If two vectors which are inclined to each other are represented in magnitude and direction by the two adjacent sides of a triangle taken in order, then their resultant is the closing side of the triangle taken in the reverse orders.
Let us consider two vectors, \(\overrightarrow { A } \) and \(\overrightarrow { B } \) inclined an angle \(\theta\) with each other as shown in figure

To find the resultant, the head of the first vector \(\overrightarrow { A } \) is connected to the tail of the second vector \(\overrightarrow { B } \). Let \(\theta\)  be the angle between them. The resultant vector \(\overrightarrow { R } \) is the line drawn connecting the tail of the first vector, \(\overrightarrow { A } \) to the head of the second vector \(\overrightarrow { B } \). Thus \(\overrightarrow { R } \) = \(\overrightarrow { A } \)+ \(\overrightarrow { B } \). The magnitude is \(\overrightarrow { R } \) (resultant) is given geometrically by the length of \(\overrightarrow { R } \)(\(\theta\)). The direction of the resultant is the angle between \(\overrightarrow { R } \) and \(\overrightarrow { A } \).
To find the magnitude of the resultant, the triangle ABN is obtained by extending OA to ON and drawing the perpendicular drop from B.

ABN is a right angled triangle. From the figure, \(\cos\ \theta={{AN}\over{B}}\ \therefore AN=B\ \cos\ \theta\)
\(\sin\ \theta={{BN}\over{B}}\ \therefore BN=B\ \sin\theta\)
For  \(\triangle OBN,\) we have OB² + ON² + BN²
\(R^2=(A+B\ \cos\theta)^2+(B\sin\theta)^2\)
\(=A^2+B^2cos^2\theta+2AB\ \cos\theta+B^2\sin^2\theta\)
\(=A^2+B^2(\cos^2\theta+\sin^2\theta)+2AB\ \cos\theta\)
\( R\ =\sqrt{A^2+B^2+2AB \cos\theta}\)
Let \(\overrightarrow { R } \) makes an angle \(\alpha\) with \(\overrightarrow{A},\) then in \(\triangle OBN,\)
\(\tan\ \alpha={{BN}\over{ON}}+{{BN}\over{OA+AN}}=\frac{B \ sin \theta}{A +B\ cos \theta}\)
\(\therefore \alpha={\tan}^{-1}\left[ {{B\sin\theta}\over{A+B\cos\theta}} \right]\)

Work-Kinetic Energy Theorem
Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.
The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs
The constant force is given by the equation,
F = ma
The third equation of motion can be written as,
\(v^{2} =u^{2}+2 a s  \)
\(a =\frac{v^{2}-u^{2}}{2 s}\)
Substituting for a in equation (2),
\(F=m\left(\frac{v^{2}-u^{2}}{2 s}\right)\)
Substituting equation (2), (1)
\(w=m\left(\frac{v^{2}}{2 s} s\right)-m\left(\frac{u^{2}}{2 s} s\right)  \)
\(w=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
The expression for kinetic energy:
The term \(\left(\frac{1}{2} m v^{2}\right)\) in the above equation is the kinetic energy of the body of mass (m) moving with velocity(v).
\(K E=\frac{1}{2} m v^{2}\)
Kinetic energy of the body is always positive. From equations (4) and (5)
\(\Delta K E =\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}  \)
\(\text {Thus, } W =\Delta K E\)
The expression on the right hand side (RHS) of equation (6) is the change in kinetic energy (\(\Delta\)KE) of the body.
This implies that the work done by the force on the body changes the kinetic energy of the body, This is called work-kinetic energy theorem.
The work-kinetic energy theorem implies the following.
1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

The angle 7.2 degree is equivalent to \(\frac{1}{8}\) radian. So \(\theta=\frac{1}{8} \mathrm{rad}\)
If S is the length of the arc between the cities of Syne and Alexandria and if R is radius of Earth, then
S = R\(\theta \)
= 500 miles
So radius of the Earth
\(\mathrm{R}=\frac{500}{\theta} \text { miles }  \)
\(\mathrm{R}=500 \frac{\text { miles }}{1 / 8} \)
R = 4000 miles
1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436 km, which is amazingly close to the correct value of 6378 km.

This is a thermodynamic process in which the volume of the system is kept constant. But pressure, temperature and internal energy continue to be variables.
The pressure-volume graph for an isochoric process is a vertical line parallel to pressure axis as shown in Figure.

