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0

When the rocket sits on the launch pad and the engine fires up, an enormous amount of energy is expended moving the exhaust gases and heating the air. No work is done to move the rocket, but that doesn't mean no energy is expended. But the kinetic / potential energy of the rocket are not changing (actually the potential energy of the rocket becomes less ...


0

Useful work=0, power DRIVING the engine =0, ALL useful power lost in the form of wasteful energy, example: exhaust gases, heat, sound and all that..


0

In your first example, a rocket on a launchpad that has just started its engines, the equation is exactly correct. The motor is generating huge force, but it has not yet moved the vehicle, so no work or power is being generated. In reality, that moment of no work or power lasts only a tiny fraction of a second. The power moves the rocket and then you can ...


1

I will try to explain this in simple english. Since the number of steps are infinite, then the distance you travel at each step is infinitly short. As such, the time teaken in each step is close to zero but not zero. Therefore the total time in taking this infinite number of steps does not necessary equal to infinite. Mathematically the total time take is ...


17

You have scrambled across one of Zeno's many paradoxes - the so-called Dichotomy paradox. The resolution lies in the fact that sum of terms in an infinite series do not necessarily add up to produce an infinity, so the basic premise is flawed. In particular, the series $$S = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + ...


0

Isn't it just a change in the reference frame and therefore the forces are equal? Update: Suppose your initial conditions: the ball ($m_1$) hits the man ($m_2$). Its velocity is equal to $u_1$. As you've said: $\m_1(v_1-u_1)/t$=$\-m_2(v_2-u_2)/t$ Now consider that an observer is moving at the velocity $u_1$. The ball looks static and the man looks like he's ...


0

You're right that the book's answer seems to have the wrong units; most likely the author meant to say $p_{max}^2$ in place of $p_{max}$. Also, both instances of $2m$ in the book's formula should really be $2m_p$. Assuming these two typos are resolved as just mentioned, it's clear what error the book is probably making. The book seems to be assuming that ...


1

An actual example in which there is a non-zero change in acceleration, that is, jerk, occurs is a spring. A spring's motion is described by a sinusoidal function. The derivative of a sinusoidal function is just another sinusoidal function. As a result, you can differentiate such a function infinitely many times, and will never have a derivative that's 0/a ...


12

http://wordpress.mrreid.org/2013/12/11/jerk-jounce-snap-crackle-and-pop/ Speaking derivatives to time: First position $x$, then velocity $v=x'=\frac{dx}{dt}$, then acceleration $a=x''=\frac{d^2x}{dt^2}$, then jerk $x'''=\frac{d^3x}{dt^3}$, then jounce/snap $x''''=\frac{d^4x}{dt^4}$, then crackle $x'''''=\frac{d^5x}{dt^5}$, then pop ...


7

Yes. Rate of change of acceleration is called jerk. Yes its dimensional formula is $[M^0, L^1, T^{-3}]$. Similarly one could also define higher time derivatives of acceleration if required for a particular problem.


3

Yes. usually we name them $a'$ . and there can be even a speed of my $a'$ that I can call that $a''$ and it goes on like that. it is only used it real life calculation that the calculation should be very precise like rocket science. and the equation of displacement (with a constant $a'$) will be : $$x =\frac16 a't^3 + \frac12 a_0t^2 + v_0t+x_0$$ (EDIT: ...


0

UPDATE: While I was typing the answer below, I see that you came to the same conclusion as I do in my answer below. From your $\dot \theta$ equation, it appears that a positive $x$ component of force acts to increase the rate of change of the angle when $0 \lt \theta \lt \pi$. But, that can't be correct. If you set $g$ and $b$ to zero and launch the ...


0

There are so many ways you can do that: put some sand on the floor (Like Olympics!) Lay some papers on the ground, and put your thrown object in the ink. then when it lands it will make a mark (actually this method were used my famous physicist Galileo Galilei ) pour down some Bean or nuts on the ground, then when object hit them it will make a mark ...


