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Note: This post is only for educational purposes. The author of this post is not liable of how the reader uses the provided information

Step by Step guide to create free anonymous wordpress blog:

Step 1 : Use VPN

The easiest way to prevent your internet service provider from logging your online data is to use a virtual private network (VPN). A VPN encrypts your data by bouncing it to different servers. Now, this doesn’t prevent your ISP from seeing the data, it just prevents it from knowing what it means. Instead of logging that you visited Quora, for example, it will see a string of symbols that seem like gibberish.

On top of this, any government agency that wants to see your data will also see nothing but gibberish. If you’re using a high-quality VPN like NordVPN or Norton, there is almost no chance that the government could decrypt this data.

Of course, the government could then go to your VPN provider and try to get the data from there, ignoring your ISP tracking data. Since VPNs are not bound by the same laws as ISPs, though, there isn’t a rule that the VPN service needs to retain your data. A good VPN service will have a “no logging” policy, which means your data is never saved — which means the government can’t get it.

Whatever you do after this step ,do not do it without VPN or Alternatively you can also use TOR, OR proxies but I would not suggest you to use TOR .First reason is every time you try to login ,you would go through robot checks multiple and secondly internet speed will be very low as compared to when you use VPN.

Encrypted browsers are also a pretty good way of keeping your internet data private. Most of these browsers — including the most popular product, Tor — mimic a VPN by bouncing your data around multiple nodes, which are actually the computers of other users.

The idea is the same here: a government agency requests the data logged through ISP tracking and the ISP can only produce gibberish. However, in this case, there is no VPN service so there is nowhere for the agency to turn to request the data.

Since Tor is free, you might think that this is better than a VPN. However, there are two very important factors to note. The first is that the “exit relay,” or the last user node your data travels through, is not encrypted with Tor. This means that somewhere out there on the internet, your data still exists. Obviously, it would be incredibly difficult for anyone to find it and match it to you, but it is not impossible.

The second reason Tor isn’t better than a VPN is that it’s only a browser. If you are doing something on the internet that’s not through a browser — such as through an app on your phone — that activity might not be encrypted.

What about using both? Interestingly, this is not advisable. The reason for this is because now you have two leakages of data: the data your VPN tracks and the output node of Tor. While a VPN with a no-logging policy seems foolproof, having your data go through two services doubles the chances of it being found.

In general, I suggest using private browsers such as Tor for specific activities only. A premium VPN with a strict no-logging policy is the best all-around solution for the average user.

When it comes to anonymity and network speed NordVPN is the best

Below I have listed some good VPN service providers:

Step 2 : Use an anonymous email account

Use ProtonMail (other alternatives Tutanota,Secure Email or Guerilla Mail) to create anonymous email account that you will be using to create hosting account.

My suggestion will be ProtonMail .

Step 3: Use a random name and never use it anywhere.

Step 4: Register domain anonymously

For free domain name ,you can use freenom or you can also use subdomain provided by hosting provider in next step.

For paid domain there are not many choices but you can use whois protection on the registered domain name to hide your details from public.

(tip: Make a believable typo in your name when registering your domain)

Step 5: Hosting Service

Now this is something that I have tried and it worked.

Make an account on infinityfree.net using the ProtonMail email and then choose the desired free domain name provided by infinityfree OR any other domain name of your choice.

You can Install 400+ applications with softaculous script installer.

I have listed some other free hosting services but I have not tried all of them with ProtonMail.

ServersFree
HelioHost
GoogieHost
Webbera
InnoxHost
ProFreeHost
Freehostia
AwardSpace
x10Hosting
110mb
5GBfree
FreeHosting
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FreeHostingEU
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Freevirtualservers
1FreeHosting
WebFreeHosting
Bravenet
FreeWebHostingArea

Some Paid Hosting services

Hostinger :

One of the better web hosting providers in general, Hostinger also offers good anonymity, and apart from Bitcoin, they also take various other cryptos, like Ripple, Ethereum, Litecoin, and several others.

Namecheap:

True to its name, Namecheap is a very affordable option and it’s a good choice for those who don’t want to spend too much. As for the cryptocurrencies, they only accept Bitcoin and Bitcoin Cash.

