Proton Motion in Electric Field

AI Thread Summary
A proton released in a vacuum between two metal sheets with a potential difference of 40 volts accelerates towards the negatively charged sheet. The kinetic energy (KE) of the proton is calculated using the formula KE = qV, resulting in a value of 6.4 x 10^-18 joules. The velocity of the proton just before it strikes the negative sheet is determined to be approximately 87548 m/s. The calculations for both KE and velocity are confirmed to be correct. This discussion highlights the relationship between electric potential, kinetic energy, and velocity in the context of charged particles in an electric field.
XTEND
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Homework Statement


2 large metal sheets are separated by a potential of 40 volts by a vaccum. When the proton particle (mass 1.67x10e-27 kg)is released close to the (+) sheet it moves closer to the (-) sheet by way of electric field. What's the KE right before the proton smacks the other sheet? And what's the velocity of before striking the (-) sheet?


Homework Equations



KE=>q*v and EK=mv^2/2

The Attempt at a Solution



KE= 1.6x10e-19(40)=6.4x10e-18

v= ?
 
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What's the formula for the kinetic energy of an object (in terms of the object's velocity)?
 
Can anyone help me with this?

I think the velocity is 87548 m/s. Is this correct?
 
XTEND said:
Can anyone help me with this?

I think the velocity is 87548 m/s. Is this correct?

Your answer is correct.
 
what about the first part for KE?
 
XTEND said:
what about the first part for KE?

If that isn't correct, you wouldn't have gotten the right answer for v. So yes, KE=qV, and your calculation is right as well.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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