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3

Tl;DR: It is mainly do to the skier's positioning on the ski and how being thrown off-balance by the avalanche affects it. Concerning the first part of the first question: 1) If all free falling objects accelerate at the same rate (this was on a fairly steep mountain section), why did we get "trapped" into the avalanche, when our acceleration already ...


3

It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...


2

The acceleration should be $$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$ with $r$ as the height above the center of mass and the Schwarzschildradius $$r_s = \frac{2\cdot G\cdot M}{c^2}$$ The force to hold the ball at rest is $$F=m\cdot a$$ As one can see it now takes an infinite force and energy to keep a body at a fixed height when $r=r_s$.


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During the acceleration phase the object's movement can be modeled with the quadratic curve $$x=x_0 + v_0t+\frac{1}{2}at^2 \qquad\text{where } x_0 \text{ is the initial position, and }v_0 \text{ is the initial velocity}$$ During the constant velocity phase, the object's movement can be modeled with the linear equation $$x=x_1 + v_1(t-t_1)$$ where $x_1$, ...


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The first equation holds good for average acceleration, but the second is the equation for uniform acceleration. The value obtained using option 1 is correct. In the time interval from 0 to 6 s, the acceleration changes (a constant value from 0 to 3 s and another constant value from 3 to 6 s). Then you cannot apply the uniform acceleration equation as ...


2

I think you are confusing the escape velocity with terminal velocity. While for the local velocity the limit is only the speed of light, terminal velocity is achieved much sooner because of the air resistance. The equation can be found here and depends on the shape, size and density of the asteroid. For particles travelling near the speed of light you have ...


2

For the sake of simplicity (at the expense of real-life accuracy), let's assume that an asteroid is already travelling at some speed $v_0$ directly toward the Earth, and it never deviates away from that direct line path (despite the revolution of the Earth around the Sun). Let's also assume the distance from Earth is very large compared to the radius of ...


2

Accelerations being equal doesn't necessarily mean that the velocities are equal, or vice versa. For example, your two cars could have the same acceleration, but if one starts before the other, the one that got going earlier wlil obviously be moving faster. An even simpler example, if one car is standing still and the other one is moving at constant speed, ...


1

Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. $$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ...


1

As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get $$\ddot x=0$$ $$\ddot y = -g$$ Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...


1

For the first part of your question, you have to realize that your net velocity (the one that you plug into the expression for centripetal force) is the vector sum of the surface velocity and your velocity relative to the surface. If you were running West as fast as the earth turns East, you would "stay in place" and the sun would appear to stop moving in ...


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Any object with mass that accelerates (is it linear or angular acceleration) produces gravitational waves, though in most occasions those will be much too small to be detected. As @CuriousOne pointed out, same happens with electromagnetic waves and accelerating charges. The gravitational waves that can be detected usually come from very massive objects (such ...


1

Suppose you take any quadratic equation: $$ y(t) = At^2 + Bt + C $$ for constant $A$, $B$ and $C$. The velocity is $dy/dt$ giving: $$ v = \frac{dy}{dt} = 2At + B $$ and the acceleration is given by differentiating again: $$ a = \frac{d^2y}{dt^2} = 2A $$ and since $A$ is a constant that means the acceleration has the constant value $a=2A$ i.e. any ...


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I would think that the difficulty is related to gravity. Without gravity, water will go everywhere in all directions.


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One promising way of implementing arbitrary physics simulations, is by programming in terms of 'constraints'. I highly recommend reading this article that covers constraints very well: http://gamedevelopment.tutsplus.com/tutorials/simulate-tearable-cloth-and-ragdolls-with-simple-verlet-integration--gamedev-519 I was very surprised when I first saw this ...



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