# Question #ed863

Jan 17, 2018

OK, we can, but it’s not very easy.

#### Explanation:

Firstly, there is the definition of an arc-second. A degree is $\frac{1}{360}$ of a circle, a minute is defined as $\frac{1}{60} t h$ of a degree and a second (or arc-second) is $\frac{1}{60} t h$ of a minute. 1 arc-second is thus ${\left(\frac{1}{3600}\right)}^{\circ}$ We need this in radians, and 1 rad = ${\left(\frac{360}{2 \pi}\right)}^{\circ}$ so an arc-second = $\frac{1}{3600} \times \frac{2 \pi}{360}$ rad so 1 arc-second = $4.848 \times {10}^{-} 6$ rad

Next there is a very simple link between arc length, s (which we will assume to be equal to the diameter of the nickel), the radius, r (the distance we need to find) and the angle in radians, $\theta$. It is just $\theta = \frac{s}{r}$ so $r = \frac{s}{\theta}$

Finally, the diameter of the nickel is 21.21mm (had to google that, I’m British) which we convert to $2.121 \times {10}^{-} 2$m as we want an answer for r in m.

$r = \frac{s}{\theta} = \frac{2.121 \times {10}^{-} 2}{4.848 \times {10}^{-} 6} = 4 , 375$ m

You could criticise this result as the diameter of the nickel is actually a chord, not an arc, but I suspect it makes little difference to your answer.