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.

 

[DrChandy]

I have a question. Something today discussed by my MCAT instructor really confused me. The statement was: if a person pushed a rock up an incline at a constant velocity, the total work performed on that rock was 0. He stated that this is due to gravity also exerting work on the rock. My question is, if you push a rock up an incline, arent you not only exerting work required to overcome gravity, but additional work to move the rock up the incline?

The important concept being emphasized in the example is the following:

When any object moves at a constant velocity (even zero velocity), there is no net force acting on it.

In this case,
{force of kinetic friction+force exerted by gravity, ie [Cos.mg]}
exactly equals
{force exerted by the person pushing the object to make it move at a constant velocity}

Since the net force equals zero (because the object is moving at a constant velocity), and work=fxd, the net work equals zero.

 

[gujuDoc]

Can someone experienced with physics explain to me the logic behind why Torque = rFSin(-) where (-)= theta????????

I've looked through numerous physics and math books to try to figure this out and still don't understand why it is equal to rFsin(-)

I know there must be some geometric rules or trig rules that I'm missing and wanted to know if one of you could explain it.

Thanks...

 

[TheSecret]

Torque = r X F, where X represents the cross product. To calculate a cross product, we use the rule A X B = AB sin (-).

For a more physical feeling of why you are multiplying by the sin of the angle between r and F, assume the formula was Torque = r (F sin (-)). Since F could be applying a force in any direction at the point in question, taking the sin allows us to only take into account the component of the force that is perpendicular to vector r, and therefore the only part of the force that is causing the rotation.

I wish I could draw you a diagram - that would make a lot more sense...but maybe you can look at a diagram in one of your books, and follow along with this explanation.

Can someone experienced with physics explain to me the logic behind why Torque = rFSin(-) where (-)= theta????????

I've looked through numerous physics and math books to try to figure this out and still don't understand why it is equal to rFsin(-)

I know there must be some geometric rules or trig rules that I'm missing and wanted to know if one of you could explain it.

Thanks...

 

[DrChandy]

Can someone experienced with physics explain to me the logic behind why Torque = rFSin(-) where (-)= theta????????

I've looked through numerous physics and math books to try to figure this out and still don't understand why it is equal to rFsin(-)

I know there must be some geometric rules or trig rules that I'm missing and wanted to know if one of you could explain it.

Thanks...

Using the angle indicated in the diagram,

Fsin(-) = the component of the force which is perpendicular to R.

By definition,

Fsin(-) x R=Torque

 

[TheSecret]

Good explanation, but I have to make one tiny correction.

Torque is actually r X F...while this seems insignificant, since it's a cross product and not simple multiplication, doing it the other way changes the direction of the force. You'll get the same numerical answer, but by the right hand rule you'll end up with a torque in the opposite direction.

Yes, very anal, but I've been burned on stupid crap like that...

Using the angle indicated in the diagram,

Fsin(-) = the component of the force which is perpendicular to R.

By definition,

Fsin(-) x R=Torque

 

[gujuDoc]

Thanks, once you mentioned cross product that explained it better.

I know I'm anal for wanting to know why it was sin theta, but it was bugging me that I couldn't figure it out. So after you pointed me in the right direction I looked at it both in the physics calc book that my roommie has and in my calculus book and see what I did not understand before.

 

[hippocampus]

does anyone know how to determine which color are the opposite of each other?

for example, plants are green cuz they reflect it. they absorb red.

so red and green = opposite.

i know it has to do with the color circle. what is the trick?

also, how does fluorescence work?

thanks

 

[Nutmeg]

does anyone know how to determine which color are the opposite of each other?

for example, plants are green cuz they reflect it. they absorb red.

so red and green = opposite.

i know it has to do with the color circle. what is the trick?

also, how does fluorescence work?

thanks

Take six colors in this order: red, orange, yellow, green, blue, purple. Put them in a wheel. It's easy to remember the order if you recall that yellow + blue = green; red + blue = purple, and yellow + red = orange. Red, Yellow, and blue are thus considered primary colors. So with opposites, it's whatever is on the opposite side of the color wheel. Yellow and purple are opposites, red and green are opposites, and blue and orange are opposites. The primary color's opposite is whichever secondary color it doesn't contribute to making. Blue goes into purple and green, but not orange; hence orange is its opposite. Et cetera.

