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6

Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$ Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$ To evaluate this we use the fact that: $$d \tau^2 = coshx~dt^2-dx^2$$ Also use $x=1$ and $dx=0$: $$d \tau^2 = cosh(1) dt^2$$ Therefore: $$u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = ...


4

If you stayed at the same radius while the earth shrinks then nothing will change other than you'll start to fall. If you stand on the surface of the new smaller radius then you will feel the increased gravitation.


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Without loss of generality, we assume that a particle is moving in a circle of radius $R$ centered at the origin and lying in the $x$-$y$ plane. Using cylindrical coordinates $(\rho, \phi, z)$, the angular velocity of the motion is $\boldsymbol\omega = \omega \hat{\mathbf z}$. The velocity is tangent to the circle and given by $\mathbf v = ...


3

The strength of the gravitational field depends on how far you are from the center of the mass, assuming the mass density is radially dependent or constant. It also depends on the total mass inside your position: $$g=\frac{GM_E}{r^2},$$ where $r$ is the distance from the center of the mass distribution. So if you stay where you are now, the gravitational ...


3

Here is one way of looking at it via a velocity-dependent potential.$^1$ The Coriolis potential is $$\tag{1} U_{\rm cor} ~=~ -m({\bf v} \times {\bf \Omega})\cdot{\bf r} ~=~-{\bf v}\cdot ({\bf \Omega}\times{\bf r} ),$$ cf. Ref. 1. The factor $2$ comes from two different terms in the corresponding force formula $$\tag{2} {\bf F}~=~\frac{\mathrm d}{\mathrm ...


3

Your first method is using the conservation of energy for the spring bullet system. You second method fails because the acceleration is not constant and so you cannot use a constant acceleration kinematic equation. On seeing $ma = -kx$ you might reorganise that the acceleration is proportional to the displacement from a point (the position of the free ...


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This is actually a famous theorem known as the Einstein Equivalence Postulate (sometimes Equivalence Principle). It's true that since Earth is spinning, acceleration in a spacecraft isn't quite the same situation we experience daily, but in general, yes, gravity is indistinguishable from uniform acceleration. Specifically, if you are in a box with no windows ...


2

All other things being equal, if a heavier object will roll at a higher speed down hill than a lighter one With the qualification "All other things being equal" your statement is not correct. The falling acceleration is the same because doubling the mass of an object doubles the force causing the acceleration (the object's weight) which means that ...


2

First draw a circular path. In order for an object to follow that path at a constant speed, there must be a force acting on it towards the centre of the circle. At any instant, that force has no component in the direction of motion, and so, if that's the only force acting on the object, its speed will stay constant. Now draw a spiral path (i.e. one in which ...


2

Let's look at the hodograph of a constant radius & constant velocity motion. Left: trajectory of one of the masses. Right: hodograph, i.e. locus of the velocity vectors. Now, let's look closer at how the velocity changes during a small time interval $\mathrm dt$. A force is needed to rotate it (difference between the brown and red arrows). If you ...


2

The first figure shows the parallel and normal components of a vector $\mathbf{r}$ relatively to a direction $\mathbf{n}$. Based of this, the second figure shows the centripetal acceleration. In case of plane circular motion $\omega R = v$.


2

I think you should take the sample velocities and divide them by the respective times after minusing the previous velocities to obtain the accelerations. If the time interval between calculating the discrete sample velocities are too small, then the above-got values may be taken as instantaneous acceleration. Now plot these over graph wrt time. Now here we ...


2

The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...


1

In addition to the already given answears this also might be of interest: When hammer and feather are dropped simultaneously they arrive at the same time, when dropped independently the hammer attracts the planet more than the feather, so you are right, the total time until impact is then smaller for the hammer. If you pick up the hammer and let it fall to ...


1

If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...


1

Astronauts do not feel weightless on the Moon where the gravitational field strength is one sixth of that on the Earth. They feel "lighter" in spite of the heavy suits that they have to wear. A spring balance would give a reading for the weight of an astronaut on the Moon which was one sixth of that on the Earth. An astronaut in orbit feels weightless and ...


1

Since acceleration is a vector you can decompose it in the coordinate system you find convenient. If you define a cartesian coordinate system whose axis are along the normal to the plane and the plane itself you see there is a component of the acceleration $g\sin\theta$ along the plane. This is why the block accelerate in this direction. Notice that along ...


1

The normal force is not playing a role in this case because the force is perpendicular to the moving direction of the box. By "not playing a role" I mean, during this motion, $\ F_N$ is only be used to balance out $\ mgcos\theta $ so that the object won't be able to move "into the inclined plane". To find the force causing the block to accelerate down, we ...


1

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


1

The $\Omega \times$ terms signify change due to the moving orientation of the item. So the velocity vector when attached to a moving body will cause a $\Omega \times v$ term in the acceleration vector.


1

A fairly simple way of treating the data is to present them as a histogram: Each data point (here 5 data points) is the quotient of the distance moved in that interval, say $\Delta y$, by the time interval $\Delta t$ and is the average velocity during that time interval: $$v_i=\frac{\Delta y_i}{\Delta t_i}$$ Where $i$ indicates interval number $i$. For ...


1

Hint: Since the car begins from rest and keeps accelerating, the equation $d=vt$ is no longer valid, because velocity is not constant. Note that you can, however, say that $d=v_0 t + {}^1/_2 a t^2 = {}^1/_2 a t^2$ and the acceleration will be the same for both time intervals. Hope that gives you a nudge in the right direction! Also, these equations may be ...


1

The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum. Given a path $y(x)$ the instantaneous radius of curvature at each x is $$ \rho = \frac{ \left(1+ \left(\frac{{\rm ...


1

The only thing I've figured out, is that the object will rotate but not moving, if the center of mass is on the line defined by the $A$ point, and the normal vector of $\vec{F⃗}$. This is not necessarily true. Look at the free body diagram. Decompose $F$ into $x$ and $y$ components. Newton now tells us that: $$ma_x=\Sigma F_x$$ $$ma_y=\Sigma F_y$$ ...


1

The net force on the string is not F. You are pulling the string forward with force F but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to F. If the masses of the string and block are m and M then for the whole system (string plus block) F=(M+m)a. The net force on the string is F '=ma=mF/(M+m). ...


1

Let two orthonormal systems $Oxyz$, $O'x'y'z'$ with a general motion (translational plus rotational) between each other and a point particle $\rm P$, see Figure. Symbol Conventions : 1.The vectors for position $\mathbf{R}$, velocity $\mathbf{U}$ and acceleration $\mathbf{A}$ of a particle with respect to $Oxyz$ expressed by coordinates of this same ...


1

In physics we recognize two different kinds of mass: inertial and gravitational. Inertial mass tells us how much an object resists a change in motion - or how much force is needed to effect an acceleration. Gravitational mass describes how much attraction (due to gravity) an object experiences as a result of gravity. Now despite very careful experiments, ...


1

if a heavier object will roll at a higher speed down hill Free fall and rolling are two different behaviors of objects. It is correct that for free fall all objects get the same acceleration ( minus friction and drag) but free fall is not the same as rolling. For going down a hill free fall can be compared to sliding, as was pointed out in the comments ...


1

As a supplement to Farcher's answer, pointing out that the constant acceleration equations fail here, we can still use $\vec{F} = m \vec{a}$ to solve for the speed of the bullet - by means of integration. We have $- k \vec{x} = m \vec{a}$, but we can rewrite $\vec{a}$. ${a} = \frac{d^2 x}{d t^2} = \frac{ d v}{d t} = \frac{dv}{dx} \frac{dx}{dt} = v ...



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