Physics question thread

[Shrike]

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All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

 

[gridiron]

Pressure = Force/Area, correct?
I was always under the assumption that the way a bullet penetrates anything is because of the high pressure it applies. But a bullet applies no force, right? The bullet is not accelerating when it hits an object. Therefore, a bullet can't apply a force or apply any pressure to an object it hits.

So, is the reason why bullets can go through objects because its kinetic energy is converted into other forms of energy, like heat, and work (i.e. ripping apart wood, stone, flesh, etc.)?

Is there some term for momentum/area? Because it seems intuitive that the more pointed the bullet, the better it will penetrate an object, so there must be some way for the area over which the energy is exerted can be considered.

Can anyone confirm this?

To answer your last question first: There is a momentum term. Consider the change in momentum as a bullet strikes its object: if the area over which it strikes is small it will penetrate to a greater extent because the force of the bullet will act over a small interval of time. If the force were spread over a large area, so the time over which the force acts is increased, the energy will be absorbed and penetration should be minimal. This is how bullet proof vests work. The material, like polyethylene plates, catches the bullet and spread the force of the bullet over a large area. Since the time over which the force acts is increased, the bullet is brought to stop. This is why when you throw an egg to the ground it breaks but if you throw it against padding like cloth it stays intact--this is a form of impulse. A bullet will penetrate an object like wood because its energy is converted into other forms but also because the force of the bullet acts on a small area over a short time. So, a bullet does apply a force because it has acceleration. Newton’s third law states that the force the bullet applies to the block is equal but opposite in magnitude to the force the wood applies to the bullet. A good example of this is seen in a football game. When a defensive player hits the opposing player very hard, sometimes the defensive player falls back as a result. Why? Because the force with which the defensive player hit is opponent is exerted back onto him in equal but opposite magnitude. Pressure does play a role in reasoning why a bullet penetrates an object but a better explanation would be using Newton’s third law, momentum and impulse. I hope this helps!!

 

[hmcalley]

I have a few questions for future MCAT takers. What topics would you like to see with greater explanation in the physics thread? Would you like more real world applications to problems in order to getter a better grasp of the material? Suggestions are welcome!

Electrostatics... (or electrodynamics... I don't know what's the difference...) Anything to do with explaining equations like W=q(delta V), F=qE... and concepts like electric field, electric potential...

(I was coming in here to look at the physics topic explanation thread but didn't see much about them and then I saw your post : )

Do people usually have a hard time on this topic because I have been staring at this electrodynamics chapter from NOVA for couple of hours now (plus after taking notes and re-reading the electrostatics chapter from Kaplan for the last day or so) and still feel like I know nothing... I don't get V and W : ( ? (I did the topical test from Kaplan and got a 56%... When I was doing the passages, I felt like I didn't know English...What the heck is going on...!!!! Sorry for venting here... at least part of my post is relevant, right?... Now going to do some more electrodynamics problems..)

 

[gridiron]

Electrostatics... (or electrodynamics... I don't know what's the difference...) Anything to do with explaining equations like W=q(delta V), F=qE... and concepts like electric field, electric potential...

(I was coming in here to look at the physics topic explanation thread but didn't see much about them and then I saw your post : )

Do people usually have a hard time on this topic because I have been staring at this electrodynamics chapter from NOVA for couple of hours now (plus after taking notes and re-reading the electrostatics chapter from Kaplan for the last day or so) and still feel like I know nothing... I don't get V and W : ( ? (I did the topical test from Kaplan and got a 56%... When I was doing the passages, I felt like I didn't know English...What the heck is going on...!!!! Sorry for venting here... at least part of my post is relevant, right?... Now going to do some more electrodynamics problems..)

Ok! I will post an explanation with real world examples in the next couple of days.

 

[UMP]

Electrostatics... (or electrodynamics... I don't know what's the difference...) Anything to do with explaining equations like W=q(delta V), F=qE... and concepts like electric field, electric potential...

(I was coming in here to look at the physics topic explanation thread but didn't see much about them and then I saw your post : )

Do people usually have a hard time on this topic because I have been staring at this electrodynamics chapter from NOVA for couple of hours now (plus after taking notes and re-reading the electrostatics chapter from Kaplan for the last day or so) and still feel like I know nothing... I don't get V and W : ( ? (I did the topical test from Kaplan and got a 56%... When I was doing the passages, I felt like I didn't know English...What the heck is going on...!!!! Sorry for venting here... at least part of my post is relevant, right?... Now going to do some more electrodynamics problems..)

topicals from Kaplan are meant to be hard... I usually got in that range too.

I highly recommend EK (which should be available through the SDN bookstore) because they have a great analogy comparing circuits to pipes and gravity (it makes sense after you read it)... basing off of that analogy everything else is right down hill 👍 And what they teach you in EK, is all you need to know for the MCAT

 

[superwillis]

Just remember.. decceleration is still acceleration!

Yeaaahh. I feel kinda ******ed for forgetting that. Thanks for the answer tho!

 

[hmcalley]

topicals from Kaplan are meant to be hard... I usually got in that range too.

I highly recommend EK (which should be available through the SDN bookstore) because they have a great analogy comparing circuits to pipes and gravity (it makes sense after you read it)... basing off of that analogy everything else is right down hill 👍 And what they teach you in EK, is all you need to know for the MCAT

It is good to know I am not the only one doing really bad in the topicals. But I am pretty worried because when I was doing the passages, I felt like I didn't understand what the heck they were asking. I guess I have to just keep practicing. Thanks for the suggestion with EK, UMP. And looking forward to BioMedEngineer's addition to the physics topics!

