The formula for half-life is:

Where A is the amount of iodine-131 initially and after 40 days, t is time, h is half-life of the isotope. Let's plug in our values to the equation:

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.
Answer:
force is decreased by a factor of 4.
Explanation:
According to the Newton's law of gravitation, the force of gravitation between the two object is inversely proportional to the square of distance between them. Now the distance is doubled, so the force between the two objects becomes one forth.
Force is decreased by a factor or 4.
<u>Answer</u>
1 1/3 m/s
<u>Explanation</u>
The momentum before collision and after collision is always conserved.
m₁v₁ = m₂v₂
(0.05×0.6) + (0.03×0) = (0.03 × v) + (0.05 × -0.2)
0.03 + 0 = 0.03v - 0.01
0.03v = 0.03+0.01
0.03v = 0.04
v = 0.04/0.03
= 4/3
= 1 1/3 m/s
Answer:
1) At the highest point of the building.
2) The same amount of energy.
3) The kinetic energy is the greatest.
4) Potential energy = 784.8[J]
5) True
Explanation:
Question 1
The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.
![E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20m%2Ag%2Ah%5C%5CE_%7Bp%7D%20%3D%202%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D%20784.8%5BJ%5D)
Question 2)
The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.
![E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5C%20E_%7Bp%7D%20%3D%202%2A9.81%2A20%5C%5C%20E_%7Bp%7D%20%3D%20392.4%5BJ%5D%5C%5C)
And the kinetic energy will be:
![E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%20%20where%3A%5C%5Cv%20%3D%20%20velocity%20%3D%2019.8%5Bm%2Fs%5D%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%5C%5C%20%20E_%7Bk%7D%3D392.04%5BJ%5D)
Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.
Question 3)
As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.
![E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DE_%7Bp%7D%3Dm%2Ag%2Ah%20%3D%202%2A9.81%2A40%5C%5C%20%20E_%7Bk%7D%20%3D%20784.8%5BJ%5D%5C%5C%20E_%7Bk%7D%20%3D%200.5%2A2%2A%2828%29%5E%7B2%7D%5C%5C%20E_%7Bk%7D%20%3D%20784%20%5BJ%5D)
Question 4)
It can be easily calculated using the following equation
![E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%3D2%2A9.81%2A40%5C%5CE_%7Bp%7D%20%3D784.8%5BJ%5D)
Question 5)
True
The potential energy at 20[m] is:
![E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]](https://tex.z-dn.net/?f=E_%7Bp%7D%3D2%2A9.81%2A20%5C%5C%20E_%7Bp%7D%3D%20392.4%5BJ%5D%5C%5CThe%20kinetic%20energy%20is%3A%5C%5CE_%7Bk%7D%3D0.5%2A2%2A%2819.8%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D392%5BJ%5D)