If you're ever unsure, you can always stare and a given color (don't move your eyes) for a long time (~1 minute) and then look at something white. You will end up seeing the color's opposite.

Fluorescence works by converting light from a higher frequency (usually UV) into light of a lower frequency. The effect is to produce that lower-frequency light at levels greater than are incident upon it. Since UV light is invisible anyways, this gives the effect of looking like the object is glowing brighter than the light that falls on it. Phosphorescence (aka "glow in the dark") is similar, except there is a time delay in the high-frequency light being absorbed and the lower frequency light being released. With fluorescence (aka "day glow") the low-frequency light is released almost immediately, but with phosphorescent stuff, the light is slowly released for an extended time period.

This functions by raising an electron to an excited state. Normally, when an electron is excited by a given frequency of light, we expect it to release that same frequency of light. But if it gets excited from a ground state to a really high state, the release of light to return to a ground state can come in two stages. The chances are that for fluorescent stuff you see, there's one drop from a greater excited state (highest energy level) to a lesser excited state (medium energy level), and that drop releases light that is visible. That can then be followed by another drop from that medium excited state to the ground state, which likely releases a photon with a frequency that is not visible.

Of course, if you don't understand orbitals, that won't make much sense. Good luck.

 

[Nutmeg]

Number 1: Three blocks (block P, block q, and block R) are all initially at rest. Block P is the most massive, and block R is the least massive. Identical horizontal forces push each block, causing them to accelerate, for five seconds. Assuming the coefficient of friction between the block and the surface is the same in all cases, which block will have the highest amount of kinetic energy?

a) Block Q
b) Block P
c) Block R
d) They all have the same kinetic energy, although they have different speeds

There's two ways to do this: a quick way and a long way. The long way is more instructive, so I'll do that first, but the short way is better on a test. Since friction is said to be the same in each case, we'll take the simplest case, where it is zero in all cases. (If this were a porblematic assumption, then the answers available would have to give us an option to the effect of "depends on the fiction coeffecient". Since they didn't give that option, it must not matter, so forget friction and save time!)

So F = Mp*Ap = Mq*Aq = Mr*Ar =Mn*An
since F is constant, with Mp [=] mass of block P, Ap [=] acceleration of block P, etc. N is a general case.

KEn = (1/2)Mn*Vn^2 in each case.

Vn = int(An)dt = An*5 (constant force and constant mass mean constant acceleration)

So KEn = (1/2) Mn*(An*5)^2 = (Mn*An)*An*25 = F*An*25

Hence, whichever has the highest acceleration will have the largest KE--in this case, that's the smallest block.

The quick way is to remember that work is the change in KE, and

W = int(F)dx

Since F is constant, whichever goes a further distance will have encountered higher work. The smallest block will intuitively go the furthest for a given force, hence greater work, hence greater KE.

Number 2: A spring-loaded dart gun is used to shoot a dart straight up into the air. The same dart is shot up a second time from the same gun, but this time the spring is compressed twice as far in comparison to the first shot. How far up does the dart go this time? Neglect friction and assume the spring obey's hooke's law.

a) Half the dist
b) twice the dist
c) four times the dist
d) eight times the dist

Thanks a bunch 😍

F = -kx

W = int(F)dx = int(-kx)dx = -(1/2)kx^2

So if we change the x to 2x, the difference in W is fourfold. Since that work becomes KE energy at the time of release from the spring, that means that the doubled compression gives 4X the kinetic energy. And since KE becomes potential energy, and PE = m*g*h, that means PE varies directly with height; the converse of that is that peak height (which is, in this frictionless world, when all kinetic energy has been converted to potential energy), that means that 4x the KE gives 4x the PE gives 4x the height.

 

[gujuDoc]

Nutmeg,

Thanks for your help on here. I was just reading the following two posts and I like the way you explain things, especially the way you use calc to explain it cuz it makes more sense that way.

By the way, in response to the color question:

The two opposites when mixed would create brown right???