I have an electric field question...

Two parallel metal plates are charged in order to create a uniform electric field between them which points up (see figure in MS doc.). A water molecule can be modeled as a simple dipole, that is having a negative end (the oxygen atom) and a positive end (the hydrogen atoms). If a water molecule is placed between the two plates, which orientation would it take to minimize its energy? (Please open MS doc to see answer choices).

I think I understand why A and B are high in energy... because in that orientation, the water molecule would experience a torque, clockwise and counter closckwise, respectively.

But I am not sure why D is higher in energy than C??? For both C and D, the net force on the water molecule is 0, right? For the water molecule in C, wouldn't it be stretched apart... and D, the molecule be compressed...

I am confused!!!

The answer given in the book doesn't help much. It says, C is correct. The positive end of the water molecule would experience a force in the same direction as the electric field, that is, up. The negative end of the molecule would experience a force in the opposite direction of the electric field, that is, down. The energy would be minimized, therefore, if the hydrogen end pointed up and the oxygen end down, so C is the answer.

 

[heymanooh1]

It is good to know I am not the only one doing really bad in the topicals. But I am pretty worried because when I was doing the passages, I felt like I didn't understand what the heck they were asking. I guess I have to just keep practicing. Thanks for the suggestion with EK, UMP. And looking forward to BioMedEngineer's addition to the physics topics!

I have an electric field question...

Two parallel metal plates are charged in order to create a uniform electric field between them which points up (see figure in MS doc.). A water molecule can be modeled as a simple dipole, that is having a negative end (the oxygen atom) and a positive end (the hydrogen atoms). If a water molecule is placed between the two plates, which orientation would it take to minimize its energy? (Please open MS doc to see answer choices).

I think I understand why A and B are high in energy... because in that orientation, the water molecule would experience a torque, clockwise and counter closckwise, respectively.

But I am not sure why D is higher in energy than C??? For both C and D, the net force on the water molecule is 0, right? For the water molecule in C, wouldn't it be stretched apart... and D, the molecule be compressed...

I am confused!!!

The answer given in the book doesn't help much. It says, C is correct. The positive end of the water molecule would experience a force in the same direction as the electric field, that is, up. The negative end of the molecule would experience a force in the opposite direction of the electric field, that is, down. The energy would be minimized, therefore, if the hydrogen end pointed up and the oxygen end down, so C is the answer.

You need to figure out how the electric dipole moment vector (p) is aligned with respect to the electric field vector (E) from the choices given.

Then use the definition of electric potential energy (U) for a dipole in an electric field. U = negative of the dot product between the vectors p and E = - p dot E = -p E cos theta, where theta is the angle between the vectors p and E.

 

[hmcalley]

You need to figure out how the electric dipole moment vector (p) is aligned with respect to the electric field vector (E) from the choices given.

Then use the definition of electric potential energy (U) for a dipole in an electric field. U = negative of the dot product between the vectors p and E = - p dot E = -p E cos theta, where theta is the angle between the vectors p and E.

I thought that any dipole that experiences an external electric field will align its dipole moment with that electric field. (This is based on the idea that when placed in an external electric field, a dipole feels no translational force because net force is equal in magnitude but in opposite direction. But, the dipole will feel nonzero torque. Torque=Fdsin theta=qEdsin theta=pEsin theta. Therefore, torque will make dipole rotate so dipole moment, ρ, aligns with the electric field.)

I don't know if my reasoning is correct... not good at this physics stuff...

If my explanation works, then eventually any molecular orientation (given the same molecule) will have the same U according to U=-pEcos theta, right? Then, won't all the choices have the same energy?

Is there a different way to explain this problem (post#606) because the book that I got the question from didn't talk about electric dipoles at all?

I still really appreciate your response though because I didn't think about the problem this way. Thanks!

 

[heymanooh1]

I thought that any dipole that experiences an external electric field will align its dipole moment with that electric field. (This is based on the idea that when placed in an external electric field, a dipole feels no translational force because net force is equal in magnitude but in opposite direction. But, the dipole will feel nonzero torque. Torque=Fdsin theta=qEdsin theta=pEsin theta. Therefore, torque will make dipole rotate so dipole moment, ρ, aligns with the electric field.)

I don't know if my reasoning is correct... not good at this physics stuff...

If my explanation works, then eventually any molecular orientation (given the same molecule) will have the same U according to U=-pEcos theta, right? Then, won't all the choices have the same energy?

Is there a different way to explain this problem (post#606) because the book that I got the question from didn't talk about electric dipoles at all?

I still really appreciate your response though because I didn't think about the problem this way. Thanks!

You are getting hung up on torque when the question is asking for the the potential energy of the dipolar molecule in a uniform electric field vector E. Why would you use the torque equation for a dipole in an electric field to calculate U?

The problem with using the torque equation is that it directly calculates the work done by E on the dipole. In order to do that, the problem would need to give an initial and final theta value since W (on a dipole by vector E) = - integral (p E sin theta d-theta) = - p E (cos theta final - cos theta initial). How can U be calculated if theta initial and final aren't given?

Like I said before, look at the orientations of the of dipole moment vector p for each choice with respect to the electric field vector. By definition, p within in a dipolar molecule goes from partial negative to partial positive charge. Since water has two atoms that are partial positive charges, the superpostion principle for vectors needs to be used to find the net directional orientation of p for water.

Once you know the vectoral orientation of p with respect to vector E for each choice, the potential energy of the dipole can be calculated using U = - p E cos theta. From the orientation of the molecules as given in the choices, two of them will have U = 0, one will have Umax and one will have Umin.