Bluehost

Bluehost, an Endurance International Group company, is a leading provider of cloud-based solutions, including web hosting services, eCommerce tools, marketing applications, and more. Built on open source technology, Bluehost designs and operates its own servers, develops innovative new internet technologies, and actively supports and participates in the open source community.Powering over 2 million websites, Bluehost offers the ultimate WordPress platform. Tuned for WordPress, we offer WordPress-centric dashboards and tools along with 1-click installation, a FREE domain name, email, FTP, and more.

explained-physics.blogspot.com

TOPIC-4

KINEMATICS

Basic concept

Any particle which is thrown into space or air such that it moves under the influence of an external force (e.g. gravity, electric forces etc.) is called a projectile. The motion of such a particle is referred to as projectile motion.

it is an example of two dimensional motion with constant acceleration.

If the force acting on the projectile is constant, then acceleration is constant. When the force is in oblique direction with the direction of initial velocity, the resultant path is parabolic.

Parabolic motion = Vertical motion + Horizontal motion

Projectile motion can be considered to be two simultaneous motions in mutually perpendicular directions which are completely independent of each other i.e. horizontal motion and vertical motion

Ground to ground projectile

Consider a projectile thrown from horizontal ground with a

velocity u making an angle 0 with the horizontal. Take the point

of projection as origin O and the path of the projectile in the first quadrant of xy - plane, as shown in the figure. The initial velocity u is resolved in the horizontal and vertical directions i.e.

\({u_x} = u\cos \theta \) \({u_y} = u\sin \theta \)

Since gravity is the only force acting on the projectile in vertically downward direction, (ignoring air resistance)

\({a_x} = 0\)

\({a_y} = - g\)

Analyzing the motion of the projectile in horizontal and vertical directions:

Horizontal direction:

Initial velocity : \({u_x} = u\cos \theta \)

Acceleration : \({a_x} = 0\)

Velocity after time t : \({v_x} = u\cos \theta \)

Vertical direction:

Initial velocity : \({u_y} = u\sin \theta \)

Acceleration : \({a_y} = - g\)

Velocity after time t : \({v_y} = u\sin \theta - gt\)

The position vector of the projectile after time t is \(\vec r = x\hat i + y\hat j = \left( {u\cos \theta \cdot t} \right)\hat i + \left( {u\sin \theta \cdot t - g{t^2}} \right)\hat j\);

Velocity after time t is \(\vec v = {v_x}\hat i + {v_y}\hat j = \left( {u\cos \theta } \right)\hat i + \left( {u\sin \theta - gt} \right)\hat j\);

Acceleration is constant ,\(\vec a = {a_x}\hat i + {a_y}\hat j = - g\hat j\)

Trajectory equation:

The path traced by the projectile is called the trajectory of the projectile.For displacement in the horizontal direction,\(x = {u_x} \cdot t\)
\(x = u\cos \theta \cdot t\)

For displacement in the vertical direction ,\(y = {u_y} \cdot t - \frac{1}{2}g{t^2}\)

\(y = u\sin \theta \cdot t - \frac{1}{2}g{t^2}\)

Substituting the value of t from eqn. (1) into eqn. (2), we get

\(y = u\sin \theta \cdot \frac{x}{{u\cos \theta }} - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}\)

\(y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)

\(y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\)

(R is the horizontal range covered by the projectile)

The equation of trajectory of the projectile is that of a parabola because the projectile covers a parabolic path.

Time of flight:

The displacement along vertical direction is zero for ground to ground projectile.

\(\left( {u\sin \theta } \right)T - \frac{1}{2}g{t^2} = 0\)

\(T = \frac{{2u\sin \theta }}{g}\)

Horizontal range:

The horizontal displacement of the projectile from the point of projection to the point it strikes the ground is called the horizontal range of the projectile.