 

[Nutmeg]

Nutmeg,

Thanks for your help on here. I was just reading the following two posts and I like the way you explain things, especially the way you use calc to explain it cuz it makes more sense that way.

By the way, in response to the color question:

The two opposites when mixed would create brown right???

Generally, yes. Brown is what happens when everything gets crazy and you start mixing all over the place. But there's also some weird differences between pigments and lights. What I described is for pigments, but when it comes to light, the three colors used (say, in your monitor or television) are red, green, and blue. Apparently it has to do with the fact that the eye has three color-sensitive cell types, corresponding to those three colors. Go figure.

 

[gujuDoc]

Generally, yes. Brown is what happens when everything gets crazy and you start mixing all over the place. But there's also some weird differences between pigments and lights. What I described is for pigments, but when it comes to light, the three colors used (say, in your monitor or television) are red, green, and blue. Apparently it has to do with the fact that the eye has three color-sensitive cell types, corresponding to those three colors. Go figure.

Oh ok. I think I may be confusing two different things.

Thanks for the clarification.

 

[Cooolguy]

i am the dark lord

 

[Righty123]

Thanks in advance!

Q: A simple circuit with three light bulbs with equal resistance are in parallel with one another. What will happen if the middle bulb burns out?
A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

 

[SteveUTexas]

Thanks in advance!

Q: A simple circuit with three light bulbs with equal resistance are in parallel with one another. What will happen if the middle bulb burns out?
A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

C.

The total resistance of the circuit is decreased so the voltage drop across the other bulbs increase making the light more intense.

 

[Righty123]

C.

The total resistance of the circuit is decreased so the voltage drop across the other bulbs increase making the light more intense.

That's what I thought as well. However, that is not the answer given.

 

[DrChandy]

Take three resistors, each 2 ohms in resistance

---------------------------------------------
In series: 2 Ohms+2 Ohms+2 Ohms=6 Ohms
Remove one resistor: 2 Ohms+2 Ohms=4 Ohms

---------------------------------------------

In Parallel: 1/2 Ohms+1/2 Ohms+1/2 Ohms=2/3 Ohms
Remove one resistor: 1/2 Ohms+1/2 Ohms=1 Ohm (greater resistance)
---------------------------------------------
Removing a resistor attached in parallel increases the resistance of the overall system. This is because the current has fewer pathways with which to travel at a given voltage.

Therefore the answer is B.

 

[Righty123]

Take three resistors, each 2 ohms in resistance

---------------------------------------------
In series: 2 Ohms+2 Ohms+2 Ohms=6 Ohms
Remove one resistor: 2 Ohms+2 Ohms=4 Ohms

---------------------------------------------

In Parallel: 1/2 Ohms+1/2 Ohms+1/2 Ohms=2/3 Ohms
Remove one resistor: 1/2 Ohms+1/2 Ohms=1 Ohm (greater resistance)
---------------------------------------------
Removing a resistor attached in parallel increases the resistance of the overall system. This is because the current has fewer pathways with which to travel at a given voltage.

Therefore the answer is B.

Sorry DrChandy that isn't the answer given either. (The source of this question is TPR).

 

[DrChandy]

Interesting... what was the answer?

 

[gujuDoc]

Thanks in advance!

Q: A simple circuit with three light bulbs with equal resistance are in parallel with one another. What will happen if the middle bulb burns out?
A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

Is the answer D???

For those of you who are our physics tutors on here.........

For series and parallel circuits.......Isn't one of them have constant Voltage but change in current and the other change in Voltage but have constant current??? Does that have something to do with the question. I took TPR, but I don't remember what question that was specifically and where it came from.

Righty,

Do you have the passage or question pg number and from which book that came from or test number??? So perhaps I can take a look at it cuz you've got me curious too now.

 

[Dr^2]

Thanks in advance!

Q: A simple circuit with three light bulbs with equal resistance are in parallel with one another. What will happen if the middle bulb burns out?
A. The other two bulbs will go out.
B. The light intensity of the other two bulbs will decrease, but they won't go out.
C. The light intensity of the other two bulbs will increase.
D. The light intensity of the other two bulbs will remain the same.