The tendency of dipoles in an electric field is to orient vector p parallel to E because it achieves the lowest energy state. And systems, whether physical, chemical or biological, always prefer minimum energy, relative to some reference value, because it's the most stable.

 

[tik-tik-clock]

Can someone please explain me this

For the vertical component, if the angle is increased from 45°, the vertical component of the initial velocity of the clown would increase

Is this simply due to the formula v=vo sin theta

 

[tik-tik-clock]

If the angle deviates from 45°, then, the horizontal distance traveled would decrease.

can someone explain me this as well please

 

[kakashi74]

Answer to your first question:
Yes.

Second question: the maximum distance an object will travel is if it is fired at 45 degrees. Anything other than 45 degrees and it will travel less than maximum. It's too complicated to explain the math behind this, but you can think of it as a balancing act between gravity and Vx. If the object takes off at too shallow of an angle, gravity will pull it down and it doesn't take long for the object to fall to the ground. If you fire it at too steep of an angle, it will stay in the air longer, but there is not enough Vx to make it travel the horizontal distance.
45 degrees is right in the middle.

 

[tik-tik-clock]

thanks kakashi, that makes a lot more sense🙂

 

[tik-tik-clock]

If the mass of a clown doubles, his initial kinetic energy, mv02/2, will:
remain the same
be reduced in half
double
quadruple

I dont understand whhy we use the equation of impulse to solve this problem:
Ft=mv

My way:

I just substituted m=2m in the equation of KE , ie .5 mv^2 and thus got the KE to be doubling, which is wrong, because the right answer is that the initial KE eneergy is halved
I understand the explanation of wen we use the impulse formula how we come up with .5 , but I am not sure why we are using the Impulse formula

 

[heymanooh1]

If the mass of a clown doubles, his initial kinetic energy, mv02/2, will:
remain the same
be reduced in half
double
quadruple

I dont understand whhy we use the equation of impulse to solve this problem:
Ft=mv

My way:

I just substituted m=2m in the equation of KE , ie .5 mv^2 and thus got the KE to be doubling, which is wrong, because the right answer is that the initial KE eneergy is halved
I understand the explanation of wen we use the impulse formula how we come up with .5 , but I am not sure why we are using the Impulse formula

For initial conditions, the clown has twice as much inertia in resisting the external force acting on it which is the gravitational field in this case.

 

[nontypicalazn]

Hi, If i have an AC power source wired parallel to a Volt meter, and then together serially wired to an Amp meter, then serially wired to an unknown load, what is the power delivered to/across/dissipated (if they make any difference) by the unknown load, when volt and amp meters are ideal? Thanks

 

[gridiron]

Hi, If i have an AC power source wired parallel to a Volt meter, and then together serially wired to an Amp meter, then serially wired to an unknown load, what is the power delivered to/across/dissipated (if they make any difference) by the unknown load, when volt and amp meters are ideal? Thanks

Hey. If the voltmeter is ideal than it will have infinite resistance. This is similar to an open circuit. (In reality this cannot be constructed). An ideal ammeter will have negligent resistance. Compared to the voltmeter, you want the ammeter to have a small resistance because if I interpreted your question correctly the ammeter is in series with an unknown load, so you want alternating current flowing through the load. Let's assume the unknown load is a resistive load. If we assume steady-state operation, the net transfer of energy will be from the power source to resistor where energy will be dissipated as thermal energy. The instantaneous rate at which energy will be dissipated across the resistor can be modeled according to the following equation:

P = i^2R
Where i is the current and R is the resistance. It is important to note that i represents the alternating current. We can model the current as:

i = Isin(wt-&#966😉
Where I is the amplitude of the driving current, and wt-φ is the phase. The phase is included in the equation because the current may not be in phase with the power source.

If you substitute the above value in the equation for i you get:
P = I^2Rsin^2(wt-&#966😉.

If you want the average energy, you should take the above equation over time. However, the sin function is a odd function. So over one cycle, the average value comes out to zero. If we take the absolute value of the sin function, or take the average power of sin^2, we get 0.5. So regardless of phase we can introduce a new equation for average power:

P avg = 0.5I^2R
or
P avg = (I/square root of 2)^2R
​

I/square root of 2 is the root mean square value of the current. Therefore, we obtain our final equation to represent the energy dissipated across the load:

P avg = I^2(rms)*R

One thing we need to define is the rms values of voltage:
V rms = V/square root of 2

This has important implications. Suppose we insert an ac voltmeter into our household electrical outlet. It should theoretically read 120 volts. However, that is the rms voltage. The maximum potential difference across the outlet is actually (square root of 2)*120 volts or about 170 volts.

I hope this helps. Good luck.

 

[nontypicalazn]

thanks BME, maybe i shouldve indicated it, but V and I here are assumed to be in rms. Don't you have to take into consideration of the capacitor and inductors in the unknown load, hence include the phase angle into the power as well, which makes P = IV cos phi ??

the question was multiple choice, i was left with 2 choices:
a. the power delivered = I*V.
c. the power delivered possibly = I*V even if there's inductor and capacitors.

also,is there such a thing as delivered power to a load in circuit vs. dissipated power by a load??

thanks

 

[gridiron]

thanks BME, maybe i shouldve indicated it, but V and I here are assumed to be in rms. Don't you have to take into consideration of the capacitor and inductors in the unknown load, hence include the phase angle into the power as well, which makes P = IV cos phi ??

the question was multiple choice, i was left with 2 choices:
a. the power delivered = I*V.
c. the power delivered possibly = I*V even if there's inductor and capacitors.

also,is there such a thing as delivered power to a load in circuit vs. dissipated power by a load??

