Physics, 22.11.2019 19:31 mattgraham366

The period of a ball moving in a circle is doubled while the radius stays the same. what happens to the velocity of the ball?

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Answer from: marjorieheckaman

The velocity of the ball becomes half when the time period is doubled.

Explanation:

Let be the time period of the ball, be radius and be the velocity of the ball performing circular motion.

Now, for a circular motion, the velocity is given as:

$v=\frac{2\pi R}{T}$

From the above relation, we can conclude that, velocity is inversely proportional to time period if the radius is constant.

This means that if time period is increased, velocity will decrease and vice-versa.

So, if time period is doubled, the velocity will reduce by half. This can be verified mathematically also given below.

$v_o=\frac{2\pi R}{T}\\v_n=\frac{2\pi R}{2T}$

Divide v_n by v_o , we get

$\frac{v_n}{v_o}=\frac{\frac{2\pi R}{2T} }{\frac{2\pi R}{T} }\\\frac{v_n}{v_o}=\frac{T}{2T}\\ \frac{v_n}{v_o}=\frac{1}{2}\\\therefore v_n=\frac{1}{2}v_o$

Here, $v_n\ and\ v_o$ are the new and old velocities.

So, it can be concluded that if the time period is doubled, velocity is reduced by half.

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.

Answer from: Quest

If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2 where t1=temperature of cold reservoir ( ° k) = 135 t2=temperature of hot reservoir ( ° k) = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58% note: this is the theoretical maximum efficiency of a perfect heat engine. real-life engines do not achieve this value.

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