# Tag Info

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"My question is, why isn't the vertical acceleration equal to $g+Ca$?" I find a very nice explanation on internet. You are missing some important facts of the situation involved. Fact 1: There are only two forces acting on stuffed animal throughout its ride these are : (a) Normal force $N$ and (b) The gravity $mg$. Fact 2: The total force acting on ...

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You can't add centripetal force and gravity, because they aren't the same type of concept. Gravity, or tension in a string, or friction between a tire and the road are real sources of force. They are there because of some physical structure and/or interaction. Centripetal force is a calculation of the force that is needed to achieve some required ...

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According to the problem, the stuffed animal has been dropped. That means the only force on it is gravity, so its acceleration after being dropped is $g$. It doesn't "remember" being dropped from a Ferris wheel - it only knows its current position and velocity; nothing else.

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I would divide the circle into 8 pieces(or into 4 pieces if you like) and note that in each segment what is the limiting speed. The rest should be straight forward. And by straight forward I mean it would basically transform into how long does it take to walk around a certain square with a constant speed $v$.

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Hint: let the particle position function be $\theta(t)$ In terms of $\theta(t)$ and $\frac {d\theta}{dt}$, calculate $v_x$ and $v_y$. Applying the upper limit gives $\frac {d\theta}{dt}$ as a function of $\theta$. Now solve $\int_0^{t'} \frac {\ dt\ }{\frac {d\theta}{dt}}=2\pi$. It is easier to do $\theta$ from $0$ to $\frac {\pi}4$, then multiply by ...

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You are considering an object (cannonball) of mass $m$. Once it is launched at speed $v$ at an angle $\theta$ from the horizontal direction, the cannonball feels only one force, its own weight $m\vec g$. $g$ is the gravity acceleration ($g\simeq9.81\,\mathrm{m.s^{-2}}$). Newton's equations state that $m\ddot{\vec r}=m\vec g$. The mass can be removed from ...

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They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

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Warning : I absolutely don't know nothing about the subject. If you are a french locutor, applyed notion of kinematical inversion (and dynamical inversion) to the Tottori earthquake are explained in this thesis (Sara Di Carli), see page $4$ for a abstract, and pages $11-29$ for details. The Haskell Model(1964) seems to be the first model. In the ...

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I believe this decomposition has a specific name "Cauchy–Helmholtz theorem (regarding the decomposition of the velocity of a point within an infinitesimal continuum particle)". You can read about this decomposition in the wikipedia page on Strain rate tensor, sections about symmetric and antisymmetric parts and shear rate and compression rate (though this ...

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Matrices It's a general property of square matrices (and 2-tensors). Any matrix $M_{ij}$ can be decomposed into a part containing the trace, and a part that is traceless. So we begin $$M_{ij} = \frac{1}{N} (\operatorname{tr} M) \delta_{ij} + \left( M_{ij} - \frac{1}{N} (\operatorname{tr} M) \delta_{ij}\right)$$ I hope that much is evident. The first part ...

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This is a variation of an old chestnut. The simplest analogy is that of a shower stall: Suppose you need to walk thru a shower (at least at my old high school, they still had a long row of showers sans walls we had to get thru). What strategy would you take to stay as dry as possible? Clearly the answer is to get out of the rain as soon as possible. ...

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You have the right ideas, but you don't dare to put them in maths ! Momentum conservation is vectorial. Here, you have a 2D system, so let's write it with 2 component vectors. I will denote $\vec{P}_{i/f}$ the initial and final momentum respectively. So the momentum conservation yields $\vec{P}_i = \vec{P}_f$ I choose to represent the $x$-axis as the ...

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Momentum conservation says $\vec{p}_A = \vec{p}_B + \vec{p}_C$, we can split this in components: $$p_{A,x} = p_{B,x} + p_{C,x} \\ p_{A,y} = p_{B,y} + p_{C,y}$$ Some of these momentums are $0$. Which one? For the block B - the only one with a motion not parallel to one of the axes' - you have to use trigonometry. To get the velocity, once you have the ...

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You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

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You should realize that the first equation you write gives you the value of $a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore $F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$Law}} ma \quad \quad \to \quad \quad a = -g\mu$ Now you know the acceleration, you can find how ...

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Remember the law $F=ma$; you already know the force from friction $F_{\mu}= -mg\mu$. Hence you can get $a=-g \mu$, and one of your unknowns is gone.

