The answer to the first one is sublimation.
components of the speed of the coin is given as
![v_x = v cos60](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos60)
![v_x = 6.4 cos60 = 3.2 m/s](https://tex.z-dn.net/?f=v_x%20%3D%206.4%20cos60%20%3D%203.2%20m%2Fs)
![v_y = vsin60](https://tex.z-dn.net/?f=v_y%20%3D%20vsin60)
![v_y = 6.4 sin60 = 5.54 m/s](https://tex.z-dn.net/?f=v_y%20%3D%206.4%20sin60%20%3D%205.54%20m%2Fs)
now the time taken by the coin to reach the plate is given by
![t = \frac{\delta x}{v_x}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Cdelta%20x%7D%7Bv_x%7D)
![t = \frac{2.1}{3.2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2.1%7D%7B3.2%7D)
![t = 0.656 s](https://tex.z-dn.net/?f=t%20%3D%200.656%20s)
now in order to find the height
![h = vy * t + \frac{1}{2} at^2](https://tex.z-dn.net/?f=h%20%3D%20vy%20%2A%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
![h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2](https://tex.z-dn.net/?f=h%20%3D%205.54%20%2A%200.656%20-%20%5Cfrac%7B1%7D%7B2%7D%2A9.8%2A%280.656%29%5E2)
![h = 1.52 m](https://tex.z-dn.net/?f=h%20%3D%201.52%20m)
so it is placed at 1.52 m height
Charge of electron = 1.6×10−¹⁹
(1.6×10−¹⁹)(1×10²) (2e)
= 3.2×10−¹⁷ J
Answer:
A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.
Explanation:
Answer:
9155 years old
Explanation:
We use the following expression for the decay of a substance:
![N = N_0\,\,e^{-k*t}](https://tex.z-dn.net/?f=N%20%3D%20N_0%5C%2C%5C%2Ce%5E%7B-k%2At%7D)
So we first estimate the value of k knowing that the half-life of the C14 is 5730 years:
![N = N_0\,\,e^{-k*t}\\N_0/2=N_0\,\,e^{-k*5730}\\1/2 = e^{-k*5730}\\ln(1/2)=-k*5730\\k= 0.00012](https://tex.z-dn.net/?f=N%20%3D%20N_0%5C%2C%5C%2Ce%5E%7B-k%2At%7D%5C%5CN_0%2F2%3DN_0%5C%2C%5C%2Ce%5E%7B-k%2A5730%7D%5C%5C1%2F2%20%3D%20e%5E%7B-k%2A5730%7D%5C%5Cln%281%2F2%29%3D-k%2A5730%5C%5Ck%3D%200.00012)
so, now we can estimate the age of the artifact by solving for"t" in the equation:
![1/3=e^{-0.00012*t}\\ln(1/3)= -0.00012*t\\t=9155. 102](https://tex.z-dn.net/?f=1%2F3%3De%5E%7B-0.00012%2At%7D%5C%5Cln%281%2F3%29%3D%20-0.00012%2At%5C%5Ct%3D9155.%20102)
which we can round to 9155 years old.