Physics Question + Answer

[The Law]

The initial potential energy is equal to KqQ/2r. The final kinetic energy is 1/2mv^2.

 

QUOTE]

 

Formula mistake: The initial potential energy is KqQ/r

 

Hmm, but the initial charge starts off at a distance equal to 2r away.. so wouldn't it's distance from the source charge be 2r? That's the only thing I'm confused about.

[avenir001]

lol it's a typo...or they got confused by their own question. choice A is right.

[Avinash]

The initial potential energy is equal to KqQ/2r. The final kinetic energy is 1/2mv^2.

 

 

Formula mistake: The initial potential energy is KqQ/r

 

The question says that the distance is 2r. I'd have to agree with Law on this one.

[srindogg@hotmail.com]

The initial potential energy is equal to KqQ/2r. The final kinetic energy is 1/2mv^2.

 

 

 

The question says that the distance is 2r. I'd have to agree with Law on this one.

 

 

My bad. Didn't read the question fully.

[avenir001]

This always confuses me. When we start looking at only half of the projectile, I'm always worried the equations aren't working. I don't see why it wouldn't though, the acceleration is constant. Usually, I'm concerned when it comes to the "t" because if we look at half the projectile, then t must be halved. I would have solved it just like avinash did - there is no t in that case. I wonder if anyone could explain why that method is incorrect.

 

it's because when u use the equation Vf^2-Vi^2 =2ad for this, Vi actually refers to the initial velocity in the vertical direction (ie Vi = Vsin(theta))...so using this equation, max height would actually be Vsin(theta)^2 / 2g..which is not among the choices...their v^2 / 2g is vague...1/8 gt^2 is a better answer..but this question is tricky if u're not careful...i remember i too got it wrong the first time i did it 8374837 yrs ago

[The Law]

it's because when u use the equation Vf^2-Vi^2 =2ad for this, Vi actually refers to the initial velocity in the vertical direction (ie Vi = Vsin(theta))...so using this equation, max height would actually be Vsin(theta)^2 / 2g..which is not among the choices...their v^2 / 2g is vague...1/8 gt^2 is a better answer..but this question is tricky if u're not careful...i remember i too got it wrong the first time i did it 8374837 yrs ago

 

Thanks avenir, I was thinking about this while I lay in bed before I took a nap and then realized that too. Merci.

[avenir001]

Thanks avenir, I was thinking about this while I lay in bed before I took a nap and then realized that too. Merci.

 

*le eyerolllllll*

[The Law]

Whaat! I did! lol... but unlike the genius Persian princess... it took me two days to get it (vs instantly). Reminds me of a famous Iraqi joke.

 

One day, the animals in the wild gathered for a festival of song, dance, and peace. They drank, ate, and shared stories amidst one another's company. The chickens, lambs, cows, horses, foxes, pigs, and a single donkey gathered around while a hawk told a joke. Following the joke, the chickens ran around laughing, the lambs, cows, horses, foxes, and pigs all also joined them in this laughing circus. The donkey, however, sat there - silently and alone. A few days later life was normal at the farm. The foxes would try to eat the chickens, the cows would get ready to provide milk to the farmer, the pigs would run around their pig pen, and the horses would await their riders. In the middle of the bustle on the farm, suddenly, a loud laughter erupted. It was startling as it came from nowhere. The animals turned off their primal instinct and went to see what the fuss was about. It was the donkey. The cow asked the donkey, "what's the matter with you, are you crazy?"

 

The donkey replied, "No, don't be stupid! I just understood the joke from our night at the festival!"

 

 

That, my dear, is me.

[The Law]

 

Can someone please explain "in series" and "parallel" to me. Sometimes it is very easy to see, but other times, I don't get it.

 

Why are f + g in series... while f + d are in parallel? Does it have to do with the fact that there is a capacitor by e, so we ignore that lane?

 

Thanks in advance.

[The Law]

You have two tanks, describe the flow of water:

 

A) Water will move to tank with least volume

Water will move from tank B to Tank A becasue the gravitational forces are greater than the pressure forces

C) Water will move from B to A b/c pressure is greater at b than a

D) Water will move from A to B because pressure is greater at a than at b

 

(answer is

 

 

 

A) Water will move to tank with least volume

Water will move from tank B to Tank A becasue the gravitational forces are greater than the pressure forces

C) Water will move from B to A b/c pressure is greater at b than a

D) Water will move from A to B because pressure forces are greater than the gravitational forces

 

Answer: D

 

*brain explodes*

[avenir001]

oooh i love these tank questions!

what u have to remember is the fluid will move in the direction of higher gravitational energy to lower gravitational energy OR higher pressure to lower pressure. so u basically have to find the difference in gravitational forces between points a & b and also the difference between pressure forces between a & b...and then decide which difference is greater (gravitational or pressure)...and that will govern the direction of fluid flow.

