If they're brought closer together
Answer:
![v=\sqrt{2gh}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2gh%7D)
Explanation:
We could use conversation of energy. Total distance the stone will cover will be
![h=\frac{1}{2} g t^{2}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B1%7D%7B2%7D%20g%20t%5E%7B2%7D)
the Final velocity will be
![v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}](https://tex.z-dn.net/?f=v%3Dgt%5C%5Cv%3Dg%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%20%7D%5C%5C%20v%3D%5Csqrt%7B2gh%7D)
We know that the Delta E + W(Work done by non-conservative
forces) = 0 (change of energy)
In here, the non-conservative force is the friction force
where f = uN (u =kinetic friction coefficient)
W= f x d = uNd ; N=mg
Delta E = 1/2 mV^2 -1/2mVi^2
umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term)
This will then give us:
1/2Vi^2-ugd = 1/2V^2
V^2 = Vi^2 - 2ugd
So plugging in our values, will give us:
V= Sqrt (5.6^2 -2.3^2)
=sqrt (26.07)
= 5.11 m/s
Answer:
According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>
<em></em>
Explanation:
For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:
v = d / t
where
- v is the speed of the spaceship
- d is the distance
- t is the time
Therefore,
d = v × t
d = (0.999 c)(4 light-years)
d = 3.996 light-years
This distance is<em> very slightly less than 4 light-years. </em>