stick, we can find the height nearest to the smallest scale in the meter stick which is the mm, thus we can exactly tell the height of the door to the nearest mm, thus the precision of the measurement is to the nearest mm.
The two precision is different because under estimation there is uncertainty in our estimation, while measuring using a meter stick one can find the value of the height to the smallest scale in the mater stick, thus the precision is in mm.
$$
= dfrac{14 text{textdegree} C}{12 hours} = 1.2 text{textdegree} C/h
$$
$$
= 10 text{textdegree} C – 2.4 text{textdegree} C
$$
$$
= 7.6 text{textdegree} C
$$
$approx8 text{textdegree} C$
$$
d=CF=5F
$$
Where d is distance, F is force and C is constant and is equal to 5.
The unit is cm/N
where “constant”=12 ,Here a=acceleration and m=mass
textbf{underline{textit{Solution}}}
$$
item Plotting the points for the given in the table,and the curve fitting all the points is drawn as the dashed black line, as in the following textbf{graph}
item The fitting curve is a straight line passing through the origin, and of slope $m=0.79$.\
item As the general equation of the straight line is given by
[ y=mx+c]
And since the fitting line is passing through origin, therefor $c=0$ and since the slope of the fitting line is 0.79, thus the equation of the straight line is
[ y=0.79x tag{1}]
item As the slope of a straight line is given by the ratio
[ y = dfrac{text{Rise}}{text{Run}}]
And, since the rise ”change in $y$-axis components of the points” is the mass, thus it has the unit of mass gm.\
And the run ”change in $x$-axis component of the points” is the volume of the alcohol, thus it has a unit of volume cm$^3$.\
Thus, the slope of the straight line has the slope of
[ y = dfrac{rm{gm}}{rm{cm}^3}]
Which the unit of density.
item Substituting the volume of the alcohol 32.5 cm$^3$ in fitting line equation (1), to get the mass of the this volume of alcohol we get
begin{align*}
M &= 0.79 times 32.5\
&= 25.675 ~ rm{gm}\
&= fbox{$25.7 ~ rm{gm}$} \
end{align*}
item See graph.
item Straight line.
item $y=0.79 x$.
item The unit of the slope is gm/cm$^3$. \
And this quantity is named density.
item 25.7 gm
$$
0.0034;mathrm{m} rightarrow 45.6;mathrm{m}rightarrow1234;mathrm{m}
$$
The volume of a rectangular bar knowing the width, length and the height of the bar can be calculated using the following formula
[ V= l times w times h]
And knowing that the dimensions of the rectangular bar are 2.347 m, 3.452 cm and 2.31 mm thus equation (1) can be used to find its volume. \
But first we need to convert the units of lengths to be the same, and since 1 m = 100 cm and 1 cm = 10 mm, hence the dimensions of the rectangular bar in cm is as follows, 234.7 , 3.452 and 0.231 cm, therefor the volume of the rectangular bar is
begin{align*}
V&= 234.7 times 0.231 times 3.452\
&= 187.2 ~ rm{cm}^3
intertext{And, as the 1 cm = 1/100 m, thus 1 cm$^3$ = 1 cm $times$ 1 cm $times$ 1 cm = 1/100 m 1/100 m $times$ 1/100 m = 10$^{-6}$ m$^3$, therefor the volume of the bar in m$^3$ is }
&= fbox{$187.2 times 10^{-6} ~ rm{m}^3$}
end{align*}
And knowing that the density is given by the following equation
[ rho = dfrac{m}{V}]
And given that the mass of the rectangular bar is 1659 gm, and we have calculated the volume of the bar to be 187.2 cm$^3$, we can find the density of the bar as follows
begin{align*}
rho &= dfrac{1659}{187.2}\
&=fbox{$8.865 ~ rm{gm/cm}^3$}
end{align*}
* $187.2 times 10^{-6}$ m$^3$
* 8.865 gm/cm$^3$
Amount of time for the drop to evaporate = $dfrac{1.7times10^{21}}{10^6text{ /s}} = 1.7times10^{15}text{ s}$
Number of years in $1.7times10^{15}$ seconds = $dfrac{1.7times10^{15}}{3600times24times365}approx 53,906,646 text{ years}$
text{Density} = dfrac{text{Mass}}{text{Volume}} = dfrac{17.6text{ g}}{2.2text{ cm}^3} = 8.0text{ g/cm}^3
$$
color{#4257b2}8.0text{ g/cm}^3
$$
Mass of water required to fill the container is $1000times 0.286 = 286$ kg
We need to round the answer to 1 significant digit
So the answer is 300 kg
text{color{#4257b2}300 kg}
$$
For an Ideal gas at a constant temperature the product of the pressure and the volume of the gas is always the same, i.e.
