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The simplest formula for the centrifugal acceleration is $$a = r\omega^2$$ Here, $r$ is the radius which is 0.25 meters in your case. $\omega$ is the angular velocity which is $2\pi$ times the frequency $f$. Your $f$ is 1500 revolutions per minute which is $1500/60=25$ revolutions per second. In the SI units, we have $$a = 0.25\times 4\pi^2 \times 25^2 = ... 4 Here are the steps you want to take. We need to find v_0. The equations are$$v_t = v_0 + g\cdot t\\ y_t = y_0 + v_0 t + \frac12 g t^2$$Two equations, two unknowns. Eliminate t, then solve for v_0 (Note that I use a Y axis that increases as you go down - just saves thinking about the sign of g). Alternatively you can use conservation of energy. ... 2 It's not really a circular reference, it's an ordinary differential equation:$$ \frac{d^2y}{dx^2}=f(y) $$In particular, it is a general 2nd order ODE, which, for some functions f(y), has known analytic solutions. One way to think of this is a mass on a spring: (source: Julius O Smith, Standorf MUS420/EE367A Supplementary: The Laplace Transform) If ... 2 No. the earth do not accelerate with 9.8 meter per sec. squared. The acceleration is quite negligible. And as the objects on the whole earth is distributed almost uniformly the net acceleration due to the objects is zero. That's why we do not notice the earth moving around like that. MATHEMATICALLY we know that | F_E | = |F_o| where F_E is force ... 2 Let's discuss a whole set of cases. You're standing on the planet. I'm guessing that you would describe the sensation as 'feeling' gravity pulling you down. You're on a skydiving trip, standing in the plane. You still 'feel' gravity. You're on the skydiving trip and you've just step out of the plane, but not had time to pick up speed. Here you don't 'feel' ... 2 There is absolutely a gravitational radiation reaction and solving for it is one of the very active fields in classical relativity theory at present. Basically, particles with nontrivial masses distort the spacetime around them; this causes them to not move on geodesics of the "background" spacetime (the spacetime that one would have found had the secondary ... 1 Acceleration due to gravity remains roughly constant near the surface of the earth. Yes, a = F/M, but as mass increases, the force exerted by gravity increases too( F\ \alpha \ m1m2\over r^2), keeping F/M or a roughly constant around the surface of the earth 1 Here's how to intuitively understand that a=g. Take a metal ball having mass 1kg and drop it. Its downward acceleration is 9.8m/s^2, right? Now take a second ball and drop it. Same thing, right? Now drop both at the same time. Same? Now connect them together (with a tiny drop of weld metal) into a single 2kg mass, and drop them. Do they suddenly slow ... 1 There is a mistake in equation (2). Its denominator should include the total mass of the system that you're considering, so the denominator should be '2m+m'. You correctly used this value for equation (1), but apparently incorrectly believed that since the position (and velocity?) of the lighter mass 'm' is zero that the value of 'm' shouldn't be included in ... 1 The key is this: how is the force applied to the rider? Gravity pulls directly down on the passenger's mass, but what keeps the passenger from heading towards the center of the Earth? When you're in the passenger seat of a sports car, and the driver floors it, you feel pushed back into the seat. But what's actually happening is that the seat is pushing you ... 1 The rider feels "forced down" because the object to which they are attached is accelerating upwards. Because the acceleration is opposite to gravity, the normal force, \mathbf{F}_{N}, being exerted on the rider must increase in magnitude (relative to the "at rest" magnitude on a horizontal track) in order to produce a net upwards acceleration. Thus, it ... 1 This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance h below and moves down at a constant 6  m/s. The bolt catches up in 3 seconds. Same problem, but the equations are simpler... 1 If we take away the Earth's gravity, and the gravity of all celestial bodies, will the force of acceleration in the rocket still be felt? Yes, you'll still feel the acceleration and you can demonstrate this even on Earth and in your car. As your car accelerates at a rate a, you experience a force F=ma with m your mass. Of course you also ... 1 I was able to determine the user status(static, slow walking, fast walking) by calculating the variance. The Va in the research was not velocity. It was my mistake to interpret it as such. It was the variance of the euclidian norm of the accelerometer data. I decreased the accelerometer update interval to 0.1s and every second I took 10 of the values and ... 1 You might want to use the idea of impulse, J, defined as$$J=\int_{t_a}^{t_b} F\,\mathrm{d}t=\Delta p=mv_f-mv_0$$In your first case, F is not time dependent, and so you have$$J=Ft_b-Ft_a=mv_f-mv_0 You should be able to solve this. In the second case, $F$ may or may not be time dependent. The equation for impulse can be changed to ...

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A constant force of 5N acts for 5 seconds on a one kilogram mass. I am assuming you can use F=ma to find a=5 m/s2. So after 5 seconds, the velocity would be v=u+at. Yes. If a constant force $F$ acts on something for 5 seconds, then that something accelerates with acceleration $a$ during those five seconds. In other words: That object accelerates ...

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A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity. The equivalence principle ...

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The velocity is downward, and the acceleration is downward. Whatever direction you choose, if you start with a velocity of zero the sign of both will be the same (if you throw the feather down, it will decelerate - so the acceleration will the "up". I don't think that is intended here). Whether the floor or the hand is zero in the coordinate system doesn't ...

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