D is the correct answer. The "light intensity" here corresponds to power. Now, P = IV, and we know that the V across each resister = V of source and is therefore constant as the resistors are in parallel.

What happens to I? As previous posters established, the *total* resistance of the system increases, meaning that the *total* current decreases:

I_tot_i = V / (R/3) = 3 V/R
I_tot_f = V / (R/2) = 2 V/R

However, the I of each bulb remains constant because I_tot_i was being divided amongst three bulbs whereas I_tot_f is divided amongst two. Therefore, the power remains constant.

BTW, I did the problem as above to reduce confusion about currents and voltage drops. The easier way would be to write

P = IV = V^2/R.

V is constant, the individual resistance of each bulb is constant, therefore power is constant.

 

[Caboose]

This may not matter, however...

I am told that KE is actually a scalar because the "v" component is speed. This is mysterious indeed.

From the equation PE = mgh I assume that since gravity has swallowed acceleration, PE is a vector. If this is so, then I would be adding a scalar + vector to find the total Energy... let me hear ya say, "that ain't right."

I like to keep these vectors and scalars straight in case it comes up in an interview; it's an embarrassing confusion from my past, but I think I can learn from it.

Caboose.

 

[QofQuimica]

This may not matter, however...

I am told that KE is actually a scalar because the "v" component is speed. This is mysterious indeed.

From the equation PE = mgh I assume that since gravity has swallowed acceleration, PE is a vector. If this is so, then I would be adding a scalar + vector to find the total Energy... let me hear ya say, "that ain't right."

I like to keep these vectors and scalars straight in case it comes up in an interview; it's an embarrassing confusion from my past, but I think I can learn from it.

Caboose.

Gravitational potential energy is a scalar, not a vector. (It cannot be resolved into components.) So there is no problem with adding it to kinetic energy to get total energy.

Edit: Kinetic energy is also a scalar because it is obtained from the product of two vectors. (It's the dot product of velocity and momentum, which both ARE vectors.)

 

[gujuDoc]

All work is scalar even if it is a product of a scalar entity and a vector entity. The logic behind work being scalar and why it is cos theta rather then sin theta can be understood by use of multivariate calculus, in which you learn through a long drawn out proof how work is defined through the use of vector calculus.

Accordingly, two rules exist, the scalar product also known as the dot product and the vector product also known as the cross product. I will not go into those things in detail here, but if you are interested, I'd look it up.

Through the long drawn out proof that is the scalar product, the end product is that the law of cosines accounts for why work is scalar and cos theta.

A scalar product generally involves multiplying two vectors, such that if you have

such that if you have ........

/
/______


Let's say the 2 slashes are attached creating an angle with that horizontal. Both are vectors. However, in work or other scalar products, to get the product of the two vectors, you'll take the......

horizontal vector * x component of the vector at an angle. This in turn results in the multplication of 2 parallel vector components and causes an increase or decrease in the size of the ultimate product.

Since the direction of the product is not changed, but the magnitude of the product vector is changed, it is considered to be scalar.

On the other hand, in the vector product, that which is used for Torque, there is directional change that causes the final product to be a vector.

If you wish to understand it more I'd look up vector and scalar products online.

However, for the purposes of the MCAT, I'd know that the scalar product results in a change in the size of ultimate vector, whereas vector product results in a 3rd vector that is perpendicular to the other two vectors. This is what the EK book states in their first chapter of physics.

 

[gujuDoc]

One more thing.......

I'd know that WORK = SCALAR so whether or not it is kinetic energy or potential energy, both are essentially form of work, so they are both scalar.

Whereas, Torque, which follows vector product is a vector and dependent on direction about the y axis.

 

[thadarknyte]

I havent taken physics since high school and I was reading this chapter from the kaplan books. It felt like i was reading chinese. Can someone explain this topic to me...and what exactly do we need to know about it. My textbook that i bought for like 2 dollars from ebay doesnt make much sense either. Thanks

 

[DrChandy]

I havent taken physics since high school and I was reading this chapter from the kaplan books. It felt like i was reading chinese. Can someone explain this topic to me...and what exactly do we need to know about it. My textbook that i bought for like 2 dollars from ebay doesnt make much sense either. Thanks

A blackbody is an object that absorbs all radiation incident upon it, (and reflects none).