thanks

You have to take into consideration the capacitor and inductor when the circuit is modeled as an RLC circuit. If it is a series RLC circuit than there will be an impedance driving the angular frequency of the circuit. The average power can than be calculated once the impedance for the circuit is evaluated. That is when the average power comes out to:

P avg = (Vrms)(Irms)cos(phi)

Cos(phi) is the power factor. Cos is an even function so cos x = cos -x and therefor the power factor is independent of the sign of the phase constant. To maximize the energy which is supplied to a resistive load, the power factor needs to be close to unity. This is not possible in application. Therefore, I would reason the answer to be c because the phase constant has to be zero in order for the power factor to be unity. However, the phase constant can only be reduced by connecting more capacitance in series but the phase can never be actually zero. So the power is more or less V(rms)I(rms). The power factor is only 1, when the phase is zero, when an ac circuit only has resistive load.

This has a very interesting application. Transmission lines carry very high voltage and very low current to minimize power loss. Another thing companies do is place more capacitors in series in order to reduce the phase constant and increase the power factor.

To answer your last question, no there is no difference between the energy supplied and dissipated by a resistive load. It is the same thing because of conservation energy.

I hope this helps and .

 

[nontypicalazn]

I understand in resistive (pure) load, power is conserved, but it seems that when the load is partially resistive and reactive, the power will be returned to the source right?

thanks for the answers BME!

 

[gridiron]

I understand in resistive (pure) load, power is conserved, but it seems that when the load is partially resistive and reactive, the power will be returned to the source right?

thanks for the answers BME!

I believe that is the case but I am not 100% sure.

 

[mcat_study]

ok, so i am really confused about capacitors, batteries, etc...
and i have constructed a chart to help me.
those of you who are great in physics, please verify it
thanks

1. charging capacitor=battery connected: V=constant; Q increase; C increase
2. discharging capacitor--> V decreasing; Q decreasing, C=constant
3. battery disconnect--> V decreasing, Q constant, C increasing
4. new higher dielectric--> V decreasing, Q constant. C increasing

we were given a similiar graph in Kaplan, but the discharging capacitor: Q=const, C decrease, V increase. i think this is incorrect???? may be it is because the teacher made a mistake or i misunderstand something.

please , help!!!!!!!!!!!!

 

[gridiron]

ok, so i am really confused about capacitors, batteries, etc...
and i have constructed a chart to help me.
those of you who are great in physics, please verify it
thanks

1. charging capacitor=battery connected: V=constant; Q increase; C increase
2. discharging capacitor--> V decreasing; Q decreasing, C=constant
3. battery disconnect--> V decreasing, Q constant, C increasing
4. new higher dielectric--> V decreasing, Q constant. C increasing

we were given a similiar graph in Kaplan, but the discharging capacitor: Q=const, C decrease, V increase. i think this is incorrect???? may be it is because the teacher made a mistake or i misunderstand something.

please , help!!!!!!!!!!!!

Hey!

I assume cases 1, 2 and 3 apply to a dielectric as well. If that is the case, than 1 and 3 are correct. When the battery is disconnected, the charge between the plates is trapped, so the charge is constant. When a dielectric is introduced, the capacitance will increase. According to the equation, Q=CV, if the capacitance increases, the voltage will increase as well and by the same factor. So, three and four are the same thing when you consider a battery which is disconnected and a dielectric is then introduced. Remember, C can only increase in the presence of a dielectric(C =kC where k is the dielectric constant) so the scenario in 3 cannot happen unless a dielectric is introduced. In a discharging capacitor, the capacitance stays the same but the charge and voltage decrease. This is because as the capacitor discharges, charge is lost from the plates and since the separation of charge is the basis of the potential difference between the plates, the voltage will decrease as well. Case 1 is true, but only when a dielectric is introduced. In a capacitor being charged by a battery, the voltage between the plates must match the voltage of the battery. Since the presence of a dielectric increases capacitance, the charge that can be stored will increase as well according to Q=CV. I hope this helps and good .

 

[mcat_study]

Hey!

According to the equation, Q=CV, if the capacitance increases, the voltage will increase as well and by the same factor. So, three and four are the same thing when you consider a battery which is disconnected and a dielectric is then introduced. :.

hi BioMed,
thanks a lot for the input 🙂

did u mean Voltage decreases when C increases? in the above case, or am i missing smth????

 

[gridiron]

hi BioMed,
thanks a lot for the input 🙂

did u mean Voltage decreases when C increases? in the above case, or am i missing smth????

Oops😳. I'm so sorry. I was typing so fast and didn't realize! Yes, the voltage will decrease when the capacitance increases according to Q=CV. Sorry for the mistake or any confusion.

 

[Whwsf]

All users may post questions about MCAT and OAT physics here. We will answer the questions as soon as we reasonably can. If you would like to know what physics topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm), though be warned, there are subjects listed there that are rarely tested, or that appear in passages only and need not be learned.

Be sure to check the Physics FAQs and Topic Writeups thread if you have a general question; eventually, many of your answers will be located there. Also, a request: to keep this thread at least somewhat neat, when replying to someone else's post please refrain from quoting anything more than what's necessary for clarity.

Acceptable topics:

• general, MCAT-level physics
• particular MCAT-level physics problems, whether your own or from study material
• what you need to know about physics for the MCAT
• how best to approach to MCAT physics passages
• how best to study MCAT physics
• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
• anything you know to be beyond the scope of the MCAT

Side note: anyone who knows how to post subscripts and superscripts in this system, please PM me the method. I also wouldn't mind knowing how to post some obvious symbols, such as Greek letters and an infinty sign. Should be a matter of changing fonts, but I don't know what's available; again, a PM would be appreciated.