0

You can draw a few equidistant lines on the paper that is on the floor and make a video of the falling projectile.


1

Your question is "Is there an infinite series of higher derivatives of position for this to work?" Answer: No. Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one. From the question: "A change is velocity is acceleration, so the value of the ...


2

You must solve for the objects initial velocity first: $$ v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s} $$ With this adjustment you should find the correct answer.


4

If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$? Using: $$v=u+at$$ you will find that the object was not released from rest...


0

The notation being followed here is cranky and misleading. The intended meaning for the exercise can be imagined this way - you are given a table which gives the instantaneous velocity $v(t)$ of the car at some specific times, all separated by fixed time increments of 0.1 units. As you mention in the question, you can obtain overall displacement by ...


0

I found my mistake. Maybe somebody else will make this dumb mistake that I made. I discretely assumed in the last set of three equations that $$E'^2 = E_1^2+E_2^2$$ Which is totally false. I should I have noticed that $$E'^2 = (E_1+E_2)^2$$ Working that out gives me the right answer.


1

As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to ...


2

It's a very strange question. In theory a ball is thrown in the shape of a parabola. You can get the graph of the thrown ball with this equation: $$y=\frac{-gx^2}{2V^2 \cos^2 (a)} + x\tan a$$ where $a$ is the angle at which the ball is thrown, $g$ is the gravitational acceleration, and $V$ is the initial velocity. If you want to calculate the initial ...


1

The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ...


1

A nice way to compare both is to invoke the definitions: $${\vec a}_{\rm avg} = \frac{\Delta {\vec v}}{\Delta t}$$ and $${\vec a}_{\rm inst} = \lim_{\Delta t \rightarrow 0}\frac{\Delta {\vec v}}{\Delta t} = \frac{d{\vec{v}}}{dt}$$ Graphically, and if you consider change over an infinitesimal time period $\Delta t \rightarrow 0$, the same definition ...


1

The instantaneous acceleration is the time derivative of the velocity vector: $$ \vec{a} = \frac{d\vec{v}}{dt} $$ If the velocity is changing then the acceleration will be non-zero.


1

If only the forces of gravity are present, all objects fall at the same rate. This is what one calls equivalence principle. In classical mechanics it shows up in the force law for two particles of gravitating mass $m_G$ and $M_G$, where $M_G$ shall denote the earth's mass. $$ m_i \cdot \vec{a} = -G \cdot \frac{m_G \cdot M_G}{|\vec{r} - \vec{r} '|^2 } \cdot ...


1

Here is an extremely simple explanation: Force = Mass x Acceleration Force / Mass = Acceleration Mass x Acceleration due to Gravity / Mass = Acceleration Acceleration due to Gravity = Acceleration For further intuition, consider this: The greater the mass, the greater the inertia. The greater the inertia, the greater the difficulty to accelerate the ...


-1

Without air resistance all objects are accelerated with gravity $$g = 9.81 \frac{m}{s^2}$$ Only the air causes a "slower" acceleration. This effect depends from the density and shape.


4

Well, if you have $2{\pi}r$ and you know the time in which you traveled it, then you could find speed. Nothing prevents this.


1

No, the baseplate will not move if the two motors are applying torques in such a way that the rotors remain at rest. To analyse this problem, it is best to consider the forces on the three parts (base disk and two rotors) individually, using free body diagrams. There are two motors, each connecting a rotor to the base plate, and the effect that each motor ...


-4

This may follow Einstein's equation and may appear to fit into classical picture, but this is taking place between two particles obeying Fermi Dirac statistics. Hence, it is a quantum phenomenon, and the direction of photon emission is arbitrary, as required by the fundamental assumption of quantum mechanics.


20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


1

Complex mass means gravitational mass + i.lambda.higgs mass. The weak coupling constants are all proportional to (higgs) mass with Higgs vacuum value (=246 GeV) as the proportionality constant. Thus mass necessarily becomes a complex number. The real part produces attractive gravity forces and the imaginary part produces repulsive "weak" forces. The factor ...