$$
tag{1} P_i V_i = P_f V_f
$$
Knowing the initial pressure of the gas is 101 kPa with an initial volume of 324 cm$^3$, and the final pressure is 404 kPa, we can calculate the final volume of the gas using equation (1).
And knowing the final volume and that the mass of the gas is 4 gm, we can find out the density of the gas using the following equation
$$
rho = dfrac{m}{V} tag{2}
$$
Where,
$m$: Is the mass of the gas.
$V$: Is the volume of the gas.
From equation (1), and substituting the knowns the final volume of the gas is \
begin{align*}
V_f &= dfrac{P_i V_i}{P_f}\ \
&= dfrac{101 times 324}{404}\ \
&= 81 cm^3
intertext{Thus, the final volume of the gas is 81 cm$^3$, textbf{underline{textit{note:}}} we don’t need to convert the unit of the pressure as they cancel out, in other words we can use whatever units of pressure and we could get the same result because in ratios the units cancels out, only the unit of the initial volume determine the unit of the final volume.}
intertext{Knowing, that the mass of the gas is 4.00 gm and the final volume is 81 cm$^{3}$ we use equation (2) to find out the density}
rho &= dfrac{4}{81}\
&=0.0494 ~ rm{gm/cm}^3
end{align*}
textbf{underline{textit{note:}}} the final volume decreased as the pressure has increase, which means that the volume and the pressure of a gas are inversely proportional, the initial density of the gas was 0.0124 gm/cm$^3$, and the the final density of the gas where the final pressure has increased is 0.0494 gm/cm$^3$, thus the density of the gas is proportional to its pressure.
textbf{underline{textit{Solution}}}
$$
From the equation of motion, one find that the maximum height of a projectile motion is given by the following equation
$$
h= dfrac{v_y^2}{2g} tag{1}
$$
Where,
$h$ Is the maximum vertical height the projectile can reach of unit m.
$v_y$ Is the $y$-component of the initial velocity which has a unit of m/s.
$g$ Is the acceleration of gravity of unit m/s$^2$.
From equation (1) one can clearly notice that the maximum height a projectile can reach depends solely on the $y$-component of initial velocity, and the acceleration of gravity.
Where, the $y$-component of the initial velocity is given by the following equation
$$
v_y = v_o sintheta tag{2}
$$
Where,
$v_o$ Is the initial speed of the projectile.
$theta$ Is the angle the projectile makes with the horizontal axis.
Thus, the maximum height in terms of initial speed of the projectile
$$
h= dfrac{v_o^2 sin^2theta}{2g} tag{3}
$$
Thus, assuming one can throw a ball with an initial speed of 20 m/s, and in a direction of $y$-axis i.e. $theta = 90^{circ}$, thus from equation (3) one can can calculate the the maximum height of the projectile he can throw.
textbf{underline{textit{Calculations}}}
$$
Substituting with initial speed of 20 m/s, and angle $theta=90^{circ}$ in equation (3) to find the maximum height, we get
$$
begin{align*}
h&= dfrac{(20)^2 (1)^2}{2times 9.8}\
&= fbox{$20.41 ~ rm{m}$}\
end{align*}
$$
* One can throw a ball to a heights up to 20 m.
* The only variable affecting the maximum height of the ball is the initial velocity of the ball in the absence of any external force.
