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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 ...

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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...

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Yes , she is very correct. Consider motion of elevator. d= vt = 6*3 = 18 So, elevator travelled 18m. Now consider bolt h= ut + 1/2 g$t^{2}$ -18 = 3u -10/2*9 u= 9m/s Now, v= u + at = 9 - 10*3 = -21 m/s so bolt falls with speed 21m/s.

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Following the link you gave in your question, torque steering as described there really happens in situations where the drive torque is applied to two wheels (front or rear), and for some reason the torque is not applied equally to both wheels. An apparent pull to the side when you accelerate your bike hard may be related to the tension in the chain, and ...

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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 ...

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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 ...

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Now, one g is equal to the acceleration due to gravity by Earth on its surface. Force by washing machine is the centrifugal force as clothes try to go along straight path but the resultant provided by the walls of container of machine which provides the required centripetal force. Now, according to Newton's third law of motion , clothes will also give equal ...

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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 $$a = 0.25\times 4\pi^2 \times 25^2 = ... 0 Assuming your friction and air resistance are not significant, we can figure it out easy without considering them and be pretty close. (I dont think you're going to go fast enough / have a large enough surface area to make air resistance significant) E = 1/2 M * V^2 Power = E / t (Wattage) V = V_0 + A*t So, the wattage will depend on what force you want ... 0 Its relativistic mass will increase to the point where it is no longer able to accelerate any further, at which point it may slow down and come to a halt. I'll assume the ball has a fictional mechanism attached that gives a constant propulsive force, forever. In that case, yes, its energy (and hence relativistic mass) will increase as its speed ... -1 the ball wouldn't come to the halt as its mass goes on increasing it would simply stop accelerating and keep on moving at the uniform motion close to the speed of light which is determined by how much force u had applied. of course it would be impossible to travel in speed of light as it would require infinite energy. 1 The velocity is downward, and the acceleration is downward. Whatever direction you choose, if you start with a velocity of zero the sign of both will be the same (if you throw the feather down, it will decelerate - so the acceleration will the "up". I don't think that is intended here). Whether the floor or the hand is zero in the coordinate system doesn't ... 2 No. the earth do not accelerate with 9.8 meter per sec. squared. The acceleration is quite negligible. And as the objects on the whole earth is distributed almost uniformly the net acceleration due to the objects is zero. That's why we do not notice the earth moving around like that. MATHEMATICALLY we know that | F_E | = |F_o| where F_E is force ... 2 It's not really a circular reference, it's an ordinary differential equation:$$ \frac{d^2y}{dx^2}=f(y) $$In particular, it is a general 2nd order ODE, which, for some functions f(y), has known analytic solutions. One way to think of this is a mass on a spring: (source: Julius O Smith, Standorf MUS420/EE367A Supplementary: The Laplace Transform) If ... 0 we talk in General Relativity [...] about curvature of space and time At least (without implying strict separability) about curvature of spacetime; right. In this context it is however possible to distinguish "free (inertial, falling) motion" from "any other motion", The latter is accordingly said to be "due to (external) influence(s)", and its ("first ... 0 The classical equations of motion should do it for you. This one for example:$$v_{final}=v_{initial}+at$$Plug in your maximum deceleration (the most negative a possible) and you other values. You can easily find the time t is will take to reach 0 m/s. If the value for the maximum deceleration is not constant, then those equations don't apply. ... 1 A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity. The equivalence principle ... 1 A constant force of 5N acts for 5 seconds on a one kilogram mass. I am assuming you can use F=ma to find a=5 m/s2. So after 5 seconds, the velocity would be v=u+at. Yes. If a constant force F acts on something for 5 seconds, then that something accelerates with acceleration a during those five seconds. In other words: That object accelerates ... 1 You might want to use the idea of impulse, J, defined as$$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$$You should be able to solve this. In the second case, F may or may not be time dependent. The equation for impulse can be changed to ... 1 I was able to determine the user status(static, slow walking, fast walking) by calculating the variance. The Va in the research was not velocity. It was my mistake to interpret it as such. It was the variance of the euclidian norm of the accelerometer data. I decreased the accelerometer update interval to 0.1s and every second I took 10 of the values and ... 0 The issue here is that there is a force acting on the axle to counter act gravity on that part of the disk, so not all parts of the disk will be accelerating uniformly at 9.81 \text{ m/s}^2. Note also that this force acts at the axis of rotation, and therefore doesn't contribute to the torque. That is why you don't need to consider it when calculating ... 1 If we take away the Earth's gravity, and the gravity of all celestial bodies, will the force of acceleration in the rocket still be felt? Yes, you'll still feel the acceleration and you can demonstrate this even on Earth and in your car. As your car accelerates at a rate a, you experience a force F=ma with m your mass. Of course you also ... 2 Let's discuss a whole set of cases. You're standing on the planet. I'm guessing that you would describe the sensation as 'feeling' gravity pulling you down. You're on a skydiving trip, standing in the plane. You still 'feel' gravity. You're on the skydiving trip and you've just step out of the plane, but not had time to pick up speed. Here you don't 'feel' ... 0 You need initial velocity. Use these two kinematics equations. So first find time using the second formula, then use the first formula to solve for initial velocity. 4 Here are the steps you want to take. We need to find v_0. The equations are$$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. ... -1 You have three knowns, and one unknown. Pick the newtonian kinematic equation that just has the 4 variables. You don't need to use two equations. 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 ... 0 In the second case, you made a mistake in the denominator. You always put the overall mass. The lighter mass is part of the system that you are trying to find the center of mass of, even when it has a zero value of acceleration. So, the denominator should be 3m. In general, we put in the denominator the mass of every body that is in the system of bodies for ... 1 Velocity is an instantaneous notion; even on a curved trajectory at any moment you have a direction and a speed in that direction. Each time you can calculate de derivative of position coordinates, you have a valid velocity. Note that with your arguments you should say as well that positions don't exist and acceleration don't exist "because they change" ... -1 The space time fabric is like a elastic/rubber sheet which can be twisted, elongated, etc. So this sheet is not straight because of depressions caused by massive objects (planets, stars, black holes). So when there is no disruption (twists, depression) in space time fabric, velocity is in straight line as the fabric is plane. But when there is a depression ... 0 If I understand the question you are partially asking what causes a skid. You only get so much friction between two surfaces. After that they slide. Even after they slide there is still friction. Static and kinetic friction The car is exerting a force. It has a negative acceleration. The car is loosing speed / momentum. The force that causes the ... 0 First you need to find the force of friction on the wheels of the car using the weight of the vehicle + the passenger. Then, calculate the acceleration. Then, calculate the force needed to accelerate the passenger by the same amount and you've found the force on your passenger. 1 Just because an object is in motion, that doesn't mean that there is a force acting on it. Newton's 1st law states that an object in motion will stay in motion unless acted on by an outside force. So the car's momentum is what keep's it going as it coasts. The forces acting on the car would be gravity, the normal force (the ground pushing up on it), and ... 0 "But what happens really is that the block is accelerating in the direction of F": that is only before the hook starts to act. Once they are in equilibrium, and as you said before, the cube will be at rest and the acceleration will be zero. It is easy to see this: ma=F-F=0 so the acceleration (or the mass, which I assume is not the case) must be zero. 2 What you are confusing here is that the force is given by F=ma, the mass times the acceleration. Both the cube and the rope will have the same acceleration if they are attached, but they will not have the same force because the masses are different. What you'll end up with is something like this for the acceleration of both objects:$$a_{sys} = ...

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