# Tag Info

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 ...

4

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 ...

3

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 ...

3

The only way to do it is to put you temporarily in free fall. But as for the room you describe, I can only think of one type. Bring a scale with you next time you go down an elevator, and watch artificial gravity reduction at work! Heh heh.

3

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 ...

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' ...

2

I've never been at a theme park where you can mount into a plane at free fall. The photo that you posted is inside a reduced gravity aircraft. So you don't modify gravity, you are just falling.

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 ...

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