If you really know your physics, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current official contributors to the this thread -- a team to which I hope to add more people:

Thread moderated by: Shrike. Shrike is a full-time instructor for The Princeton Review; he has taken the MCAT twice for no good reason, scoring 14 on the physical sciences section each time. He majored in mathematics, minored in physics, and spent several years accumulating unused school experience (in economics and law).

Also answering questions: Xanthines, a Kaplan instructor. He scored 13 on the PS section of the MCAT and 34 overall.

I hav e a physics question i got from 1001 questions in physics from Examkrackers.... question number 38.... how is the answer B??

 

[QofQuimica]

I hav e a physics question i got from 1001 questions in physics from Examkrackers.... question number 38.... how is the answer B??

Whwsf, we don't have the EK books, so we can't answer a question like that. You'll have to ask the actual question.

 

[Whwsf]

ok.... here is the question..
A particle starts from rest and accelerates at 10 m/s^2. How long does it take for the particle to reach 45 meters??

i thought it was 4.5 seconds but it is 3 seconds.
why and how. thank you

 

[gridiron]

ok.... here is the question..
A particle starts from rest and accelerates at 10 m/s^2. How long does it take for the particle to reach 45 meters??

i thought it was 4.5 seconds but it is 3 seconds.
why and how. thank you

Hey. You have to use what the Princeton Review terms as "The Big Five" Equations for kinematics. They are as follows:

1.) d = 0.5(vinitial+v)*t
where d=distance, v=velocity and t is time
2.) vfinal = vinitial + at
where a is acceleration
3.) d = vinitial*t + 0.5at^2
4.) d = vt - 0.5at^2
5.) vfinal^2 = vinitial^2 + 2ad

These are the five equations you must know for the MCAT. Notice for each there is a missing quantity:

1.) a
2.) d
3.) vfinal
4.) vinitial
5.) t

This is how you decide which equation to use in a particular situation--if a question does not ask for a particular variable, then you do not use that particular equation from the list with the missing quantity. So, if a question does not ask for a then you use 1. In our case, we are not asked for vfinal so we use 3.

d = vinitial*t + 0.5at^2
If you plug in the initial values you gave, 45 m and 10 m/s^2, you should get 3.

I hope this helps and good.

 

[Whwsf]

thanx..... i was told you can solve any physics problems wuth equations 2,3,5... i never saw 1 or 4 before... please elaborate on them... thanx..

 

[kevster2001]

1. it's easier to view the equation as (Vi+Vf)/2 * t . basically yo'ure finding the average velocity and multiplying it by time to find the distance traveled

4. is just d = vfinal*t - 0.5at^2 so basically you know the final velocity, the acceleration and teh time it took to reach that final velocity. with those 3 you can find how far it traveled to get there without knowing how fast it was going initially

 

[neurodoc]

It's really very simple. The equation is: s=1/2at^2. Therefore t=square root of90/10, or the square eoor of 9=3 seconds.

 

[commuter9]

I can't figure out this TPR problem about a J-shaped tube. The top of the tube is open and I need to find out the gauge pressure at point P. Point P is 25 meters below the top of the J, but only 5 meters below the top of the enclosed part. I thought that I would use 5 meters as my height, but the answers say to use 25 meters and not worry about the shape. This doesn't make sense when the portion above point P is enclosed (and only has a height of 5 meters). Why would we take the height of the open part in consideration?

I'm not sure if I explained this very well since I'm realizing the picture is hard to describe. Maybe someone will know what I mean.

 

[gridiron]

I can't figure out this TPR problem about a J-shaped tube. The top of the tube is open and I need to find out the gauge pressure at point P. Point P is 25 meters below the top of the J, but only 5 meters below the top of the enclosed part. I thought that I would use 5 meters as my height, but the answers say to use 25 meters and not worry about the shape. This doesn't make sense when the portion above point P is enclosed (and only has a height of 5 meters). Why would we take the height of the open part in consideration?

I'm not sure if I explained this very well since I'm realizing the picture is hard to describe. Maybe someone will know what I mean.

Hey! If I understand your question correctly, you want to find the pressure at a point 25m below the exposed part of the tube? The reason you use the height of 25m and not 5m, which according to your description is the distance of the point from the closed end of the tube, is because the pressure gradient is caused by the exposed part of the tubing. For example, say you fill the J tube with mercury and hook it up to a pipe with flowing gas. Due to the pressure being exerted by the gas flow, the height of the mercury will change. But how? The gas flow exerts pressure downward on the exposed part of the tube, P=F/A. This will push the mercury and cause for a pressure gradient to develop. Due to this gradient, you get difference in height. The closed part of the tube, which is essentially a vaccum, exerts no force on the mercury and it does not affect the difference height from the initial to final position. What is important to note about these manometer problems is what causes the height difference and what you are solving for. You are solving for pressure but in the derivation of the equation (P = rho*g*h) you are taking into consideration the cause of the pressure gradient and the height change in response to the gradient. In other words, to solve for the pressure, gauge or total, you take the difference in height from the perspective of where the pressure change exists. In this case, it is due to the exposed end of the tube. To illustrate this another way, take a u-tube manometer filled with mercury which is exposed at both ends. If you were to hook this up to a pipe with gas flow, how would you find the change in height? This time, a pressure gradient exists at both ends of the tube. Depending where the pressure is greater, you will see the greater height change (p=rho*g*h). To find the overall difference in height, you would take the difference in height of the mercury between both ends of the tube--because at one end the greater pressure will push the mercury down more. In essence, to solve these types of problems, you look at what is causing the difference in height. In your case, the open end of the tube is the reason--that is why you use the distance from the open end and not the sealed end. I hope this helps and good .