1

Yes it would be greater than 360°. Because what you basically do is you calculate $\frac{TraveledDistance}{circumference}$ . In your example this quotient is 0.5, but it can be also larger than 1.


1

Yes, absolutely. This is in some sense parallel to the distance v displacement question in linear motion. It's true that, at the end of it all, the angle difference between the final and initial configuration can't be more than 360 degrees or $2 \pi$ radians or whatever. But the angle that was traveled through--which is really the important thing for your ...


0

Yes. Two particles at a given position, one moving and the other at rest are having the same kinematic state. But dynamic state takes in the velocity. When you need to specify both position and velocity - it becomes kinetic state. After all kinematics + dynamics = Kinetics.


4

Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can. No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third ...


0

It's like that because the rotation without sliding is caused by static friction, not the kinetic one. Thus $T \le f N$. On the other hand, while if the cylinder is sliding, we take the kinetic friction.


1

Yes you can use the motor to slow and stop the disk, but you have to have (1) Alignment of the motor and its load angular momentum with the disk's angular momentum, and (2) enough motor load (moment of inertia) and motor speed to fully transfer the disk's angular momentum into the motor and load. The process of momentum transfer is done all the time on ...


1

If the motor is fixed on a moving rotating disc there can only be one result. If you attach the rotor from the engine to the platform, then either the rotor or the platform (or the engine) will tear itself apart. It doesn't matter if that platform is moving or not. See my drawing; if I glued the rotor disc to the platform, it cannot rotate. If it is forced ...


3

Have a look here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function There are functions which are identically zero for negative arguments, non-zero for positive arguments and smooth everywhere.


26

Yes, if an object is stationary, then starts moving in the positive direction, all derivatives of the position can become positive when the object starts moving. But so what? It's not a paradox. It's just a mathematical statement. All the derivatives are positive. No big deal. It's hard to tell why you find this confusing. You might be thinking that there ...


1

You are differentiating with respect to $t$, and by the chain rule $${d \over dt}(f(x)) = {df \over dx}{dx \over dt}={df \over dx} \dot{x}$$ You have just missed the $\dot x$ part resulting from proper application of the chain rule!


14

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV). ...


5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...


0

which should not be possible. Indeed, uniform coordinate acceleration $a$ is inconsistent with special relativity however, uniform proper acceleration $\alpha$ is consistent. The proper acceleration is the acceleration of the object according to an attached accelerometer. For 1D motion, the relationship between $\alpha$ and $a$ is given by $$\alpha = ...


13

Have a look at the article by Phil Gibbs on the relativistic rocket. This describes the motion of a rocket that is accelerating with a constant acceleration. In this context constant acceleration means the crew of the rocket feel a constant acceleration. Technically the rocket has a constant four-acceleration. Anyhow, the velocity of the rocket as observed ...


1

Look at sparknotes.com/physics/specialrelativity/dynamics/…, you can see $dE/dx=F$ - if your force is constant, it is the energy that increases constantly. $E=\gamma(v)m_0c^2$, you can deduce the $v$. Beacause of laziness I used mathomatic, and it gives me something like this: $v=c\sqrt{1-\frac{m_0 c^2}{(F\cdot x + m_0 c^2)^2}}$ If you check it for x=0 and ...


6

Is there some other formula ... which ... does not allow the speed ... to surpass the speed of light? That would be the equations of special relativity mentioned by sahin in a comment. Image from Loodog? Another factor you have to take into account with classical mechanics is to work out how a constant force can be applied to your object over 11 ...


0

When a projectile is launched at an initial velocity $V_0$ and an angle $\theta$, the projectile has both a horizontal ($x$-) component and vertical ($y$-) component, as shown below: From the diagram, you can see that $v_x=v_0 \mathrm{cos\theta}$ and $v_y=v_0 \mathrm{sin\theta}$. If we try to solve for the horizontal range of your projectile, we can use ...



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