Because energy cannot be destroyed, blackbodies have to emit the energy incident upon it somehow, and do so in the form of thermal energy (which is a form of radiation). This thermal energy that is emitted is a property of the temperature of the blackbody and is not dependent on the structure/nature of the blackbody. The hotter the blackbody gets, the shorter the wavelength of radiation emitted.

 

[dermchick]

I'd appreciate help with this question:

Two cars, X & Y, are travelling in the same direction, weighing 1000kg each and initially 120m apart and travelling at 30m/s and 10m/s respectively, with car Y ahead of X. How long before car X overtakes Y.

Thanks--please explain?

 

[DrChandy]

I'd appreciate help with this question:

Two cars, X & Y, are travelling in the same direction, weighing 1000kg each and initially 120m apart and travelling at 30m/s and 10m/s respectively, with car Y ahead of X. How long before car X overtakes Y.

Thanks--please explain?

Use (xf-xi)=(Vi)t+(1/2)at^2

------------------------------------

Where (xf-xi)=120m
Vi=20m/s (from 30m/s - 10m/s)
a=0 (cars are moving at constant speed)

------------------------------------

Solve for the equation to get t=6s

------------------------------------

Note that mass has no relevance in solving the problem

 

[appleciousrx03]

Hey There I was hoping that someone could help me with a physics problem...

A heat engine utilizes a heat source of 560 degrees C and has an ideal (Carnot) efficiency of 30 percent. To increase the efficience to 48 percent, what must the temperature (in degrees Kelvin) of the heat source?

 

[DrChandy]

Hey There I was hoping that someone could help me with a physics problem...

A heat engine utilizes a heat source of 560 degrees C and has an ideal (Carnot) efficiency of 30 percent. To increase the efficience to 48 percent, what must the temperature (in degrees Kelvin) of the heat source?

The carnot engine efficiency is defined by:

[(Th-Tc)]/Th=efficiency

Where Th=hot temperature, Tc=cold temperature, and efficiency is equal to a fraction

-----------------------------------------
Under ideal efficiency conditions,

Th=560 degrees
Tc=unknown
efficiency=0.3

Solve for Tc under ideal conditions.
-----------------------------------------
Now with Tc known, use
[(Th-Tc)]/Th=efficiency
where

Th=unknown
Tc=calculated value from ideal condition
efficiency=0.48

Solve for Th to get the new value.
Convert Th from Centigrade to Kelvin.

-----------------------------------------

 

[NontradICUdoc]

For the dopler effect, when the source is moving towards the detector the denominator is v-Vs and when the source if moving away from the detector the equationis v+Vs. So, if the source is stationary and the detector is moving is it true that
1) When the detector is moving Towards the source numerator is v+Vd
2) When the detector is moving away from the source the numerator is v-Vd

Or is it the other way around?

 

[QofQuimica]

For the dopler effect, when the source is moving towards the detector the denominator is v-Vs and when the source if moving away from the detector the equationis v+Vs. So, if the source is stationary and the detector is moving is it true that
1) When the detector is moving Towards the source numerator is v+Vd
2) When the detector is moving away from the source the numerator is v-Vd

Or is it the other way around?

You've got it right. The biggest thing you need to remember for Doppler Law problems is that the detector goes in the numerator, and the source goes in the denominator. Always consider the source and detector separately. If D is moving toward S, *or* S is moving toward D, you want F' to be greater in either case. This is done by making your numerator positive (using a + sign for D) or your denominator negative (using a - sign for S) respectively. Conversely, if D is moving away from S, or S is moving away from D, that will make F' smaller in either of those cases. Therefore, you want a negative sign in the numerator for D, or a positive sign in the denominator for S. I always remember that the detector goes on top because D comes before S in the alphabet. It's a stupid mnemonic, but it works. 😛

 

[NontradICUdoc]

Unlike resistors in series, you need to calculate the current through each individually correct? Whereas in series, the current through 1 is the current through them all.

