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You do not need the kinetic energy. Working with the total energy $\gamma m c^2$ produces the same result. Assuming both the total initial energy $\bar E_0 = \gamma_0 m c^2$ and the additional energy $E_i$ are known, write $\gamma_1 mc^2 = \frac{mc^2}{\sqrt{1-\beta_1^2}} = \bar E_0 +E_i$ for $\beta_1 = \frac{v_1}{c}$, then $$\sqrt{1-\beta_1^2} = ... 4 So for starters you need to make sure the center of gravity is in the middle of the tray (so in you scenario of lighter/heavier cups, balance it out). This is assuming that you make as many right turns as left turns. If this is not the case, say 70% of your turns are right turns, then you would want to slightly offset the orientation so that it's not quite ... 3 It is independent of the mass of the planet if you assume the bearings are frictionless. Also assume that the tyres do not make dents in the ground. However, in reality, the bearings have friction. Additionally, there is rolling resistance as the tyre makes small deformations in the ground as it is rolling. This is why you see a characteristic "W" shape in ... 3 The equation for your curve is given by:$$ \frac{dv}{dt} = \frac{F(v)}{m} $$where F(v) is the net force on the car, which is a function of the velocity. we solve the equation by integrating to get:$$ \int \frac{dv}{F(v)} = \frac{t}{m} $$The trouble is that the net force F(v) is a complicated function that doesn't generally have a simple analytic ... 3 You don't need acceleration. Just calculate change in potential energy with altitude and make sure your initial Kinetic energy is sufficient that your final velocity is greater than zero. You cannot use your formula without integration. As for actual formula it would be$$g= GM/r^2$$where M is mass of earth and r is varying distance. 2 It depends on various things: the energy E of each neutrino, how likely each of them is to interact with a particle in your body (expressed as a cross-section \sigma, which depends on E), and how large an acceleration a you consider "significant". Further let m be your mass, 75\;\rm kg. I've pilfered some rough numbers from here, namely ... 2 1) why a_c has to have opposite sign. a_c is the centrifugal acceleration in the rotating frame experienced by the tennis ball.$$\vec{a_c} = (\vec{\omega} \times \vec{r}) \times \vec{\omega}$$You’re answer for \vec{a_c} gives the correct sign and magnitude. If you draw the vectors for \vec{\omega} and \vec{r} and apply the right hand rule ... 2 This is how you do the calculation. The elapsed time on an observer's clock is called the proper time, \tau, and it is calculated by integrating the metric:$$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$In this case we'll assume all motion is radial so d\theta = ... 2 If we assume all other things being equal other than the downward force due to gravity, the vehicle on Earth would be capable of greater acceleration. The ability of the tires to grip the surface on which they are resting depends on the downward force keeping them in contact (Coefficient of friction). That will pretty much relate the gravitational ... 2 The average velocity is given by$$ \bar v=\frac{1}{T}\int_0^T v(t)\mathrm dt=\frac{1}{T}(v_1t_1+v_2t_2) $$where t_1 is the time spend on the first interval, t_2 is the time spend on the second one, and T=t_1+t_2. Using$$ v_1t_1=v_2t_2=d $$you get$$ \bar v=2\frac{v_1v_2}{v_1+v_2} $$I believe you can take it from here. 1 The acceleration at the point of reflection is actually quite complicated. It is caused by the elastic forces of the surface and the ball and has a complicated time dependence. However, the timespan in which the ball touches the ground is very short (especially if the ground and the ball are very rigid), therefore we can simplify the actual acceleration ... 1 Let's imagine you walking down a tall mountain when the sun is setting. Let's say it is cold at the top of the mountain, so walking down the mountain tends to make you warmer. But also the sun is setting so you might get colder. How do you take both effects into account? Well the change in temperature you feel (LHS) is the change in temperature at the ... 1 Newton's 2nd Law says that F=ma. This law says NOTHING about the physical properties of the object that you are accelerating. Thus, the answer is "yes", the two objects will accelerate at the same rate, so if they start at the same velocity, they will continue having matching velocities as long as they experience the same acceleration. 1 No. If you want an answer beyond a simple yes or no have a look at my answer to How does "curved space" explain gravitational attraction?. The acceleration is caused by the curvature of spacetime. This curvature does also cause time to run more slowly as you approach a spherical mass, but the change in the rate taht time runs does not cause the ... 1 This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle.$$ \frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx $$For a constant force and mass \int F\cdot dx = F (x_A - ... 1 Firstly it's worth noting that such a discontinuity can never be 100 % real. To go from acceleration a to b instantaneously (\Delta t = 0) would require an instantaneous change in the net force responsible for the accelerations and that isn't possible in the material world. Secondly, I think you are over-thinking your problem. Just write the ... 1 If you take your final expression$$ x(t) = \underbrace{\left(x_0 + \frac{b - a}{2}\tau^2\right)}_{x_0^*} + \underbrace{\left(v_0 + \tau(a - b)\right)}_{v_0^*} t + \frac{b}{2} t^2, \quad \text{with}\ t>\tau,  then $x_0^*$ and $v_0^*$ would be the position and velocity at $t=0$, however this is only meaningful if $\tau<0$ (however in that case $x_0$ ...

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The Equivalence Principle of General Relativity holds that acceleration and gravity can be described identically. With an accelerometer, you can tell whether or not you are accelerating in empty space, regardless of whether another object is available to act as a reference point. Under acceleration, your weight will change just as though you were ...

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If you have no means to change the inclination of the tray while you are driving, then the answer is: flat. You'll make as many left turns as right turns, so you can't favor left tilt vs right tilt. And you will speed up as well as slow down, so you can't favor forward tilt vs backward tilt. The distribution of cups ... which cup goes where ... ...

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