# Tag Info

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Apply conservation of energy and find the velocity. KE = -U 0.5Mv2 + 0.5mv2 = mgh –Mgh (M is the heavier mass) 0.5v2 (M+m) = (M-m)gh Calculate v and then finally apply v2 =u2 +2as (s is displacement)

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The mechanism of a speedometer determines the travel speed of the vehicle indirectly by measuring the rotation speed of the transmission output or wheels. One turn of the wheel causes the vehicle to cover a distance equal to the circumference of the wheel, so there is a direct relation between the wheel's angular speed and the vehicle's linear speed; ...

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Yes. With low air pressure, the axle is closer to the road: the radius of rotation is reduced. Some cars have systems that warn you when your tire pressure is low. The way they work is by measuring the rotation rate of the tires. If the rotation rate exceeds a certain limit, the dashboard light glows. With under-pressured tires, your speedometer will ...

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No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ...

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Start with: $${\bf F}_{\rm net} = m\,\ddot{{\bf r}}$$ with: $${\bf r} = x(\theta)\hat{\bf x} + y(\theta)\hat{\bf y}$$ and a dot denoting a derivative with respect to time. Then using the tangent vector: $$\hat{\bf T} = \frac{\frac{{\rm d}{\bf r}}{{\rm d}{\theta}}}{\left|\frac{{\rm d}{\bf r}}{{\rm d}{\theta}}\right|}$$ to define the normal vector: $$\hat{\bf ... 0 While they have the same speed, their velocity keeps changing direction so that one particle always moves towards one of the others. Now, as no particle is heading away in a straight line from any other particle, the relative velocity between any two particles is one that favours them to keep getting nearer. The hexagon formed by the particles will therefore ... 1 Sure, easily. We don't have to invoke any specific force like friction or gravity--just imagine a block at rest on the ground, and forces act either to the right or to the left. I could have four forces each of F = 1N on the right, and one force on the left of F = 4N, and that would balance. The question is akin to asking "can we add up five different ... 1 If a object weighs 6 lbs and there are 5 1 lb forces pushing up, then it will not move. The ground pushes up, causing a zero net force. 0 I am sorry to say this but surface area is not proportional to the force of friction. Friction is given by the coefficient of friction (static or kinetic) multiplied by an objects normal force. What may ne useful is the theory that the coefficient of friction is equal to the tangent of the angle at which an object starts to move on a plane. Also something ... 0 The rotational motion theory, the kinetic motion theory, the friction theory, and added to further too in special conditions 2 As always, a diagram clears things up quickly: We are going to assume that there is no lift on the arrow (which is wrong - arrows do not fly like regular projectiles but that is not the point of this question). From thie diagram we can see the approach we need to take (I am not going to show the details of the steps, just give you some direction). the ... 1 No its SPEED will remain same, velocity has a direction, it will change, both while entering and leaving. On the curve, frictional force will balance the centrifugal force. Assuming constant coefficient of friction throughout the track and constant force provided to engine. 3 First of all, you need to be careful when saying the train does not accelerate. In Physics, acceleration is the rate of change of velocity. As velocity is vector, if the velocity of the train changes direction, as it does on the curved portion of the track, then the velocity is changing over time (changing in direction). Therefore, there must be an ... 8 The train does accelerate as it goes around a curve. Velocity is a vector, with magnitude and direction. Speed is the magnitude. The train changes direction. Acceleration is caused by a force. If the force causes a change in direction with no change in speed, it must be perpendicular to the direction of velocity. For example, A planet in a circular orbit ... 0 As it turns out, adding mass (while keeping the dimensions of the car fixed) will make the car go faster. This sounds contrary to what all physics students are taught, but the reason is while friction scales with mass, air resistance doesn't. That's why a thirty-foot rock will fall faster than a thirty-foot parachute in an atmosphere. It's also why the ... 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 ... 2 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} + ...

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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 ...

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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 ...

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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 ...

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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 ...

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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.

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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: ...

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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 ...

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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 ...

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

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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 ...

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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.

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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...

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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 ...

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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.

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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 ...

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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 ...

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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 ...

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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 ...

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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.

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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.

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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.

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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 ...

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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.

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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 ...

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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 ...

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