# Tag Info

## New answers tagged velocity

0

The second law states that the net force on an object (F) is equal to the rate of change (that is, the derivative with respect to time) of its linear momentum p (p = m*v, where v is the velocity) in an inertial reference frame. When F is zero, due to the above equality, dp/dt is also zero. This means that d(mv)/dt = 0 --> mv = const --> v = const. ...

2

Are Lorentz transformations more adequate representations of motion, than the more intuitive velocities? Yes. The non-associativity that bothers you simply arises because there is no group of three dimensional boosts. Confined to one dimension, boosts form a rather lovely one parameter subgroup of the Lorentz group $SO^+(1,3)$. So everything "works", ...

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Making what Terry pointed out more explicitly: Since there are no external forces on the system then momentum is conserved. This means that: $$m_\text{boy}\times v_\text{boy}=m_\text{board}\times v_\text{board}.$$ Edit: To clarify, $v=\Vert\boldsymbol v\Vert.$ And the speeds are measured with respect to the system's center of mass.

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Because these particles would be photons- photons have a momentum of h/wavelength, so what would happen is that due to momentum conservation, one would have 2 times the momentum, and have 1/2 the wavelength (possibly absorbing the other). It would have twice the frequency, so twice the energy. Therefore, the photons speed remains the same.

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In theory, no - nothing detectable, that is. As Olin said, there would be some more wind resistance with a longer line, but this wouldn't be a huge factor. The change in potential energy would be the same for both harnesses, because the change in height for both would be the same. Since the kinetic energy of the rider is originally from the original ...

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Only the weight force acts in the beggining, otherwise the block does not slide. Decomposing the weight force : P = Pcos(θ) + Psin(θ) ;Pcos(θ)=48N; Psin(θ)=14 Normal Force=N; N=Pcos(θ); Fat= μk.N=0.3333x48=16; Find the total work of the forces and use the work energy theorem to find the velocity at L=5 Work of weight=Psin(θ)xL=14x5=70J; Total Work= 70J ...

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It's a little bit hard to follow your reasoning. Let me try to give the method I would use - simple balance of energy. If the car reaches a velocity $v$ after time $t$, the power of the engine will have been used mostly for five components: Kinetic energy of car: $\frac12 m v^2$ Rotational kinetic energy of tires: $4 \times \frac12 I_t \omega_t^2$ ...

2

The only thing that the device "knows" when it is hit, is the force with which it gets hit, and the duration of that hit. Transfer of momentum $m\Delta v = F\Delta t$. So what matters is the momentum of the hammer's head - or more specifically, the momentum that you are able to transfer. Ultimately it comes down to giving the most momentum to the head of the ...