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It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)

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From this data, you don't actually know how the speed varies between steps. If you have more information about the acceleration then you could change the model from this but I would propose the following... Assume a linear change in the speed between steps, you could take a simple graphical approach. Plot the speed (in km/s) on a vertical axis, against the ...

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Well, this is an easy problem in kinematics. "The rule" for solving it is drawing a graphic of the (differences of) time versus the speed. You will get some points. Join these points to obtain a trapezoid histogram. So, what do you think it is the area beneath this curve? It has the dimensions of a length, so... Now, you want an average speed, that is a ...

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Hints: $30 \,km \, s^{-1}$ is about $\dfrac{1}{10000}$ times the speed of light and more precisely $\dfrac{30000}{299792458}$ times Distance of $4.2$ light years Number of years neded is ...

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1 year is approximately $y = 3.15 \times 10^7 s$, and light travels at $c = 3.00\times 10^5 km/s$, so convert the speed of the space probe in km/s into ly/s and then into ly/y. Note that $\frac{km}{s} = \frac{km}{s} \left(\frac{3.15 \times 10^7s}{1 y}\right) \times \left(\frac{ly}{3.00\times 10^5 km/s \times 3.15 \times 10^7s}\right) = \frac{1}{30.0\times ... 0 Note that the speed of the criminals relative to the river ($27\,\mathrm{mph}$) is larger than the speed of the stone relative to them ($11\,\mathrm{mph}$). Then, the resulting velocity would be expected to be in the negative direction 1 All correct. Your values for$ \dot{x}_P$and$ \dot{x}_C$are negative because you defined$ \dot{x}_Rto be positive (both boats are moving against the river). No problem there, this is completely up to you, but to satisfy the requirement of the question you are required to state that motion of the river has been designated by you as positive direction ... 0 Start with \begin{align} \alpha(t) = \frac{d\omega}{dt}(t). \end{align} Integrate both sides fromt_i$to$t_f; \begin{align} \int_{t_i}^{t_f}dt\,\alpha(t) = \int_{t_i}^{t_f}dt\, \frac{d\omega}{dt}(t) = \omega(t_f) - \omega (t_i) \end{align} The second equality on the right follows from the fundamental theorem of calculus which basically says that if ... 0 First let us convert to SI units. U = 50km/hr *(1000/3600) V = 80km/hr *(1000/3600) Since 1km=1000m,1hr=60minutes*60seconds. Now we use the formula V = U + at a = (V-U)/t a = (80-50)/(20*3.6) a = 0.4167 m/(s^2) a = 1.5 km/hr 1 Changing the "wind" to a "current" in an attempt to save the question, the OP's solution is still not quite correct. By subtracting the current's west component from the boat's overall, west velocity over the ground, the OP has correctly found the west component of the boat's velocity through the water. However, he has forgotten that the boat must cancel ... 3 This question can't be answered because important information is not given. We have no idea how much influence, if any, the wind has on the boat's velocity. In the limiting case, if the boat's wind resistance is 0, then the wind has no effect. The question also asks what the boat would do in still water, but we were never told how still or not the water ... 0 For each case think about the total kinetic energy and number of cars that are available to absorb all of it during impact and it should be apparent which case has more energy (damage) per car. (crushing, deformation, sounds are evidence of energy conversion, so a tough rigid wall doesn't absorb any) A more challenging problem to intuition is (A) two cars ... 1 It's not missing, it's in the\ddot{R}(t)$matrix. It doesn't show up on its own when you do the calculation with matrices instead just vectors however. Vector equations The first equation should be written $$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$ (just reverse where you have$A$and$B$) since given the position in frame$B\$ you ...

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You may be overthinking it. In general when you need to solve for coefficients you first need to ask what relates the coefficients? In this case it doesn't seem you've written precisely what equation you want. In your post, you've written the parabola in terms of x. Do you mean you want y position as a function of x position? This isn't the standard way to ...

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There's a more straightforward calculation. In order to travel eastward, the plane's velocity must have a southward component of 60kph to cancel the wind from the south. Since the plane's speed is 100kph, we have the eastward component (in kph) is just: $$v_E = \sqrt{100^2 - 60^2} = 80$$ Thus, you are correct; the time required to travel 189km eastward ...

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