contestada

If you had carried out the algebra using variables before plugging numbers into your expressions, you would have found that (vf)α=−2qαΔVmα−−−−−−−√, where ΔV is measured in volts. To verify that this expression for (vf)α has the correct units of velocity, you need to perform some unit analysis. Begin by finding the equivalent of a volt in terms of basic SI units. What is a volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (C)? Express your answer using the symb

Respuesta :

aleperezme

Answer:

Step-by-step explanation:

Volt is energy per electric charge

[tex]1V=1J/C[/tex]

and the energy units in terms of meters, seconds and kilograms are:

[tex]1J= 1kg\frac{m^{2}}{s^{2}}[/tex]

So the unit of volt is

[tex]1V= kg \frac{m^{2}}{Cs^{2}}[/tex]

So the velocity expression (it can not be read completly) but i guess that it could be as follows

[tex]vf_{a} = -2\sqrt{\frac{q_{a}\Delta V}{m_{a}}}[/tex]

So the unit analysis would be

[tex]vf_{a} = \sqrt{\frac{C kg\frac{m^{2}}{Cs^{2}}}{kg}}}[/tex]

note that the coulomb unit of numerator is cancelled from the denominator. The same occurs for kg unit.

[tex]vf_{a} = \sqrt{\frac{m^{2}}{s^{2}}}[/tex]

Applying square root

[tex]vf_{a} = \frac{m}{s}[/tex]

Otras preguntas