-now to measure the gravitational differential between points a & b, u measure from the bottom (to points a & ...then take the difference.

-to measure pressure differential between points a & b, u measure from the top (to points a & ...then take the difference.

-compare the two differentials and whichever one's bigger decides the direction of flow.

[avenir001]

 

Can someone please explain "in series" and "parallel" to me. Sometimes it is very easy to see, but other times, I don't get it.

 

Why are f + g in series... while f + d are in parallel? Does it have to do with the fact that there is a capacitor by e, so we ignore that lane?

 

Thanks in advance.

 

ya exactly...it's cuz of the capacitor...also if u just imagine the current flowing, then when it gets to the junction between resistors d & f, the current will divide (ie some of it will go thru resistor d and the rest thru resistor f, depending on their relative strengths)...whenever u have current dividing like this, it means u have resistors in parallel. but all current going thru f will also pass thru g..whenever u have resistors in series, the same current will pass thru all.

[The Law]

oooh i love these tank questions!

what u have to remember is the fluid will move in the direction of higher gravitational energy to lower gravitational energy OR higher pressure to lower pressure. so u basically have to find the difference in gravitational forces between points a & b and also the difference between pressure forces between a & b...and then decide which difference is greater (gravitational or pressure)...and that will govern the direction of fluid flow.

-now to measure the gravitational differential between points a & b, u measure from the bottom (to points a & ...then take the difference.

-to measure pressure differential between points a & b, u measure from the top (to points a & ...then take the difference.

-compare the two differentials and whichever one's bigger decides the direction of flow.

 

If only EK could explain like you! Thanks.

[The Law]

ya exactly...it's cuz of the capacitor...also if u just imagine the current flowing, then when it gets to the junction between resistors d & f, the current will divide (ie some of it will go thru resistor d and the rest thru resistor f, depending on their relative strengths)...whenever u have current dividing like this, it means u have resistors in parallel. but all current going thru f will also pass thru g..whenever u have resistors in series, the same current will pass thru all.

 

Let's say we remove F and G and E....

 

Would C and G be in parallel?

 

Now, let's say we remove the capacitor...

 

Would F and G now be parallel? E and G would be parallel too, oui?

 

Do we always ignore the lanes with the capacitor?

 

*must kill physics*

[The Law]

 

Can someone please explain "in series" and "parallel" to me. Sometimes it is very easy to see, but other times, I don't get it.

 

Why are f + g in series... while f + d are in parallel? Does it have to do with the fact that there is a capacitor by e, so we ignore that lane?

 

Thanks in advance.

 

I have a question about this still.

 

Let's say the resistance of each resistor is:

a = 4

b = 2

c = 12

d = 4

f = 4

g = 2

e = 2

battery is 90v

 

I tried this question, and ALMOST got it right... but just came short. What is the current through resistor d?

 

A) 1A

2A

C) 3.75A

D) 5.25A

 

First, I run a tally of all the resistances. Initially we have 4(a) + 2(... then the current splits. First it splits with d and f, the resistance of this is equal 1/4 + 1/4 = 2. Both of these are in series with g so 2 + 2 brings our total effective resistance to 4. Then, this is in parallel with c 12 so we get a final resistance of 3.

 

Now, our running tally is 4 + 2 + 3 = 9.

 

Using the voltage of the battery, we can get the current in the circuit. I=V/R so 90V/9 ohms = 10 A current.

 

So, I understand the 10A flows directly past a and b. It then must split up with c and d/f/g. What I am confused about is when dealing with this split up in current, we have to find the combined effective resistance of d/f/g and compare it with c. I did that and found that d/f/g have an effective resistance of 4, while c has a resistance of 12.

 

So, the 10 A will flow through d/f/g three times as much as through c. So I set the amount flowing through c equal to x:

 

10 = 3x + x

x = 2.5 <--flowing through c

 

This means 7.5 A has to flow through d/f/g. Now, this is the part that got me. They said that 1/2 of the 7.5 will flow through d and half will flow through f. Why is this the case? When we were dealing with the previous situation, I had to calculate the total effective resistance of d/f/g before I could figure this out... how come now I can just ignore g when figuring out how much current will flow through d vs f? I have a feeling it has to do with the fact that d might be in series with g, but I'm not entirely sure.

