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Use the directions up the ramp and perpendicular to the ramp as the simplest coordinate system. Find the component of the vertical force of gravity that acts down the ramp. Find the component of the horizontal force F that acts up the ramp. Find an expression for the net force up the ramp, and equate to the mass of the cart times the acceleration of the ...

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Let me present a slightly different perspective to Alfred's answer, although I'm basically saying the same thing. I suspect you've got hung up on the idea that velocity causes the relativistic effects like time dilation, but the underlying cause is something different. All the weird effects in SR are caused by a fundamental symmetry of spacetime, which is ...

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We need to untangle this a bit but first: the cause of time dilation is the geometry of spacetime which is such that there is an invariant speed c. Now, remember that velocity or speed is not a property of an object; there is no absolute rest. Further, consider the case of three objects in uniform relative motion with respect to each other. If I choose ...

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Consider a mass-spring system executing simple harmonic motion. If I draw the displacement, velocity and time graph, it would look something like this: You may see that when t=1 second, velocity is maximum and acceleration is zero. Another way to explain this is by using the definition of acceleration. Acceleration = change in velocity/time = gradient ...

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As taught in calculus, to find the local extrema of a function, one solves for the values of the argument where the derivative of the function is zero. So, at the maxima and minima of velocity, the (time) derivative of velocity is zero. But acceleration is the (time) derivative of the velocity. Do you see the answer now?

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When you hit the obstacle (if you don't destroy it) your speed goes down to zero in a quite small time. That gives you the acceleration. It starts when you start hitting and it ends when you come to a complete stop. This acceleration is due to the force that the obstacle generates on the car. Think about: $\displaystyle a=\frac{\Delta v}{\Delta ... 3 You should realize that the first equation you write gives you the value of$a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore$F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$ Law}} ma \quad \quad \to \quad \quad a = -g\mu$Now you know the acceleration, you can find how ... 1 Remember the law$F=ma$; you already know the force from friction$F_{\mu}= -mg\mu$. Hence you can get$a=-g \mu$, and one of your unknowns is gone. 1 However, the problem says that the floor pushes the man upwards with a stronger force than his legs. This can not be true. The floor pushes the man upwards with as much force as his legs exert downwards. However, this happens to be more than his weight, which is why he accelerates upwards. 0 @AlfredCentauri The condition of locality conventionally only requires a locally constant gravitational field during the time the experiment runs. For a stationary rocket in the gravitational field of e.g. the earth, this condition is perfectly satisfied, without requiring the gravitational field to be uniform throughout space. Clearly, the stars are not ... 1 If a net force is towards the positive x direction, then the$x$component of velocity of the object is increasing; period. Now, knowing this, are any of the statements (a) through (d) true? a) It can be moving in the negative x direction Sure, it could be moving in the negative$x$direction. All that is required is that the velocity is ... 1 The only thing that a net force in the positive$x$direction means is that the net acceleration is in that direction. It tells you nothing about its position or velocity. a) Imagine placing the$x$-axis vertically and letting the positive$x$direction be downward. Now imagine that the constant force is gravity and the particle is a ball. If you throw the ... 1 a), b), c), d) All yes. Imagine a ball that is moving at 10 meters per second to the left (negative x direction). Now, a force in the positive x-direction will slow it down initially, and finally make it go to the right, constantly accelerating to the right. This is all independent of the y-direction, in which it might also be moving. A force in the ... 2 This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate. Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For ... 1 When I read about this question, I was thinking, well, "a" can't be zero because it should represent your deceleration from 900 miles/hour. The resulting F would be the force you experience when you decelerate. (The deceleration depends on how you hit the obstacle, how long did it take you to stop/slow, across what distance, and the degree of deformity of ... 4 Another way to think about Newton's second law (and the way he originally defined it) is$F=\dfrac{d\rho}{dt}$, where$\rho=mv$is momentum and$\dfrac{d\rho}{dt}$is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of ... 6 You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a$F = ma$calculation where$a$is the acceleration of your gas pedal. Instead in the collision you have a force$F$resulting from the collision and you ... 0 F=M x A= F 25 x 25=6.25 For the explanation is,the formula must be F=M x A=F For example 25 kg. is the mass and .25.00.00 is the acceleration of the speed limit.Now you must multiply the two 25 and the answer is 6.25 1 In one sentence: More mass means stronger attraction and less buoyancy (they fall faster), but the effect is negligible in most cases. 6 Suppose you pick two people at random. From one, you pluck a single hair from their head. Is it possible to tell who had the hair plucked by weighing the people? Technically, plucking a hair makes a person very slightly lighter, so you get a tiny bit of information about who had the hair plucked by weighing the people. But the information is very slight ... 0 What if I don't look at other stars, but the microwave background radiation, the "echo of the Big Bang", as it is sometimes called. Doesn't that give me a clue about my "absolute" motion, i.e. relative to "the universe"? 2 First, it turns out that there are no uniform gravitational fields so the equivalence principle holds only locally. But, for the sake of argument, let's assume that a uniform gravitational field can exist. Now, consider the situation where an astronaut is in a rocket and the rocket's accelerometer reads a constant, non-zero value. According to the ... 0 The stars you see at night are blue-shifted partly because they gain energy as they fall down Earth's potential well. Sorry, but Einstein wins this round :) 0 English mathematician Sir Isaac Newton published Principia, which hypothesizes the inverse-square law of universal gravitation. He deduced that the forces which keep the planets in their orbs must be reciprocally as the squares of their distances from the centers about which they revolve. If he was reasoning this way about forces (F), he was also doing so ... 1 "Focus" is an inconvenient word if you're thinking of changing the potential, because if you do then the orbits are no longer conics and the word kind of loses its meaning. That aside, let me see if I understood your question correctly: Given a gravitational potential that's spherically symmetric around a central point$\mathbf{r}_0$, and which has a ... 0 The velocity of an orbit around some central object can be easily calculated for a circular orbit. Let us assume that there is some central Force$F=c\cdot r^\alpha$, where$c$and$\alpha$are some constants (for gravity$c=Gm_1m_2$and$\alpha=-2$). For a stable orbit, this central force must be equal to the necessary centripetal force (not balance the ... 0 The best way to explain it (and even the way Kepler's second law can be derived) is by conservation of angular momentum. The latter is given by $$\mathbf{L}=\mathbf{r}\times m\mathbf{v},$$ where$\mathbf{r}$is the position vector and$\mathbf{v}$is velocity. Since this quantity has to be conserved for the motion of the object at all times, assuming an ... 2 First of all, you're overloading your time variable. Let the delay of the launch of the 2nd projectile be$t_d$while the time variable is$t$. As you've already correctly written, the equation for the displacement of the 1st projectile is (for$t \ge 0$): $$s_1 = ut + \dfrac{at^2}{2}$$ Now, for the 2nd projection, we have (for$t' \ge 0$):$\$s_2 = ut' ...

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