 

[mrjbb]

To answer your last question first: There is a momentum term. Consider the change in momentum as a bullet strikes its object: if the area over which it strikes is small it will penetrate to a greater extent because the force of the bullet will act over a small interval of time. If the force were spread over a large area, so the time over which the force acts is increased, the energy will be absorbed and penetration should be minimal. This is how bullet proof vests work. The material, like polyethylene plates, catches the bullet and spread the force of the bullet over a large area. Since the time over which the force acts is increased, the bullet is brought to stop. This is why when you throw an egg to the ground it breaks but if you throw it against padding like cloth it stays intact--this is a form of impulse. A bullet will penetrate an object like wood because its energy is converted into other forms but also because the force of the bullet acts on a small area over a short time. So, a bullet does apply a force because it has acceleration. Newton’s third law states that the force the bullet applies to the block is equal but opposite in magnitude to the force the wood applies to the bullet. A good example of this is seen in a football game. When a defensive player hits the opposing player very hard, sometimes the defensive player falls back as a result. Why? Because the force with which the defensive player hit is opponent is exerted back onto him in equal but opposite magnitude. Pressure does play a role in reasoning why a bullet penetrates an object but a better explanation would be using Newton’s third law, momentum and impulse. I hope this helps!!

Thanks for clearing that up! This is completely true and logical. A bullet that is traveling towards a piece of wood is going to penetrate it because the bullet's force and energy is concentrated towards a certain and limited area. But when a bullet hits a bullet proof vest the vest absorbs the energy as "BioMedEngineer" said and spreads it throught the vest, therefore the bullet traveling towards the vest at the same speed as the bullet traveling towards the wood does not penetrate. Basically when the bullet came in contact with the vest it was forced not to be concentrated on one spot by the vest itself. When a bullet hits a human, or an animal for example there is nothing there to prevent it from penetrating unless the bullet hits a spot on the surface of the skin where it was some what protected by something that could've distributed it's force evenly throught the object between the bullet and the skin. I.E. a bullet hits a metal compass or something of that nature which was in between the body and the skin therefore the bullet did not penetrate the skins surface. Just my 2 centz..

 

[lisichka]

hello i would like to get these facts straight. is this true that....

if a gas in isothermal system is compressed, then it follows
U=Q-W, U=0J then Q=W, and since the work is done on the system W- and Q-, which means heat flows out of system. this is the one that goes against logic, since i always thought that if gas is compressed, Q must flow into system, but oh well.... 🙁

in adiabatic compression
U=-W, and if W- then U is positive and internal inergy of sytem increases and temperature increases.

in isobaric compression W-, Q increases, U increases, temp increases.
please let me know if there are some bumps in the above summary.
thank you,
if there is anything else in thermo i am missing, please let me know.

thanks a bunch,
😍
lisichka

 

[lisichka]

Pic of circuit attached.....
Please help me with this circuit.....
hello, i have a question about an alternative solution........ please read on

what is R? if Req is 4.5 V

i know the right solutiion--->that we first have to find Rtotal for parallel resistors,
Rt=3R/R+3
and then substitute Rtot+3=4.5ohm and find R. BUT.... my question is why can't i first add in series the two 3 ohm resistors, and then solve for R.--->
like 3+3=6, then 1/6 + 1/R= 1/4.5 why is this wrong??

do i just have to memorize that when we have the above setup, we first wrok out parallel resistances first, and then series resistors, to find Req???

thanks. 🙂

 

[gridiron]

hello i would like to get these facts straight. is this true that....

if a gas in isothermal system is compressed, then it follows
U=Q-W, U=0J then Q=W, and since the work is done on the system W- and Q-, which means heat flows out of system. this is the one that goes against logic, since i always thought that if gas is compressed, Q must flow into system, but oh well.... 🙁

in adiabatic compression
U=-W, and if W- then U is positive and internal inergy of sytem increases and temperature increases.

in isobaric compression W-, Q increases, U increases, temp increases.
please let me know if there are some bumps in the above summary.
thank you,
if there is anything else in thermo i am missing, please let me know.

thanks a bunch,
😍
lisichka

Hey! When you are using the first law of thermodynamics to analyze a system, you must note that heat and work are inexact differentials and internal energy is an exact differential. So your first law actually is:

du = del Q - del W
And if you integrate between two points you get:
delta U = Q -W

If this problem is in context of the MCAT, you can ignore the first part of the explanation, but if this is for a thermo class the first part is essential. Some books might right the above equation as:
delta U = Q+W

It all depends on the convention you use. I usually use the first form, Q-W, where W is the work done by the gas as it expands. So if the gas expands, than the internal energy decreases and if the gas is compressed the internal energy increases.

So, if the isothermal compression is reversible, than delta U = 0 because the initial and final temperatures are the same. That leaves you with:
Q = -W
Remember, with this convention, the integral is taken from volume 1 to volume 2 where volume 2>volume 1. In this case, the gas is compressed, so your final answer will be +W. That means heat flows into the system.

For reversible adiabatic compression, Q=0 since the heat flow is zero. This leaves:
delta U = -W
Now work is done on the gas, and the internal energy increases.

Isobaric compression occurs at constant pressure. So your first law is:
delta U = Q - W

Work is done on the gas, so the internal energy increases.

There are other cases as well like: isochoric process, isothermal expansion, polytropic process. However, the extent to which you will need to know those depends on whether you are studying for the MCAT or a thermo class. Good .