 

I was initially confused with why C was in parallel with d/f AND g... but then I figured the current splits up when it gets to the junction between d and f... and since g is part of the circuit with f, we have to include it in the resistance measurement <-- is this reasoning right?

 

So using that reasoning, I said d and f are in parallel... but wouldn't the fact that f and g being in series with each other affect the amount going through d? The only thing I could figure that would change this situation is if d is in fact - also in series with g. I think it is now, but I'm not sure!

 

If someone clears this up, I'd be VERY HAPPY! Sorry for the long winded question.

 

 

By the way, for this interested, this is question 837 in EK 1001 questions set.

[avenir001]

just take out the branch with the capacitor and redraw the circuit in a simpler way. in fact, d & f are in parallel with each other..and together they're in series with g....and the whole thing is then in parallel with c.

the 7.5A current going into the d/f junction will divide in half cuz d & f have equal resistances so they each get half the current...then the current coming out of d/f will again recombine (to become 7.5A) and pass thru g.

[avenir001]

Let's say we remove F and G and E....

 

Would C and G be in parallel?

 

Now, let's say we remove the capacitor...

 

Would F and G now be parallel? E and G would be parallel too, oui?

 

Do we always ignore the lanes with the capacitor?

 

*must kill physics*

 

if u remove f, d, and e...c & g are in parallel

hmmm...let's leave the capacitor in there ...i think it possibly becomes a complicated circuit if u take it out...some circuits are too complicated to be classified as parallel or series...they're kind of mixed..u wont get those on the mcat.

[ploughboy]

let's leave the capacitor in there

 

Let's not! You were right in the previous post...

 

This is a DC circuit, so the resistance of the (perfect) capacitor at steady state is infinite. So we can ignore the cap and resistor e.

 

The circuit can then be simplified into three equivalent resistors:

 

The first one is a+b = 6 ohm

 

The second one is just resistor c = 12 ohm

 

The third one is the combination of d and f in parallel, all of which is in series with resistor g. So that becomes 1/(1/4 + 1/4) + 2 = 4 ohms.

 

 

Ok, this is hard to describe without diagrams, but I'll try my best. The simplification above can *itself* be simplified to the 6 ohm equivalent resistor in series with the parallel combination of the 12 and 4 ohm resistors. In other words, a 6 ohm resistor in series with a 3 ohm resistor ( since 1/(1/12 + 1/4) = 3).

 

So, 60 volts of your power supply drops across the 6 ohm resistor, and the remaining 30 volts drops across the 3 ohm equivalent resistor.

 

Now remember that the 3 ohm equivalent resistor is actually a parallel combination of a 12 ohm real resistor and a 4 ohm equivalent resistor. Each arm of the parallel path must drop the 30 volts.

 

The 4 ohm equivalent arm is actually two 4 ohm resistors in parallel followed by a series 2 ohm resistor (it is d || f + g), which is the same thing as a 2 ohm resistor in series with another 2 ohm resistor. So half of the 30 volts drops across the d||f combination, while the other half drops across resistor g.

 

So 15 volts (30/2) drops across d||f, which has an equivalent resistance of 2 ohm. By Ohm's law, 7.5 amperes of current flows through d||f. 3.75 amps flows through each arm (edit: since the elements have equal resistance in this case).

 

Clear as mud? I'm so sorry, I wish I could draw a diagram....

[avenir001]

Let's not! You were right in the previous post...

 

This is a DC circuit, so the resistance of the (perfect) capacitor at steady state is infinite. So we can ignore the cap and resistor e.

 

 

oh i know, i know...but i think in the previous question law was doing a thought expt asking what if we just pretend the capacitor doesn't exist, but the wire and resistor e are still there...in which case, i found it confusing to keep track of which resistors are in series/parallel with what

 

btw, nice to have u visit us here

[The Law]

Indeed, thanks for the post ploughboy and avenir (la genieuse - is that even a french word? lol).

 

Anyway, so would I be right in saying that BOTH F and D are in series with G? The reason I ask is because I figured out that 7.5 amps was flowing through the G||F + D part of the circuit... but was confused about how to split up the 7.5 amps. The dumb thing I did was I found F + D's effective resistance first (as they are in series) and then used this resistance as being in parallel with G. Upon doing this, I split up the current based on the resistances.

