Via half-life equation we have:
Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes. By plugging those values in we get:
There is 6.25 grams left of Ra-229 after 12 minutes.
5. Which of the following is NOT a course goal?
1 answer:
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Answer:
Explanation:
Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence
The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.
The kinetic energy is calculated by the following equation.
m= mass of the cars=26500kg
V=speed
E=kinetic energy
taking into account the above, the following equation is inferred
1= the cars are separated
2= the cars are togheter
E1=E2
where
m= mass of each car
V1= 1.2m/s
Va=3.5,m/S
m= mass of each car
V=speed (in m/s) of the three coupled cars after the first couples with the other two
Solving
the speed of the three coupled cars after the first couples with the other two is 2.245m/s
In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
