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slamgirl [31]
2 years ago
6

A 2,493-kg car is moving down a road with a slope (grade of 14% at a constant speed of 13 m/s. what is the direction and magnitu

de of the frictional force? (define positive x in the forward direction, i.e., down the slope?
Physics
1 answer:
sweet [91]2 years ago
3 0
To calculate for the frictional force, we calculate first for the normal force (forcer perpendicular to the surface)
                               Fn = (2,493 kg) (9.8 m/s²) = 24243.51 N
Then, we multiply this value with the frictional constant which is not given in this problem. If the frictional coefficient is equal to x then, the frictional force is equal to 24243.51x. 
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_{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:

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_{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}.

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