\(R = {u_x} \cdot T\)

\(R = u\cos \theta \cdot \frac{{2u\sin \theta }}{g}\)

\(R = \frac{{{u^2}\sin 2\theta }}{g}\)

Maximum height:

Applying \({v^2} = {u^2} + 2as\) in the vertical direction between the point of projection and the topmost point, we get \({0^2} = {u^2}{\sin ^2}\theta - 2gH\)

\(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

Resultant velocity, at any instant t:

\(\vec v = {v_x}\hat i + {v_y}\hat j = \left( {u\cos \theta } \right)\hat i + \left( {u\sin \theta - gt} \right)\hat j\)

\(\left| {\vec v} \right| = \sqrt {{u^2}{{\cos }^2}\theta + {{\left( {u\sin \theta - gt} \right)}^2}} \)

\(\tan \alpha = \frac{{{v_y}}}{{{v_x}}} = \frac{{u\sin \theta - gt}}{{u\cos \theta }}\)

\(\alpha \) is the angle made by the velocity vector of the projectile with the horizontal at any time instant t

General result

For maximum range \(\theta = {45^ \circ }\) \({R_{\max }} = \frac{{{u^2}}}{g}\)

In this situation \({H_{\max }} = \frac{{{u^2}{{\sin }^2}45}}{{2g}} = \frac{{{u^2}}}{{4g}}\)

\({H_{\max }} = \frac{{{R_{\max }}}}{4}\)

We get the same range for two angles of projection \(\alpha \) and ( \({90 - \alpha }\ ).But in each of the two cases, maximum height attained by the particle is different.

\(R = \frac{{2{u^2}\sin \alpha \cos \alpha }}{g} = \frac{{2{u^2}\sin \left( {90 - \alpha } \right)\cos \left( {90 - \alpha } \right)}}{g}\)

If R = H i.e. \(\frac{{{u^2}\sin 2\theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

\(\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

\(\therefore \tan \theta = 4\);\(\theta = {\tan ^{ - 1}}4\)

Range can also be expressed as \(R = \frac{{{u^2}\sin 2\theta }}{g} = \frac{{2u\sin \theta \cdot u\cos \theta }}{g} = \frac{{2{u_x}{u_y}}}{g}\)

Change in momentum

Initial velocity \({{\vec u}_i} = u\cos \theta \hat i + u\sin \theta \hat j\)

Final velocity \({{\vec u}_f} = u\cos \theta \hat i - u\sin \theta \hat j\)

Change in velocity from the point of projection to the point where the projectile strikes the ground.

\(\Delta \vec u = {{\vec u}_f} - {{\vec u}_i} = - 2u\sin \theta \hat j\)

Change in momentum from the point of projection to the point where the projectile strikes the ground.

\(\Delta \vec P = {{\vec P}_f} - {{\vec P}_i} = m\left( {{{\vec u}_f} - {{\vec u}_i}} \right) = m\left( { - 2u\sin \theta } \right)\hat j = - 2mu\sin \theta \hat j\)

where m is the mass of the projectile

Velocity at the highest point of the projectile is \(u\cos \theta \hat i\). Change in momentum from the point of projection to the highest point =\( - mu\sin \theta \hat j\)

TOPIC-3

KINEMATICS

Motion under gravity

A body thrown vertically upwards or vertically downwards or dropped from a height will move in a straight vertical line.
If air resistance is ignored, the 'body will be subjected to acceleration due to gravitational force exerted by the earth, which is denoted by g. The value of g on the earth is 9.8\(m/{s^2}\) in the downward direction.
For small heights, the value of g is constant, we can use equations of uniformly accelerated motion.
We shall take upward direction as positive & down direction as negative, as our convention.

Motion of a particle projected vertically upward from the ground

Consider a particle projected vertically upward from the ground with velocity u.

Taking upward direction positive

\(u = u\)

\(a = - g\)

\(\therefore \) At any time t, velocity \(v = u + at\) and displacement \(s = ut - \frac{1}{2}g{t^2}\)

To find time of ascent (\({t_a}\)) , apply \(v = u + at\) between the point of projection and the highest point,

\(\begin{array}{l}v = 0\\u = u\\a = - g\end{array}\)

\(\therefore {t_a} = \frac{u}{g}\)

To find total time of flight (T). apply \(s = ut + \frac{1}{2}a{t^2}\) between the point of projection and the time instant when the particle is again at point of projection

\(\begin{array}{l}s = 0\\u = u\\a = - g\end{array}\)

\(\therefore T = \frac{{2u}}{g}\)

Time of descent (\({t_d}\) ) between the highest point and back to the point of projection is also \(\frac{u}{g}\)

\(\therefore {t_d} = \frac{u}{g}\)

