Answer:
The mass of Uranium present in a 1.2mg sample is 
Explanation:
The ration between Uranium mass and total sample mass is:
For a sample of mass 1.2 mg, the amount of uranium is:

The centripetal acceleration a is 4.32
10^-4 m/s^2.
<u>Explanation:</u>
The speed is constant and computing the speed from the distance and time for one full lap.
Given, distance = 400 mm = 0.4 m, Time = 100 s.
Computing the v = 0.4 m / 100 s
v = 4
10^-3 m/s.
radius of the circular end r = 37 mm = 0.037 m.
centripetal acceleration a = v^2 / r
= (4
10^-3)^2 / 0.037
a = 4.32
10^-4 m/s^2.
Answer:
Velocity = 4.33[m/s]
Explanation:
The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.
![E_{M}=E_{p} + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2} \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]](https://tex.z-dn.net/?f=E_%7BM%7D%3DE_%7Bp%7D%20%20%2B%20E_%7Bk%7D%20%5C%5CE_%7Bp%7D%20%3D%20potential%20energy%20%5BJ%5D%5C%5CE_%7Bk%7D%20%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%20%3D%204%2A9.81%2A0.5%3D19.62%5BJ%5D%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2A4%2A%283%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D18%5BJ%5D%5C%5CTherefore%5C%5CE_%7BM%7D%20%3D18%2B19.62%5C%5CE_%7BM%7D%3D37.62%5BJ%5D)
All this energy will become kinetic energy and we can find the velocity.
![37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]](https://tex.z-dn.net/?f=37.62%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B37.62%2A2%7D%7B4%7D%20%7D%20%5C%5Cv%3D4.33%5Bm%2Fs%5D)
Answer: The part of the microscope that is the circular area is the APERTURE
I hope this helped!