Physics what is the aim of this experiment
1 answer:
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Answer:
The product of the decay its Sulfur-32
Explanation:
Phosphorus-32 ( lets write it , where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:
where its the electron, and
the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).
So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:
.
.
The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its
.
and the product of such decay its Sulfur-32
Answer:
a) Check Explanation
b) Check Explanation
c) The rate at which water is being pumped into the tank = 2.631 cm³/min
Explanation:
Let the rate of flow of water into the tank be k cm³/min
a) The image of the conical tank is presented in the attached image
Note, the radius and height of a cone are related through the similar triangles principle.
As shown in the attached image, it is evident that
r/h = 3/10
r = 3h/10 = 0.3 h
b) The quantities given in the problem.
- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3
- total height of the tank, H = 10 cm
- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm
- rate at which water is leaking from the tank = 1.5 cm³/min
- water is being pumped into the tank at constant rate of k cm³/min
- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min
c) volume of the tank at any time = πr²h/3
Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)
dV/dt = k - 1.5
V = πr²h/3 and r = 0.3 h, r² = 0.09 h²
V = 0.03πh³
dV/dt = (dV/dh) × (dh/dt)
dV/dh = 0.09π h²
dV/dt = 0.09π h² (dh/dt)
dV/dt = k - 1.5
0.09π h² (dh/dt) = k - 1.5
But at h = 2 cm, (dh/dt) = 1.0 cm/min
0.09π h² (dh/dt) = k - 1.5
0.09π 2² (1) = k - 1.5
k - 1.5 = 1.131
k = 1.5 + 1.131 = 2.631 cm³/min