The equation of state for an isochoric process  is given by
\(P=\left( \cfrac { \mu R }{ V } \right) T\)   ...(1)
Where \(\left( \cfrac { \mu R }{ V } \right) \)= constant
It is that the pressure is directly proportional to temperature. This implies that the P-T graph for an isochoric process is a straight line passing through origin.
If a gas goes from state (P_i,T_i) to (P_f, T_f) at constant volume, then the system satisfies the following equation
\(\cfrac { { P }_{ i } }{ { T }_{ i } } =\cfrac { { P }_{ f } }{ { T }_{ f } } \)   ...(2)
For an isochoric processes, \(\triangle \)V = 0 and W = 0. Then the first law becomes
\(\triangle \)U = Q   ....(3)
Implying that the heat supplied is used to increase only the internal energy. As a result the temperature increases and pressure also increases.
Suppose a system loses heat to the surroundings through conducting walls by keeping the volume constant, then its internal energy decreases. As a result the temperature decreases; the pressure also decreases.
1. When food is cooked by closing with a lid as shown in figure.
When food is being cooked in this closed position, after a certain time you can observe the lid is being pushed upwards by the water steam. This is because when the lid is closed, the volume is kept constant. As the heat continuously supplied, the pressure increases and water steam tries to push the lid upward's.

2. In automobiles the petrol engine undergoes four processes. First the piston is adiabatically compressed to some volume as shown in the Figure (a). In the second process (Figure (b)), the volume of the air-fuel mixture is kept constant and heat is being added. As a result the temperature and pressure are increased. This is an isochoric process. For a third stroke (Figure (c)) there will be an adiabatic expansion and fourth stroke again isochoric process by keeping the piston immoveable (Figure (d)).

Let us consider a massless spring with stiffness constant or force constant k attached to a ceiling as shown in figure. Let the length of the spring before loading mass m be L. If the block of mass m is attached to the other end of spring, then the spring elongates by a length l. Let F₁ be the restoring force due to stretching of spring. Due to mass m, the gravitational force acts vertically downward. We can draw free body diagram for this system as shown in figure. When the system is under equilibrium,
\(\mathrm{F}_{1}+\mathrm{mg}=0\)  .....(1)
But the spring elongates by small displacement l, therefore,
\(F_{1} \propto l \Rightarrow F_{1}=-\mathrm{k} l\)   ......(2)
Substituting equation in equation, we get
\(-\mathrm{k} l+\mathrm{mg} =0  \)
\(\mathrm{mg} =\mathrm{k} l  \)     or
\(\frac{m}{k} =\frac{l}{g} \)    .....(3)
Suppose we apply a very small external force on the mass such that the mass further displaces downward by a displacement y, then it will oscillate up and down. Now, the restoring force due to this stretching of spring (total extension of spring is y+l) is
\(\mathrm{F}_{2} \propto(y+l)  \)
\(\mathrm{F}_{2}=-\mathrm{k}(\mathrm{y}+l)=-\mathrm{ky}-\mathrm{k} l\)  ......(4)
Since, the mass moves up and down with acceleration \(\frac{d^{2} y}{d t^{2}},\) by drawing the free body diagram for this case, we get
\(-\mathrm{k} y-\mathrm{k} l+m g=\mathrm{m} \frac{d^{2} y}{d t^{2}}\)   .....(5)
The net force acting on the mass due to this stretching is
\(\mathrm{F}=\mathrm{F}_{2}+m g  \)
\(\mathrm{~F}=-\mathrm{k} y-\mathrm{k} l+m g\)    .....(6)
The gravitational force opposes the restoring force. Substituting equation in equation, we get
\(\mathrm{F}=-\mathrm{k} y-\mathrm{k} l+\mathrm{k} l=-\mathrm{k} y\)    .....(7)
Applying Newton's law, we get
\(m \frac{d^{2} y}{d t^{2}} =-\mathrm{k} y  \)
\(\frac{d^{2} y}{d t^{2}} =-\frac{k}{m} y\)      ......(8)
The above equation is in the form of simple harmonic differential equation. Therefore, we get the time period as
\(\mathrm{T}=2 \pi \sqrt{\frac{m}{k}} \text {second }\)   ........(9)
The time period can be rewritten using equation
\(\mathrm{T}=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{l}{g}} \text { second }\)   ......(10)
The acceleration due to gravity g can be computed from the formula
\(g=4 \pi^{2}\left(\frac{l}{T^{2}}\right) m s^{-2}\)    ......(11)

(a) Closed organ pipes:
Closed organ is a pipe with one end closed and the other end open. If one of a pipe is closed, the wave reflected at this closed end is 180^o out of phase with the incoming wave. Thus there is no displacement of the

particles at the closed end. Therefore, nodes are formed at the closed end and anti-nodes are formed at open end.
Let us consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the open end and node at closed end. From the figure let L be the length of the tube and λ be the wavelength of the wave produced. For the fundamental mode of vibration, we have,
\(L={\lambda_1\over4}(or)\lambda_1=4L\)
The frequency of the note emitted is
\(f_1={v\over \lambda_1}={v\over 4L}\)
which is called the fundamental note.
The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.
The figure (2) shows the second mode of vibration having two nodes and two anti-nodes

