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    Electric Field for the circular path of a positively charged particle

    how i visualize fields is with a positive test charge, so in this environment wherever the test charge goes is the electric field lines.. if the electric field lines show positive test charges going towards that one area where it looks like they're attracted to a negative then in reality won't...
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    Electric Field for the circular path of a positively charged particle

    Here is picture. Answers is A. My attempt was that I thought if i were to place a positive test charge then it would go from top to bottom if there was a positive charge in the center it was avoiding and a positively charged particle at the top, but an electron at the bottom so it would avoid...
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    Some motion graph questions

    1. Homework Statement picture 1: picture 2: 2. Homework Equations 3. The Attempt at a Solution For the first picture I was thinking answer B because at the start velocity increases, but as terminal velocity is hit the velocity doesn't increase as much.. why is it graph A? For the...
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    Speed of ball vs Acceleration of the ball?

    @haruspex but wouldn't acceleration due to gravity at the start be greater than whatever drag forces existed? if acceleration is positive then velocity also must be positive.
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    Speed of ball vs Acceleration of the ball?

    1. Homework Statement 2. Homework Equations v = s/t a = v/t 3. The Attempt at a Solution My thinking was that through this small distance of air since velocity would increase due to the change in position changing at an increasing rate and thus since velocity is increasing then...
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    Why is the work zero here?

    @Young physicist Okay, so work is the transfer of energy due to force given in the equation W = F x D Obvciously yes if you apply force on a wall and lets say you're on rollerskates and you move backwards that DOES NOT mean you did work as lets say you applied 5N OF of force on the wall, but...
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    Efficiency of an Electric motor pulling a truck up an inclined plane

    @CWatters But the answer for this question as shown in the picture is "B" which is 24kJ and my working proves it too... what makes you say 24kJ is not correct?
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    Why is the work zero here?

    1. Homework Statement Is the work zero here because its a frictionless surface? If, so why? If not then whats the reason and why? 2. Homework Equations 3. The Attempt at a Solution I was originally thinking 1/2mv^2 .. but KE is zero right, but having it be mgh didn't make sense..
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    Efficiency of an Electric motor pulling a truck up an inclined plane

    @jbriggs444 I figured it out.. so seeing 2/3 thats just an efficiency.. knowing that 48 is the total work out that means 48 is 2 and 1 is 24 thus 3 is 72 so then 2/3 is also 48/72 and then the energy dissapated from 72kJ to 48kJ is 24kJ so there we go. my explanation sucks, but i know it now..
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    Efficiency of an Electric motor pulling a truck up an inclined plane

    1. Homework Statement See picture. 2. Homework Equations 3. The Attempt at a Solution I tried at first just doing (1/3)*(48kJ) to get the energy dissipated, but that would i=give me 16kJ
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