 

[QofQuimica]

Unlike resistors in series, you need to calculate the current through each individually correct? Whereas in series, the current through 1 is the current through them all.

If you're asking about resistors in parallel, then yes, you're correct. Resistors in series all have the same current but different voltages. Resistors in parallel have the same voltages across them but different currents. (This is assuming that the resistors have different resistance values.)

 

[suckermc]

heres a question on hydralics.

a friend of mine and I were watching a discovery channel program about the biting power of animals. they had a meter used to record how much pressure or force of an animal bite (as the animal chomps on it) and compared it to a human bite.

the way they recorded the human bite was to have a bronze human skull with the jaws powered by hydralics. in one exp they compared a lion bite (360 psi) and used the same force for the human model to chomp on the bone.

we both agreed it wasnt a fair representation, but finally heres the question:

he said the hydralics made the bite go too fast. i say hydralics arent used for speed. they are used to provide useful work with less force over a greater distance, using a incompressable liquid. is this right, or is there a speed? do i understand hydralics or can someone explain it to me?

 

[Orthodoc40]

I don't know why this subject continues to give me trouble. It seems difficult for me to figure out which equation to use when. Particularly questions where a minimum amount of info is given, and you need to figure out time of flight. How do you figure out time of flight from v = sqrt/2gh?

 

[QofQuimica]

heres a question on hydralics.

a friend of mine and I were watching a discovery channel program about the biting power of animals. they had a meter used to record how much pressure or force of an animal bite (as the animal chomps on it) and compared it to a human bite.

the way they recorded the human bite was to have a bronze human skull with the jaws powered by hydralics. in one exp they compared a lion bite (360 psi) and used the same force for the human model to chomp on the bone.

we both agreed it wasnt a fair representation, but finally heres the question:

he said the hydralics made the bite go too fast. i say hydralics arent used for speed. they are used to provide useful work with less force over a greater distance, using a incompressable liquid. is this right, or is there a speed? do i understand hydralics or can someone explain it to me?

With the caveat that I didn't see the show and so I may not know the whole story, I agree with you that from what you wrote it sounds like they were measuring the pressure of the bite, not the speed or power of it. I think it would be tough to accurately measure bite speed with just a skull, because you wouldn't know anything about the muscles that are controlling the jaw. In other words, it sounds like your friend is wanting to measure the power of the bite, and that would depend on time. If the guys on TV were doing that, it sounds pretty suspect to me, too.

BTW, you are right that hydraulic lifts do not depend on speed. They rely on Pascal's principle, which assumes that when you apply a pressure to a liquid, the pressure is transmitted undiminished throughout the entire liquid, and does not change the liquid's volume. In other words, a liquid is not compressible the way that a gas is.

 

[QofQuimica]

I don't know why this subject continues to give me trouble. It seems difficult for me to figure out which equation to use when. Particularly questions where a minimum amount of info is given, and you need to figure out time of flight. How do you figure out time of flight from v = sqrt/2gh?

I suggest that instead of looking first at the equations, start by writing down what you know and what you want. In most cases, there will be an equation that contains all of those variables. If there isn't, you will have to solve for something else that is unknown first before you can get the unknown that you want. I also find it helpful to draw a picture before starting a physics problem. It doesn't have to be beautifully artistic (mine sure aren't!), but a sketch will help you understand conceptually what is going on in the problem, and what you need to do to approach that problem. In general, it is always better to understand a problem conceptually than it is to plug-and-chug. The MCAT rewards understanding over blind calculating by asking questions that have a slightly different twist on various topics that you've likely never seen before.

In answer to your question, it would be more efficient to use one of the other equations that contains time in it instead of using the one you gave. You will eventually get the right answer if you solve for v first and then plug that into one of the other equations, but it will take you longer and give you more opportunities to make a careless error. Always try to do as little calculation as possible on MCAT problems in order to save time and avoid errors.

 

[suckermc]

heres the question:

a projectile launched over level ground reaches max height in 10 seconds. what is the range if it was launched with a speed of 200m/s?

my question:

i understand the total flight time is 20 seconds.
so v initial is v=at or 100m/s. but the answer says that this is equal to vcos theta of 30 degrees. how do i know that if the angle is not given?