4L = 3λ₂ 
\(L={3⋋_2\over 2}or\ ⋋_2={4L\over 3}\)
The frequency for this
\(f_1={v\over \lambda_2}={3v\over 4L}=3f_1\)
is called first over tone, since here, the frequency is three times the fundamental frequency it is called third harmonic.
The Figure (3) shows third mode of vibration having three nodes and three anti-nodes.

We have 4L= 5⋋₃
\(L={5\lambda_3\over4}\ or\ \lambda_3={4L\over 5}\)
The frequency
\(f_3={v\over \lambda _3}={5v\over 4L}=5f_1\)
is called second over tone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and frequency of the n^th harmonic is f _n = (2n + 1)f₁. Therefore, the frequencies of harmonics are in the ratio
f₁:f₂:f₃:f₄:.... = 1:3:5:7:....
(b) Open organ pipe: 
Consider the picture of a flute, shown in Figure. It is a pipe with both the ends open. At both open ends, anti-nodes are formed. Let us consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe

From Figure (5), if L be the length of the tube, the wavelength of the wave produced is given by
\(L={\lambda_1\over 2}or\ {\lambda_1=2L}\)
The frequency of the note emitted is
\(f_1={v\over \lambda_1}={v\over 2L}\)
which is called the fundamental note.
The frequencies higher than fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.

The Figure (6) shows the second mode of vibration in open pipes. It has two mode of vibration in open pipes. It has two nodes and three anti-nodes, and therefore,
L= ⋋₂ or ⋋₂= L
The frequency
\(f_2={v\over \lambda_2}={v\over L}=2\times{v\over 2L}=2f_1\)
is called first over tone. Since n = 2 here, it is called the second harmonic.

The Figure (7) above shows the third mode of vibration having three nodes and four anti-nodes
\(L={3\over2}\lambda,\ or\ \lambda_3={2L\over 3}\)
The frequency
\(f_3={v\over \lambda_3}={3v\over 2L}=3f_1\)
is called second over tone. Since n = 3 here, it is called the third harmonic. 
Hence, the open organ pipe has all the harmonics and frequency of n^th harmonic is f _n = nf₁, Therefore, the frequencies of harmonics are in the ratio
f₁ : f₂ : f₃ : f₄ :... = 1 : 2 : 3 : 4 : ....

As F \(=G{m_1m_2\over r^2}\)
\(G={F.r^2\over m_1m_2}\)
\([G]={{MLT}^{-2}.L^2\over MM}={M}^{-1}L^3{T}^{-2}\)
\(\therefore\) a = -1, b ~ 3, c = -2

\(\therefore\) \(n_2=n_1{\left[ {M_1 \over M_2} \right]}^{a}{\left[ {L_1 \over L_2} \right]}^{b}{\left[ {T_1 \over T_2} \right]}^{c}\)
\(=6.67\times{10}^{-8}{\left[ {{1\over 1000}} \right]}^{-1}{\left[ {{1\over 100}} \right]}^{3}\left[ {1\over 1} \right]^{-2}=6.67\times{10}^{}-11\)
Hence in SI units,   G = 6.67\(\times\)10^-11Nm² kg^-2

n\(\propto\) I^aT^bm^c, [I] = [M^oL¹T^o]
[T] = [M¹L¹T^-2] (force)
[M] = [M¹L^-1T^o]
[M^o L^oT^-1] = [M^oL¹T^o]^a [M¹L¹T^-2]^b [M^oL^-1T^o]^C
b + c = 0
a + b - c = 0
-2b = -1 \(\Rightarrow\) b = \(1\over2\)
c =\(-{1\over2}a=1\)
n\(\propto\) \({1\over l}{\sqrt{T\over m}}\)

Initial velocity 'u' =0
Acceleration 'a' = 0.2 ms^-2
Time (t) = 3 minutes = 180 seconds.
Speed of a bus, v = u + at
= 0 + 0.2\(\times\) 180
= 36 m/s.
Displacement s = ut+\(\frac { 1 }{ 2 } { at }^{ 2 }\)
\(=0\times 180+\frac { 1 }{ 2 } \times 0.2\times ({ 180) }^{ 2 }\)
\(=0+\frac { 1 }{ 2 } \times 0.2\times 32400\)
= 3240 m
=3.24 km