1

If i understood your question correctly, you seem to not know the displacement of the car but you still need to increase the velocity of the car. Velocity of the car has to be expressed in terms of a quantity. I suggest you use velocity as a function of time, $$v_f = v_i + at$$ $a = g \sin \theta$ as you told correctly. So $$v_f = v_i + g \sin \theta ... 1 You can cast your problem in terms of what is the velocity of the car at any instant of time after it started. The answer is v = a_{\rm eff} t from the first kinematical equation. In the same time, the car would've traveled a distance s = (1/2)a_{\rm eff} t^2 from the second kinematical equation. The other option is - if your inclined plane is of any ... 0 You should integrate the velocity to get the distance$$s(4)=x_0-x(4)=\int_{x(0)}^{x(4)}dx=\int_0 ^4vdt=\int_0 ^4( 2 + 3t^2 )dt=2t+t^3|_0^4=2*4+4^3 = 72 \,\text{m}$$The initial position doesn't matter since de distance s is the difference between the initial position and the position at a given time (in your case, t=4). I don't know what is that ... 0 If you have acceleration as a function of speed use the first equation. If you have speed as a function of position use the second equation.$$t(v)=\int_{v_1}^{v}\frac{1}{a(v)} \, \mathrm{d}vt(x) = \int^x_{x_1}\frac{1}{v(x)}\, \mathrm{d}x$$It seems you have acceleration (force) as a function of speed with a(v) = C_0 + C_1 v. The first integral then ... 3 Wormholes don't just exist by themselves, they have to be created by a form of matter called exotic matter (which almost certainly doesn't exist, but let's gloss over this). To construct a wormhole you need to gather up some exotic matter and arrange it in a particular configuration. So if the exotic matter is moving the wormhole will move along with it. ... 2 There does not need to be a force on an object for it to move, only for it to accelerate, as can be seen from Newton's second law:$$F=m \cdot a$$I think your confusion arises from forgetting to take into account frictional forces. In practice, a moving object will slow down because of friction: the net force is not zero! Therefore you need to apply an ... 0 That is because when you apply force on any object momentum is transferred which means velocity is transferred to an object at rest. now we know rate of change of velocity is proportional to force applied so if velocity does not change, so rate of change of momentum is zero and so force applied is also zero and hence the object will continue to move. however ... 3 If the velocity decreases by a little bit, it goes into an elliptical orbit with the apogee the same as the original orbit. The orbit will be stable if no other changes happen. Only when the velocity decreases to the point where the elliptical orbit intersects the Earth's atmosphere will the object crash into the Earth. 3 The flaw in your reasoning is that you believe that pressure and velocity are said to be inversely proportional, p=K/v, by the law. Instead, Bernoulli's law says that these two variables are in inverse relationship (but not "proportionality") which means that one of them is a decreasing function of the other, and you wrote what the function is. p=K/v is ... 2 babou, this is really an extended comment to your own answer. I think your answer is pretty much spot on, but I would simply the reasoning a bit. What kills you is when the distance between different parts of your body changes. You give the example of the separation between your head and feet changing by more than 5% (something exploited by hangmen over the ... 1 Try this method for solving the majority of Momentum/Impulse Problems with these two simple equations. Michel Van Biezen on Youtube teaches this method. Sure beats m1v1 + m2v2 (initial) = m1v1 + m2v2 (final). Given: Girl- Mass of 45.5kg; Velocity +1.47m/s Plank- Mass of 140kg Questions: QA Find the VELOCITY of Plank (that is girl + plank) on ice. QB ... 2 It really is the stress that kills you. Velocity, acceleration & jerk are all fine as long as they are spatially uniform. It is a postulate of general relativity that you can not even detect acceleration due to a uniform gravitational field, no matter how intense. However if the there are spatially non uniform forces applied to your body, then there will ... 2 If you fly head first into a wall, at the moment of impact the top of your head will accelerate very rapidly while the est of your body continues to travel at a constant speed until they make contact with the wall. Clearly having different parts of your body rapidly accelerate in different direction will lead to some very large forces on various parts of ... 5 Acceleration does not kill us any more than speed. If your head and feet do not move at the same velocity long enough, whatever the cause, you are in trouble. Velocity does not kill us when the whole body has the same velocity. Similarly, I doubt acceleration kills us when all parts of the body accelerate, but without having to transmit forces. It is said ... 2 The problem is, there isn't just one way in Newtonian mechanics to kill someone. You can cause as little or as much acceleration as you want. A few things worth analyzing are: Whiplash. If you're under constant acceleration and you reach a steady state (and aren't dead yet), a change in acceleration (jerk) could cause a whip effect. The Earth-Sun system. ... 0 The very definition of a frequency is the rate at which something occurs over time. Therefore by shortening the time interval in which something occurs you are increasing its frequency, and by increasing the time interval you decrease the frequency. Hence, Frequency is inversely proportional to time. 1 First, the context is a function of time that is periodic which means that it is repetitive with repetition period T.$$g(t) = g(t + T)$$So, if one sampled the function every T seconds, one would get the same value each time. Now, we have the period of time T which tells how long it takes for the signal to go through one cycle. The inverse ... 3 To answer the question "same wavelength, different frequencies, which arrives first?": naturally, the one with biggest speed, which is proportional to frequency AND wavelength according to the formula:$$v = \lambda f$$So, for the same wavelength \lambda, the one with bigger frequency f will have bigger speed v, thus arriving earlier. For the ... 3 The theory of relativity tells us what is the answer to this question. See Wikipedia, as John Rennie recommends, http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_case:_parallel_velocities If in the formula$$v_{rel} = \frac{v_1 + v_2}{1 + v_1 v_2/c^2}$$you set v_1 = v_2 = 0.9 \, c, you get v_{rel} = 1.8/1.81, i.e. slightly less than ... 1 Hint to the question (v2): For a velocity-dependent force {\bf F} (such as e.g. the Lorentz force), the relationship between force {\bf F} and potential U is$$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.  See e.g. Goldstein, Classical Mechanics, Chapter 1. See also e.g. this and this Phys.SE ...

2

Different parts of the blades have different speeds, but all parts of the blades have the same instantaneous angular speed, that is all parts travel through the same angular displacement in a given amount of time. This is always true for a rigid body (when the angular speed is measured about the rotation axis). As ACuriousMind pointed out in a comment, you ...

1

Since gravity is constantly acting on both spheres once released, both objects have constant downward directed acceleration and, thus, have zero velocity only for infinitesimal time. That is to say, the velocity of both objects is not constant at any time between their release and their impact with the ground. Assuming the objects have different upward ...

0

I will offer a very short and simple answer. The object does not get more mass, as no matter is added on it. However the object at different speeds acts like it has different mass because it has more energy and bigger momentum. When the speed is small compared to that of light's the difference is negligible. Nevertheless when it's speed approaches the speed ...

2

First, to clarify: A body does not "gain mass" upon acceleration, it "gains mass" at high speeds. That is, whether or not the body's velocity is changing is not relevant, only its speed relative to the observer is important. That being said, a body doesn't actually "gain mass" when it moves at a high velocity. The mass of a body is always the same, and is ...

0

David Mermin came up with a beautiful thought experiment, which can be found in his book Boojums all the way through: Suppose that an observer in an inertial frame $S$ watches a train move at constant velocity $v$. The length of the moving train, as measured in $S$, is equal to $L$. Now, at time $t=0$, a photon and a massive particle (with speed $w < c$) ...

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Your reading is correct, aside from the "Wind Resistance". It is a "drag co-efficient" and is a "fudge factor" so that $A\,C$ is the area of the "effective face" that the falling object presents to the fluid. It measures how much different the effective area is from the actual cross-sectional area. Another way to think of it is as a "tweak" factor to mop up ...

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