 

[QofQuimica]

heres the question:

a projectile launched over level ground reaches max height in 10 seconds. what is the range if it was launched with a speed of 200m/s?

my question:

i understand the total flight time is 20 seconds.
so v initial is v=at or 100m/s. but the answer says that this is equal to vcos theta of 30 degrees. how do i know that if the angle is not given?

Ok, I think part of your problem is that you're not keeping your x and y variables separate. You figured out that the v(initial) in the y-direction is 100 m/s. (In projectile motion, acceleration in the x-direction should be zero assuming no air resistance. It's -10 in the y-direction only.) That's equal to 1/2 the total v(initial) you were given in the problem. You should also know that v(initial) in the y direction is equal to v(initial)sin(theta). (Draw yourself a diagram and resolve the vector if you're not convinced of this.) Now, what angle has a sin equal to 1/2? Thirty degrees. That's where they got that angle from. So now, knowing this, you can take the cos(30 degrees) and multiply that by 200 to get the v(initial) in the x-direction, which you need to solve the problem. (You are being asked to solve for x, the range, not y, the height, of the projectile motion.)

 

[suckermc]

Ok, I think part of your problem is that you're not keeping your x and y variables separate. You figured out that the v(initial) in the y-direction is 100 m/s. (In projectile motion, acceleration in the x-direction should be zero assuming no air resistance. It's -10 in the y-direction only.) That's equal to 1/2 the total v(initial) you were given in the problem. You should also know that v(initial) in the y direction is equal to v(initial)sin(theta). (Draw yourself a diagram and resolve the vector if you're not convinced of this.) Now, what angle has a sin equal to 1/2? Thirty degrees. That's where they got that angle from. So now, knowing this, you can take the cos(30 degrees) and multiply that by 200 to get the v(initial) in the x-direction, which you need to solve the problem. (You are being asked to solve for x, the range, not y, the height, of the projectile motion.)

hey doc

i understand what you said, but i assumed that all angles would be given on the mcat. i understand i will have to memmorize some.

usually when the question says "at an angle to the horizontal " i assume i have to resolve it into vector components. if it doesnt , i just assume i can solve in the usual way. but is it because it is a range prob that i have to use the angles? do you see what i am getting at?

 

[QofQuimica]

hey doc

i understand what you said, but i assumed that all angles would be given on the mcat. i understand i will have to memmorize some.

usually when the question says "at an angle to the horizontal " i assume i have to resolve it into vector components. if it doesnt , i just assume i can solve in the usual way. but is it because it is a range prob that i have to use the angles? do you see what i am getting at?

You have to resolve it because the question told you it's a projectile motion. You should assume that any projectile moves in a parabola (i.e., theta is between 0 and 90 degrees). I re-read the problem again, and the way I know that it isn't shot directly horizontally is that they tell you it reaches maximum height after x amount of time. They wouldn't say that if it had been shot out horizontally at 0 degrees, because the maximum y in that case would be at t=0. Likewise, it couldn't have been shot out vertically at a theta of 90 degrees, because if it had, the range would be zero. (In other words, it would fall right back down where it came from, and your problem wouldn't be very interesting.)

You definitely should memorize the sines and cosines of 0, 30, 45, 60, and 90 degrees for the MCAT. They may or may not give those values to you on the test, but I wouldn't count on it. If you get another angle besides one of these (unlikely to happen, but just in case), round it to the nearest one that you know. For example, if the problem told you that theta is 49 degrees, treat it like a problem where theta is 45 degrees, and go on from there.

FWIW, I agree with you that the form of your problem is unusual, because MCAT problems don't commonly require you to come up with the value of theta. But the MCAT does like to ask you about common science topics in unusual ways, so it wouldn't totally surprise me. Always start any physics problem by considering what is going on. Draw yourself a sketch; I find that to be very helpful. Don't ever knee-jerk plug-and-chug into an equation without having an understanding of what's going on in the problem first. Often, you can avoid performing calculations at all, or at least eliminate 1 or 2 wrong answers, by thinking logically about what kind of answer you expect to have.

