Physics, 30.01.2020 15:03 mmelody5

If an object is moving in a circle at constant speed, what does newton’s second law inform us about the motion? 1. xf = 0 since v for the object is constant. 2. xf points in the opposite direction to the object’s velocity. 3. xf points in the direction the object is moving. 4. xf points directly outward, away from the center of the circle. 5. xf points toward the center of the circle.

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Answer from: cheyfaye4173

XF points toward the center of the circle

This situation is characteristic of the uniform circular motion (figure attached), where $\vec{V}$ is the velocity vector, whose direction is perpendicular to the radius of the trajectory.

The other thing that can be observed is that its acceleration $\vec{a}$ is directed toward the center of the circumference (that's why it's called centripetal acceleration).

Now, according to Newton's 2nd law, the force $\vec{F}$ is directly proportional and in the same direction as the acceleration:

$\vec{F}=m.\vec{a}$

Therefore the net force XF resulting from the movement points towards the center of the circle and is called Centripetal Force.

Therefore the correct option is 5.

If an object is moving in a circle at constant speed, what does newton’s second law inform us about

Answer from: Quest

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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