$$
begin{align*}
c_0&=3 times 10^8 mathrm{m/s}\
r_E&=149,600,000 mathrm{km}=1.496 times 10^{11} mathrm{m}\
r_J&= 778, 000, 000 mathrm{km}= 7.78 times 10^{11} mathrm{m}
end{align*}
$$
Unknown:
$$
begin{align*}
t_E&=? \
t_J&=?\
\
&c_0 tag{speed of light in vacuum} \
&r_E , r_J tag {distance from Sun to planet}\
&t_E, t_J tag{Time it takes light to reach Earth/Jupiter}
end{align*}
$$
$$
v=s/t rightarrow t=s/v
$$
$$
begin{align*}
v& rightarrow c_0\
s& rightarrow r_E, r_J \
t & rightarrow t_E, t_J \
\
t_E&=dfrac{r_E}{c_0} \
&=dfrac{1.496 times 10^{11} mathrm{m}}{3times10^8 mathrm{s}}\
&=498.7 mathrm{s}\
\
t_J&=dfrac{r_J}{c_0} \
&=dfrac{7.78 times 10^{11} mathrm{m}}{3times10^8 mathrm{s}}\
&=2 593 mathrm{s}
end{align*}
$$
$$
t_E=498.7 mathrm{s}
$$
Time to become dark on Jupiter is:
$$
t_J=2593 mathrm{s}
$$
In this short text we will provide a brief overview of the development of electricity from antiquity to the development of classical electrodynamics and part of modern physics. Many scientists have participated in the development of this area, but we will mention the most important ones.
The very emergence of electricity was not studied until the advent of the natural sciences in the Renaissance. The beginning of scientific observations in the field of electricity occurred in 1600 when W. Gilbert described the direction of action of electric force, although he did not yet know that it was a force, but only described the action of rubbed amber.
In 1729, Stephen Gray made great strides in experimenting with electricity. It divides materials into conductors and insulators.
In the 18th century, the development of machines for generating and storing electricity began, which accelerated the study of the same. In 1749, W. Watson, through a series of experiments, came to the conclusion that electricity is not generated but transmitted. during this time there was a great misunderstanding about the idea of the existence of one or two kinds of electricity. This dispute was resolved only in the 20th century.
A well-known and very influential scientist in the field of electricity is B. Franklin. He is best known for his work with lightning, for which he theorized that it was a natural phenomenon of electric discharge, which was later proven. His great invention that is still used today is the lightning rod.
$$
F = k cdot dfrac {Q cdot q}{r^2}
$$
He also determined how to measure the charge using a torsion balance.
Mathematical theory of electricity and magnetism
developed by Pierre-Simon Laplace, Jean-Baptiste Biot, Siméon Denis Poisson, Augustin-Louis
Cauchy, Carl Friedrich Gauss, Wilhelm Eduard Weber.
A new era of electricity research was made possible by L. Galavani, who discovered a new source of electricity with friction as the only one to date. His research on electricity in the animal world, especially in frog legs, greatly inspired the next giant, A. Volt.
A. Volta, inspired by Galvani’s experiments, discovered that in these experiments the main role in the production of electricity was played by metal utensils and the solution between them and not by the living organism itself. In 1800, the Volta made the first source of electricity, which was essentially the first battery. The production of direct currents has led to a more precise definition of the concepts of current strength,
electrical voltage and electrical resistance.
In 1811, Poisson introduced the notion of electric potential, which inspired the work of G. S. Ohm. His greatest contribution is certainly Ohm’s law of electrical resistance which says that resistance is equal to the ratio of Voltage and current.
$$
R= dfrac{U}{I}
$$
the same force as if it were a magnet.
The connection between electricity and magnetism was explained in more detail by Ampere. His significant contribution is the mathematical formulation of the relationship between current strength and magnetic field. Ampère introduced the notion of electrodynamics, then the notion of an electric circuit, an electric one
currents and the term galvanometer. It is named after the unit of electric current, Ampere
A great turn in the understanding of electricity was caused by miraculous experimental discoveries M.
Faradaya. From experimenting with electricity in various media Faraday sees that the essence of electrical and magnetic phenomena takes place in processes in the space between electric charges. He introduced the concept of electric and magnetic forces, which is still used today
by which he influenced modern field theory.
Faraday’s discoveries and experiments paved the way for the development of the theory of electrodynamics.
J. C. Maxwell, inspired by Faraday’s understandings, develops a system of equations that interrelate changes in electrical and magnetic fields, and combine optics with electromagnetism. Maxwell’s great achievement was the formulation of the so-called the classical theory of electromagnetic radiation, which for the first time linked electricity, magnetism and light as different manifestations of the same phenomenon. In Maxwell’s painting, light is a transverse traveling wave, an electromagnetic wave propagating at the speed of light.
J. J. Thomson using cathode ray tubes showed in 1897 that all atoms contain
tiny negatively charged subatomic particles called electrons.