 

[gridiron]

Pic of circuit attached.....
Please help me with this circuit.....
hello, i have a question about an alternative solution........ please read on

what is R? if Req is 4.5 V

i know the right solutiion--->that we first have to find Rtotal for parallel resistors,
Rt=3R/R+3
and then substitute Rtot+3=4.5ohm and find R. BUT.... my question is why can't i first add in series the two 3 ohm resistors, and then solve for R.--->
like 3+3=6, then 1/6 + 1/R= 1/4.5 why is this wrong??

do i just have to memorize that when we have the above setup, we first wrok out parallel resistances first, and then series resistors, to find Req???

thanks. 🙂

Hey! The reason you cannot add the two resistors in series is because they are not in series. When you are given a circuit you have to look at the nodes where all the wires are connected. It helps if you follow the current from the voltage source. The current will first pass through the top 3 ohm resistor and then encounter a node, or break point. It has two options: it can travel to the other 3 ohm resistor or the unknown resistance. Depending on which resistance is greater, it will split into two different currents. That is why the unknown resistance and the 3 ohm resistance is parallel. The top 3 ohm resistor is in series to the equivalent resistance between the bottom 3 ohm resistor and the unknown resistor. So, you have to find the equivalent resistance between the 3 ohm and unknow resistance:

They are in parallel so:
1/req = 1/R + 1/3
Which turns out to be: (R+3)/3R

This is in series with the top 3 ohm resistor so:
Req = 3 + (R+3)/3R
If Req= 4.5 than, R = 3

You don't memorize the setup to determine if resistors are in series or parallel. You have to take into consideration the circuit diagram and the nodes of the circuit. If, as you follow the current through the circuit, the current splits at two points of the wire, than the resistors are in parallel. If you can follow the initial current completely from the start to finish, than the resistors are in series. I hope this helps and good .

 

[lisichka]

There are other cases as well like: isochoric process, isothermal expansion, polytropic process. However, the extent to which you will need to know those depends on whether you are studying for the MCAT or a thermo class. Good .

BioMedEngineer,
thank you so much for such a thourough answer. I am studying for MCAT, but could you also expand a bit on some high-yield thermo examples for MCAT like adiabatic, isothermal expansion,as for [isochoric and polytropic (we do not need to know them>???)] .
again many thanks.😍

 

[gridiron]

BioMedEngineer,
thank you so much for such a thourough answer. I am studying for MCAT, but could you also expand a bit on some high-yield thermo examples for MCAT like adiabatic, isothermal expansion,as for [isochoric and polytropic (we do not need to know them>???)] .
again many thanks.😍

Hey! You do not need to know polytropic process of an ideal gas. However, you will need to know adiabatic expansion, isothermal expansion, and isochoric processes. Adiabatic expansion is the expansion of the gas when there is no heat exchange with the system. The pressure decreases but the volume increases. Isothermal expansion is the expansion of a gas at constant temperature. The pressure decreases and the volume increases. Finally a isochoric process is at constant volume. During this time, there is no work done on or by the gas. During isochoric cooling, the pressure of the gas drops but during isochoric heating the pressure of the gas goes up. For the MCAT make sure you understand these concepts. I hope this helps and good .

 

[lisichka]

Hey! You do not need to know polytropic process of an ideal gas. However, you will need to know adiabatic expansion, isothermal expansion, and isochoric processes. Adiabatic expansion is the expansion of the gas when there is no heat exchange with the system. The pressure decreases but the volume increases. Isothermal expansion is the expansion of a gas at constant temperature. The pressure decreases and the volume increases. Finally a isochoric process is at constant volume. During this time, there is no work done on or by the gas. During isochoric cooling, the pressure of the gas drops but during isochoric heating the pressure of the gas goes up. For the MCAT make sure you understand these concepts. I hope this helps and good .

thanks😀

 

[pguin]

Hey! You do not need to know polytropic process of an ideal gas. However, you will need to know adiabatic expansion, isothermal expansion, and isochoric processes. Adiabatic expansion is the expansion of the gas when there is no heat exchange with the system. The pressure decreases but the volume increases. Isothermal expansion is the expansion of a gas at constant temperature. The pressure decreases and the volume increases. Finally a isochoric process is at constant volume. During this time, there is no work done on or by the gas. During isochoric cooling, the pressure of the gas drops but during isochoric heating the pressure of the gas goes up. For the MCAT make sure you understand these concepts. I hope this helps and good .

Hah...i took the MCAT and did very well but I dont think I've ever seen those words before in my life!!! sounds like PV=nRT...i think

 

[commuter9]

Hey! If I understand your question correctly, you want to find the pressure at a point 25m below the exposed part of the tube? The reason you use the height of 25m and not 5m, which according to your description is the distance of the point from the closed end of the tube, is because the pressure gradient is caused by the exposed part of the tubing. For example, say you fill the J tube with mercury and hook it up to a pipe with flowing gas. Due to the pressure being exerted by the gas flow, the height of the mercury will change. But how? The gas flow exerts pressure downward on the exposed part of the tube, P=F/A. This will push the mercury and cause for a pressure gradient to develop. Due to this gradient, you get difference in height. The closed part of the tube, which is essentially a vaccum, exerts no force on the mercury and it does not affect the difference height from the initial to final position. What is important to note about these manometer problems is what causes the height difference and what you are solving for. You are solving for pressure but in the derivation of the equation (P = rho*g*h) you are taking into consideration the cause of the pressure gradient and the height change in response to the gradient. In other words, to solve for the pressure, gauge or total, you take the difference in height from the perspective of where the pressure change exists. In this case, it is due to the exposed end of the tube. To illustrate this another way, take a u-tube manometer filled with mercury which is exposed at both ends. If you were to hook this up to a pipe with gas flow, how would you find the change in height? This time, a pressure gradient exists at both ends of the tube. Depending where the pressure is greater, you will see the greater height change (p=rho*g*h). To find the overall difference in height, you would take the difference in height of the mercury between both ends of the tube--because at one end the greater pressure will push the mercury down more. In essence, to solve these types of problems, you look at what is causing the difference in height. In your case, the open end of the tube is the reason--that is why you use the distance from the open end and not the sealed end. I hope this helps and good .

