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Tasya [4]
3 years ago
8

What is the voltage, V2, in units of Volts, across resistor R2 in the circuit shown below where VS = 4V, R1 = 14 Ohms and R2 = 3

5 Ohms?

Physics
1 answer:
svetlana [45]3 years ago
3 0
<h2>Correct answer:</h2>

\boxed{v_{out}=2,85V}

<h2>Explanation:</h2>

We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage v_{out} that is a fraction of its input voltage v_{in}. So we can use the formula:

v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}, \ where \ v_{in}=v_{s}=4V \\ \\ \therefore v_{out}=\frac{35}{14+35}(4) \\ \\ \therefore v_{out}=2,85V

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Answer:

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Explanation:

Phosphorus-32 ( lets write it _{15}^{32}P, 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:

_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}

where e^- its the electron, and \bar{\nu_e} 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:

_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}.

_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}.

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

_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}.

and the product of such decay its Sulfur-32

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determine the maximum angle theta for which the light rays incident on the end of the optical fiber of radius 1 mm are subhect t
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Answer:

Explanation:

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