For maximum height attained ( \({h_{\max }}\) ) apply \({v^2} = {u^2} + 2as\) between the point of projection and the topmost point,

\(\begin{array}{l}v = 0\\u = u\\a = - g\end{array}\)

\(\therefore {h_{\max }} = \frac{{{u^2}}}{{2g}}\)

The particle will return back to the point of projection with same speed as the speed of projection but in the opposite direction.
Motion under gravity is symmetric

Consider a particle projected from A. B is a point at height h and C is the topmost point.
In the above case speeds\({u_1}\) and \({u_2}\) are equal,\({t_{BC}} = {t_{CB}}\)\({t_{AB}} = {t_{BA}}\)

Motion of a particle projected downwards from height h above surface of earth

Suppose a particle is projected downwards from height h above the surface of the earth with speed u. To find the time taken by it to strike the surface of the earth, taking upward direction as positive,

\(\begin{array}{l}u = - u\\a = - g\\s = - h\end{array}\)

\(s = ut + \frac{1}{2}a{t^2}\)

solve the quadratic and get the positive value of t.

Motion of a particle protected vertically upwards from height h above surface of earth

\( - h = ut - \frac{1}{2}g{t^2}\), solve the quadratic and get the positive value of t.

Motion of a particle dropped from a height h above surface of earth

Solve using \({v^2} = {u^2} + 2as\) and \(s = ut + \frac{1}{2}a{t^2}\), taking

u = 0,

Velocity with which it strikes the surface will be \(\sqrt {2gh} \) and the time it will take to strike the surface will be \(\sqrt {\frac{{2h}}{g}} \).

TOPIC-2

KINEMATICS

Acceleration
Acceleration is defined as the rate of change of velocity with time.
Acceleration is a vector quantity.
SI unit of acceleration is \(m/{s^2}\) .

Average acceleration

Average acceleration is defined as the ratio of change in velocity over a time interval to the time interval.
If a particle moving along a straight line has velocity \({V_1}\) at an instant \({t_1}\) and velocity \({V_2}\) at instant \({t_2}\) ,then average acceleration during time interval \({t_2} - {t_1}\) is given by \({a_{avg}} = \frac{{\Delta v}}{{\Delta t}} = \frac{{{V_2} - {V_1}}}{{{t_2} - {t_1}}}\)

Summary of equations for uniformly accelerated motion.

\(v = u + at\) . . . Eq. I
\(s = ut + \frac{1}{2}a{t^2}\) . . . Eq. II
\(x - {x_0} = ut + \frac{1}{2}a{t^2}\) . . . Eq. III
\({v^2} = {u^2} + 2as\) . . . Eq. IV
\({v^2} = {u^2} + 2a\left( {x - {x_0}} \right)\) . . . Eq. V
\(s = \left( {\frac{{u + v}}{2}} \right) \times t\) . . . Eq. VI

Where

u - Initial velocity or instantaneous velocity at time t = 0

v - Instantaneous velocity at time instant t

a - uniform acceleration

s- Displacement at time t

t - Time instant

\({{x_0}}\) - Initial position or position at t = 0.

x-Position at time t

\({S_{{n^{th}}}}\)-Displacement in \({{n^{th}}}\) second

Equations for uniformly accelerated motion in vector form.

\(\vec v = \vec u + \vec at\)
\(\vec s = \vec ut + \frac{1}{2}\vec a{t^2}\)
\(\vec v \bullet \vec v = \vec u \bullet \vec u + 2\vec a \bullet \vec s\)
\(\vec s = \left( {\frac{{\vec u + \vec v}}{2}} \right) \times t\)

TOPIC-1

KINEMATICS

Kinematics :

Kinematics is the study of motion of physical bodies without going into the cause of the motion.
Kinematics deals with physical quantities like distance, displacement, speed, velocity, acceleration etc.

Motion and Rest :

Motion is a combined property of the object under study and observer.
If the position of the object under study changes with time, as seen by the observer, the object is said to be in motion from the frame of reference of the observer.
If position of the object does not change with time, as seen by observer, object is said to be at rest from the frame of reference of the observer.
Rest and motion of an object under study depend on the frame of reference of the observer. For eg. A book kept on a table may be at rest for all students sitting in the class. But the same book will be in motion, as seen by an observer on a moving bus. Thus absolute rest and absolute motion are meaningless.
In most cases, if attributes of motion of an object are given without specifying the frame of the observer, it is to be assumed that the object under consideration is being observed by an observer who is at rest with respect to the earth.