Hey thanks BioMedEngineer. What you said does make sense and you did it w/o a picture which is pretty impressive.

 

[commuter9]

I'm confused about how this problem was solved.

An electron is accelerated through a distance of .1m by a potential difference of 10,000 volts. What is the electron's energy as it strikes the anode?
A. 100ev
B. 1,000 eV
C. 10,000 eV
D. 1 J

The answer is C. This question is supposedly asking about change in KE and so W=qV is used. This is what I don't get. They say q = e so it is just equal to one. Why isn't it equal to X coulombs?

 

[heymanooh1]

I'm confused about how this problem was solved.
An electron is accelerated through a distance of .1m by a potential difference of 10,000 volts. What is the electron's energy as it strikes the anode?
A. 100ev
B. 1,000 eV
C. 10,000 eV
D. 1 J
The answer is C. This question is supposedly asking about change in KE and so W=qV is used. This is what I don't get. They say q = e so it is just equal to one. Why isn't it equal to X coulombs?

One of the conversion factors you need is 1 eV = 1.6 x 10 ^-19 J.

The change in energy, delta U, is given as the potential difference, delta V, between two points multiplied by the charge of the particle, q. Therefore, delta U = q x (delta V) = Work done on the particle W.

delta U has units of Joules, J. delta V has units of Joules per coulomb, delta V = J/C

W = delta U = [1.6 x 10 ^-19] x [10, 0000 J/C] = 1.6 x 10^-15 J

Using the 1 eV conversion factor,
delta U = [1.6 x 10^-15 J] x [1eV/(1.6 x 10^-19 J)] = 10,000 eV = W.

The change in electric potential, delta V, is work done per unit charge, not work done per coulomb.

Also, U = delta P.E. + delta K.E. = 0 because of energy conservation, keeping in mind that energy is on a relative scale. Therefore there's always a reference point which will be either stated or implied based on the information given in the passage.

 

[lisichka]

Hi,
another question

when we lift weight the same distance and lower it, it takes the same amount of work, since W=F*d and force, since F=mg what about power, it should be different because it is energy/time, and it should be smaller for up then down? no?

but why in real life it is always harder to lift something up than lower it? can you please explain it in the MCAT context somehow??? is it because of air resistance, i think not, because it acts up, and when we lift the block, we are acting with it, not agaist it, right? gravity pul;, but it is acounted for by mg=F

hmmm, i have no idea, makes me nervous

thank you very much😍 ahead of time 🙂

 

[novawildcat]

Hi,
another question

when we lift weight the same distance and lower it, it takes the same amount of work, since W=F*d and force, since F=mg what about power, it should be different because it is energy/time, and it should be smaller for up then down? no?

but why in real life it is always harder to lift something up than lower it? can you please explain it in the MCAT context somehow??? is it because of air resistance, i think not, because it acts up, and when we lift the block, we are acting with it, not agaist it, right? gravity pul;, but it is acounted for by mg=F

hmmm, i have no idea, makes me nervous

thank you very much😍 ahead of time 🙂

The force of gravity applied to an object in the gravitational field of the earth is a conservative force. Another way to state Newton's 2nd law is the fact that the Work of a conseravtive force=-delta PE (PE is the potential energy). When you are lifting an object up you are increasing the the potential energy of the object which takes energy. Systems like to be in the lowest state of energy as possible which it is why it seems hard to lift the object up. When you let the object lower, you let it decrease its potential energy which is why it seems easier to lower it.

 

[lisichka]

The force of gravity applied to an object in the gravitational field of the earth is a conservative force. Another way to state Newton's 2nd law is the fact that the Work of a conseravtive force=-delta PE (PE is the potential energy). When you are lifting an object up you are increasing the the potential energy of the object which takes energy. Systems like to be in the lowest state of energy as possible which it is why it seems hard to lift the object up. When you let the object lower, you let it decrease its potential energy which is why it seems easier to lower it.

thank you 🙂 novawildcat!

 

[heymanooh1]

Hi,
another question

when we lift weight the same distance and lower it, it takes the same amount of work, since W=F*d and force, since F=mg what about power, it should be different because it is energy/time, and it should be smaller for up then down? no?

but why in real life it is always harder to lift something up than lower it? can you please explain it in the MCAT context somehow??? is it because of air resistance, i think not, because it acts up, and when we lift the block, we are acting with it, not agaist it, right? gravity pul;, but it is acounted for by mg=F

hmmm, i have no idea, makes me nervous

thank you very much😍 ahead of time 🙂

It's "harder" to lift upwards since you are working against the direction of the field, gravitational in this case, which points down. Air resistance is not necessarily acting upwards, it depends on the direction of motion of the body/particle. For example, if something is being lowered downwards, the force of air resistance would be acting in the opposite direction. Think of it being analogous to kinetic friction, as a body slides along a surface, where the force of kinetic friction is opposite to that of the direction of motion.

Draw free body diagrams to keep track of the direction of the forces at work.