Electron in an electirc field.

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To determine the potential difference required to accelerate an electron to a speed of 1.80×10^6 m/s, the relevant equation is E = 0.5mv^2 = eV. The user initially miscalculated by using the kinetic energy formula without converting the energy from Joules to Volts. After realizing the error, they acknowledged the need for proper unit conversion to find the correct potential difference. The discussion highlights the importance of unit consistency in physics calculations. Understanding these concepts is crucial for solving problems involving charged particles in electric fields.
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Homework Statement



Across what potential difference does an electron have to be accelerated in order to reach speed v=1.80×106 m/s?

Homework Equations



I checked to see if relativity was important and it was not
E=.5 m v^2=eV


The Attempt at a Solution




I don't know if I am mistaken in what I am trying to find but I thought the potential difference in this case was .5mv^2 But when I plug in 9.109x10^-31 kg in for the mass and 1.8x10^6 m.s for the velocity I don't get the right answer. I do not know what I am doing wrong.
 
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I figured out what I was doing wrong. It was in Joules and I needed to convert to Volts.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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