Recent content by annjolino

excellent thank you

Homework Statement rearrange the formula d=Vft-1/2at^2 to solve for a Homework Equations d = Vft-1/2at^2 d+ 1/2at^2 = Vft 1/2at^2 = Vft-d 1/2a = Vft-d / t^2 then does a = 2 (Vft-d / t^2) The Attempt at a Solution

part c of the question is, if the acceleration of a charged particle between the plates is a when it is halfway between the plates, why will the particle acceleration experience the same acceleration if it is only one quarter the distance between the plates? defend your answer... Ok so i...

i just noticed a little hint in the question 'NB 1Vm-1 is equivalent to 1NC-1 so from the Vm-1 does that mean that the F=V/d? could it be that simple

does the F in this question equal the force of the electric field or does the F mean the E int the equation Electric field = force/charge... i know that might osund a bit weird but the way i read the question F = E= f/c... i think i am going the long way about this but so far i have arranged...

Hmmm bad day with spelling i guess...:rolleyes: work done = force x distance

Homework Statement a certain parrallel plate capacitator has a plate separation d and a potential difference between the plates of V. If the electric field between the plates is uniform a) What is the strength of the electric field F Homework Equations Coloumbs Law F=k(q1q2/d^2)...

thank you so much i can make sense of this when its explained to me... its a little tough to work out by myself...:biggrin:

ok so this is what i got hopefully its the last for this... i am assuming that Vx is the velocity in the horizontal direction as we have said throughout the problem... so here goes Vo^2 - Vx^2/2 = gh 18ms-10.32ms/2 = 9.8ms x h 108.7488/9.8ms = h h= 11.1m please let that be right...

Quick question are the units for the gravitational force N ? seeing how in the equation there are 2 kg / m^2

YAY! thank you so much for your help your an angel o:)

ok now when i put the values in the new equation -Tc= E x Th - Th -Tc= (0.62 x 600K) - 600K -Tc = -228K could i then make them both positive as they would cancel each other out?? Tc= 228K

so then would it be closer to Tc= E x Th - Th

Thank you that feet really threw me off... i can understand what formula you were using i couldn't understand how to rearrange that KE+PE equation to get what i wanted.. no the question only states that it was thrown from the ground at the 55* angle so i don't thnk we have to factor any extra...

Homework Statement ideal heat engine takes in heat Qin at a temp Th. It exhausts heat Qout a) how much work is done by the engine b) What is the effiency of the engine c) what is the exhaust temp of the engine Homework Equations W = Qin - Qout Effiency = Qout/Qin E = Th-Tc/Th...