Position :

For a particle moving along a straight line, position of the particle can be specified with only one coordinate. A coordinate system is chosen by choosing some reference point as the origin. The origin is assigned the number zero. Most situations can be analysed by setting up an appropriate coordinate system. In order to do so, following are the essential requirements:
Choice of origin
Choice of coordinate axis
Choice of positive direction of axis. All these parameters constitute a reference frame. In any physics problem, the reference frame must be specified.
In the figure below, point O is the chosen origin, X - axis is the chosen coordinate axis and rightward direction is chosen as the positive direction.

Similarly, if the motion of a particle is 2 - dimensional or 3 - dimensional, the coordinate axes will comprise of x, y and z axes and position will include x, y, and z coordinates.

Displacement and distance :

Displacement is a vector quantity. It is the change in position vector. Distance is the total length of the actual path covered. Distance is a scalar quantity.
Suppose a particle travels from point A to point B as shown in the fig below along a zig -zag path in a finite time interval.

Coordinates of A are( \({x_1},{y_1}\) )and that of B are ( \({x_2},{y_2}\) ) . Position vector of A is\({{\vec r}_A} = {x_1}\hat i + {y_1}\hat j\) ,

position of vector of B is \({{\vec r}_B} = {x_2}\hat i + {y_2}\hat j\). Distance will be equal to the total length of the actual path covered by the particle. Displacement will be \(\vec S = {{\vec r}_B} - {{\vec r}_A} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j\).

The distance covered will always be greater than or equal to the magnitude of the displacement.
Displacement and distance are equal in magnitude in case the particle is travelling along a straight line without change in direction.
SI unit of distance and displacement is meters.
In simple language, displacement can be said to be the shortest line joining the initial and final positions of a body in motion, irrespective of path followed and it is directed from initial position to final position.
Change in position vector is displacement and change in displacement vector is also displacement.

Average speed :

\(AvgSpeed = \frac{{TotalDis\tan ceTravelled}}{{TotalTime}}\) ,We define average speed of a particle as the ratio of the total distance travelled to the total time taken.
SI units of speed is m / s

Instantaneous speed :

Speed of a particle at a particular instant is called instantaneous speed.
The speedometer of a vehicle indicates the instantaneous speed. The speedometer reading on a crowded city road continuously changes, indicating instantaneous speed is continuously changing.

Velocity :

Velocity is defined as rate of change of displacement with time.Velocity is a vector quantity. SI unit of velocity is m/s.
\(AvgVelocity = \frac{{TotalDisplacement}}{{TotalTime}}\) , Average velocity is defined as the ratio of the Total time displacement covered to the total time taken.
Just as distance is always greater than or equal to the magnitude of displacement, average speed is greater than or equal to the magnitude of average velocity. Average speed and the magnitude of average velocity are equal when particle is travelling in a straight line without change in direction.

Instantaneous velocity :

Suppose a particle moves from position x at time t to position \(x + \Delta x\) at time \(t + \Delta t\) . Then, the average velocity of the particle over time interval \(\Delta t\) is \(\frac{{\Delta x}}{{\Delta t}}\).
Making \(\Delta t\)infinitely small,\(\frac{{\Delta x}}{{\Delta t}}\) gives the velocity of the particle at instant t and can be written as\(v = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta x}}{{\Delta t}} = \frac{{dx}}{{dt}}\),where v is the instantaneous velocity of the particle at time instant t.
The magnitudes of instantaneous velocity and instantaneous speed are always equal.

Uniform motion :

Motion of a body in a straight line with uniform velocity is called uniform motion.\(\frac{{ds}}{{dt}} = v\), but v is constant in uniform motion.\(\therefore \int {ds = } \int {vdt = v\int {dt = vt} } \)

\(\therefore \) s = vt,where s is the displacement, t is the time interval, v is uniform velocity.

Uniform motion can also be said to be motion in which equal displacements are covered in equal intervals of time, however small the time intervals may be.

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