 

[kevin86]

Question about fluids. When asking for gauge of pressure of an object submerged under 2 layers of different density fluids, why do u add up the density in the equation (d1 +d2)gy d1 and d2 being the two densities. It seems kind of weird if say d2 >d1 and if the whole tank was filled with p2 liquid with the object position not changed, the gauge pressure would only be dp2gy. In the case of only density d2, the pressure is smaller than the case of denisty d1 + d2. But that makes no sense having less pressure if d2 is more dense than d1 + d2 under the same volume.

 

[QofQuimica]

Question about fluids. When asking for gauge of pressure of an object submerged under 2 layers of different density fluids, why do u add up the density in the equation (d1 +d2)gy d1 and d2 being the two densities. It seems kind of weird if say d2 >d1 and if the whole tank was filled with p2 liquid with the object position not changed, the gauge pressure would only be dp2gy. In the case of only density d2, the pressure is smaller than the case of denisty d1 + d2. But that makes no sense having less pressure if d2 is more dense than d1 + d2 under the same volume.

Hmm, that seems kind of weird to me, too. Because the pressure should also be different depending on how deep each of the layers is. If the d1 layer is only 1 meter deep, but the d2 layer is ten meters deep, they shouldn't contribute equally to the total pressure. In other words, I think the two layers should also have their own separate y-values, unless by sheer coincidence both layers are exactly the same thickness. Can you double-check that formula?

 

[Caboose]

Huge picture of this
I object to the final equilibration step.
30g of H2O@ 50 C are mixed with 1g of steam @110 C. After final equilibration, the temp is...? (Heat condensation=540cal/g, specific heat of steam=0.5cal/g, sp.heat H2O = 1cal/g)
The answer is about 70 C
The last calculation is: deltaQ of H2O = deltaQ of steam (30)(1g)(Tfinal-T) = (1)(0.5)(100-Tfinal) I'm truckin' along, on top of the world like Leonardo until the part about the specific heat = 0.5 comes in. See, I think that by the time the steam reaches 100C I've already converted it into water... so why would I use the specific heat of steam at the end again?

Also - why is one temperature (final - initial) and the steam one (initial - final)? Just a sign deal?

Caboose.

 

[QofQuimica]

Huge picture of this
I object to the final equilibration step.
30g of H2O@ 50 C are mixed with 1g of steam @110 C. After final equilibration, the temp is...? (Heat condensation=540cal/g, specific heat of steam=0.5cal/g, sp.heat H2O = 1cal/g)
The answer is about 70 C
The last calculation is: deltaQ of H2O = deltaQ of steam (30)(1g)(Tfinal-T) = (1)(0.5)(100-Tfinal) I'm truckin' along, on top of the world like Leonardo until the part about the specific heat = 0.5 comes in. See, I think that by the time the steam reaches 100C I've already converted it into water... so why would I use the specific heat of steam at the end again?

Also - why is one temperature (final - initial) and the steam one (initial - final)? Just a sign deal?

Caboose.

You're right....once you've converted the steam to water using Q = mH, then you shouldn't have to use the steam's c again. The amount of energy needed to convert the steam to water at 100 deg. C uses 545 cal, which would raise the temp of the water up to 68 deg. C from 50 deg. (What they're calling "T" in that calculation is approximately equal to 68 degrees.) Then I would set it up as follows:

(30)(1)(Tf-68) = (1)(1)(100-Tf)

This would come out to a Tf of 69 point something, which is in line with their calculation. I think they just made a mistake by putting the 0.5 in there for the c. Luckily the c of steam is not too different than the c of water.

Yes, you subtract the Tf from 100 because it's a smaller number than 100. You subtract 68 from Tf because Tf is larger than 68.

 

[Caboose]

You're rad.
Caboose.

 

[kevin86]

I have a question about which way the magnetic field curls for a electron. the the direction of an electron is one direction, it means current is in the other, so then i should point thumb in the other direction and curl right?

 

[Lests55]

I don't know how to deal with capacitors and resistors that are in the same circuit. Unfortunately we did not cover this in my physics class (of course we covered them severly, but not together). Any thoughts on this?

Thanks