 

I had done: G || F + d

 

Obviously this was wrong since the 7.5 amps splits to G and F evenly... so just want to make sure that I understand it properly now. G and F are both in parallel with d?

 

This was kind of like when I tried to figure out the current of the entire circuit. To do this, I had to get the circuit to its simplest form (4 + 2 + 3). To do that, I said (sorry my notation is awful, but think it gets the point across): C || (d || f + g).

 

C is not in series with g (obviously, since what flows through it never flows through g)... so hear, we did have to add g to d||f. On the other hand, what flows through d eventually DOES flow through g (so they must be in series) and that's why I got the question wrong (I think).

 

Hopefully I am understanding properly now!

 

Also, pb - I didn't use your technique when I initially solved it, but it brings up an interesting point. So when we simplify it to all being in series 4 + 2 + 3... we can also split up the voltage across the corresponding components? When you returned it to being more complicated, you said 30 V goes across the 3 ohm branch of the circuit (i.e. 1/3 of the voltage since 3 ohm's is 1/3 of the total resistance)... so does this mean the more resistant resistors get proportionally more voltage?

 

Thanks a lot for everyone's input!

[avenir001]

Anyway, so would I be right in saying that BOTH F and D are in series with G? The reason I ask is because I figured out that 7.5 amps was flowing through the G||F + D part of the circuit... but was confused about how to split up the 7.5 amps. The dumb thing I did was I found F + D's effective resistance first (as they are in series) and then used this resistance as being in parallel with G. Upon doing this, I split up the current based on the resistances.

 

I had done: G || F + d

 

Obviously this was wrong since the 7.5 amps splits to G and F evenly... so just want to make sure that I understand it properly now. G and F are both in parallel with d?

 

uhh no...G is in series with (D ll F).

 

This was kind of like when I tried to figure out the current of the entire circuit. To do this' date=' I had to get the circuit to its simplest form (4 + 2 + 3). To do that, I said (sorry my notation is awful, but think it gets the point across): C || (d || f + g).

 

C is not in series with g (obviously, since what flows through it never flows through g)... so hear, we did have to add g to d||f. On the other hand, what flows through d eventually DOES flow through g (so they must be in series) and that's why I got the question wrong (I think).

 

Hopefully I am understanding properly now! [/quote']

 

yes, good.

 

Also' date=' pb - I didn't use your technique when I initially solved it, but it brings up an interesting point. So when we simplify it to all being in series 4 + 2 + 3... we can also split up the voltage across the corresponding components? When you returned it to being more complicated, you said 30 V goes across the 3 ohm branch of the circuit (i.e. 1/3 of the voltage since 3 ohm's is 1/3 of the total resistance)... so does this mean the more resistant resistors get proportionally more voltage? [/quote']

 

i'm not pb, but yes...the stronger the resistor, the more voltage drops across it...remember, V = IR.

 

so u start with 90V and u keep dropping voltage across each resistor (depending on their relative strengths) until u're left with zero voltage at the 'end' of the circuit.

current at the end on the other hand must add up to what u started with (ie charge is conserved)

 

ps. i drew a diagram i'll email u.

[The Law]

uhh no...G is in series with (D ll F).

 

 

 

yes, good.

 

 

 

i'm not pb, but yes...the stronger the resistor, the more voltage drops across it...remember, V = IR.

 

so u start with 90V and u keep dropping voltage across each resistor (depending on their relative strengths) until u're left with zero voltage at the 'end' of the circuit.

current at the end on the other hand must add up to what u started with (ie charge is conserved)

 

ps. i drew a diagram i'll email u.

 

 

Alright, gotcha! diagram was indeed "pure gold"

[AndyDude]

Good going Law, I just stumbled upon the same problem as I near the end of Examkrackers 1001. lol.

[The Law]

Good going Law, I just stumbled upon the same problem as I near the end of Examkrackers 1001. lol.

 

That was a tricky problem set, but I think it really tests most of the concepts.

 

After optics, it's on to bio/orgo and verbal for me. Grrrrrreaaat.

[The Law]

A lens is manufactured in such a way to allow the object and the image to be the same distance from the lens. If the lens is not flat, the only way this could be true is if the lens were:

 

A) a diverging lens with the object at the focal distance

a diverging lens with the object at twice the focal distance

C) a converging lens with the object at the focal distance

D) a converging lens with the object at twice the focal distance

 

 

Answer:_D_

 

 

If someone could explain how to solve this, it would be awesome. I read it in EK, and it kind of makes sense, but I wonder if someone has a strategy to solve this type of problem.