Physics question thread

I am confused about standing sound waves in tubes. I have hard time visualizing and learning since I did not learn it in Gen phy. Any suggestions to help absorb the concepts?

Also, why is it that voltage is constant when resistors are added in series to a circuit while currentl goes down? I realize V= IR but why not V go down and current stay the same?

Thanks for all your help!

I have not touched my physics book for a year
but i do remember ohm's law = V = IR
Voltage is the potential difference you apply across two ends of a circuit. Think it as a water flowing from the top of a mountain. More the height is more potential engery it has, and it has more tendency to flow down. But if you put something (rock) in the way, you will decrease the flow downward, but water has same tendency to flow. IF you make analogy, you can see Voltage has same tendency to flow electrons, but since resistor (rocks) resisting electron (water) flow, your current(amp) will be decreased

dill_punjabi said:

I have not touched my physics book for a year
but i do remember ohm's law = V = IR
Voltage is the potential difference you apply across two ends of a circuit. Think it as a water flowing from the top of a mountain. More the height is more potential engery it has, and it has more tendency to flow down. But if you put something (rock) in the way, you will decrease the flow downward, but water has same tendency to flow. IF you make analogy, you can see Voltage has same tendency to flow electrons, but since resistor (rocks) resisting electron (water) flow, your current(amp) will be decreased

Click to expand...

Thanks for your response. However, I don't know if this is a completely right analogy as adding resistors in parallel increases current, which does not really make sense according to the analogy you provided. Does anyone know how to approach this concept mathematically using equations? Thank you again!

pezzang said:

I am confused about standing sound waves in tubes. I have hard time visualizing and learning since I did not learn it in Gen phy. Any suggestions to help absorb the concepts?

Also, why is it that voltage is constant when resistors are added in series to a circuit while currentl goes down? I realize V= IR but why not V go down and current stay the same?

Thanks for all your help!

Click to expand...

See my earlier post on standing waves.

When 2 UNEQUAL resistors are added in series, the current is the same through both resisitors. The flow of current is similar to fluid flow in a pipe. If there are no junctions, the current is the same. V = IR, and
since I is constant when resistors are connected in series, there will be a larger voltage drop across a larger resistor.

When 2 UNEQUAL resistors are connected in parallel, we can use Kirchhoff's loop rule (i.e. the sum of
voltage changes around any loop is zero...since by definition voltage is potencial energy per unit charge,
Kirchoff's loop rule is just another statement of conservation of energy) to see that voltage drop across both resistors is the same. Since I = V/R, the larger resistor will have a smaller current flowing through it.

A fast train (50 m/s) is moving directly toward Samuel, who is standing near the tracks. The train is emitting a whistling sound at 420 Hz. The speed of sound is 350 m/s at the outdoor temp. 31 degree C. ( I tried using the EK physics method to solve it and I couldn't. Can someone please help me out..I'll greatly appreciate it)
[/COLOR]
1. What frequency does Samuel hear?
2. What if Samuel whistles at 420 Hz and a passenger on the train could hear him, what frequency would she hear?
3. After the train passes Samuel, what frequency does he hear from the whistle?

a. 360 Hz
b. 367.5 Hz
c. 480 Hz
d. 490 Hz

V4viet said:

A fast train (50 m/s) is moving directly toward Samuel, who is standing near the tracks. The train is emitting a whistling sound at 420 Hz. The speed of sound is 350 m/s at the outdoor temp. 31 degree C. ( I tried using the EK physics method to solve it and I couldn't. Can someone please help me out..I'll greatly appreciate it)
[/color]
1. What frequency does Samuel hear?
2. What if Samuel whistles at 420 Hz and a passenger on the train could hear him, what frequency would she hear?
3. After the train passes Samuel, what frequency does he hear from the whistle?

a. 360 Hz
b. 367.5 Hz
c. 480 Hz
d. 490 Hz

Click to expand...

DOPPLER EFFECT fL = fS ((v + vL) / ( v – vS ))


· fL is frequency of sound waves encountered by the listener (i.e. detected frequency)

· fS is frequency of sound waves emitted by the source (i.e. emitted frequency)

· v is speed of the sound wave (in this case, through air)

· vL is speed of the listener relative to still air
o vL is positive (+) if the listener is moving towards the source
o vL is negative (-) if the listener is moving away from the source
o vL = 0 if the listener is not moving relative to still air

· vS is speed of the source relative to still air
o vS is positive (+) if the source is moving towards the listener
o vS is negative (-) if the source is moving away from the listener
o vS = 0 if the source is not moving relative to still air

· When source and detector approach each other, the detected frequency is higher than the emitted one (because v + vL > v – vS even if vL = 0 or vS = 0)

&#183; When source and detector recede from each other, the detected frequency is lower than the emitted one (because v - vL < v + vS even if vL = 0 or vS = 0)



1) fL = 420 ((350+ 0) / ( 350 &#8211; 50 )) = 490 Hz
In this case you can eliminate any answer choices less than 420
even before using the formula because source (i.e. train in this case) and detector (i.e. Samuel in this case) are approaching each other.

2) fL = 420 ((350+ 50) / ( 350 &#8211; 0 )) = 480 Hz
In this case you can eliminate any answer choices less than 420
even before using the formula because source (i.e. train in this case) and detector (i.e. Samuel in this case) are approaching each other.

3) fL = 420 ((350+ 0) / ( 350 &#8211; (&#8211;50) )) = 367.5 Hz
In this case you can eliminate any answer choices greater than 420 even before using the formula because source (i.e. train in this case) and detector (i.e. Samuel in this case) are receding from each other.


You do the math

Thank you so much for taking the time out to help me with my problems.. I really appreciate it..I was really freaking out before.

V4viet said:

Thank you so much for taking the time out to help me with my problems.. I really appreciate it..I was really freaking out before.

Click to expand...

You are welcome.

Correction to the comment for case 2: In this case you can eliminate any answer choices less than 420 even before using the formula because source (i.e. Samuel's mouth) and detector (i.e. the ear of the person on the train) are approaching each other.

Hi, I've had a lot of trouble with questions dealing with this very important concept that keeps coming up in problems. The book says that if an object is floating, then bouyant force (Fb) is equal to the weight of the object.

One example problem I had difficulty with involves a cube with volume of 1000cc, weighing 16N in air. When half of it is submerged in an unknown liquid, it weighs only 10 N. The question asks for the density of the liquid. However, I would like to know the free energy diagram and the forces that are acting on the cube. In my opinion, since the density of the object is 1.6g/1000cc (density is mass/volume, the object has mass of 1.6g, and volume of 1000cc), while the answer says the liquid density is 1.2 g/cc, then the object is less dense than the liquid, and it should float. And for a floating object, Fb=weight of the object. Therefore, if the diagram consists of Fb acting up, and mg acting down, then there should be no net force on the object, right? So what does it mean when it says it has an apparent weight of 10N? Since Wactual-Fb=Wapparent, and Fb=actual weight of the object, then why isn't Wapparent zero? And if there is a Wapparent, why is the object not sinking? I am having trouble visualizing what is going on in this problem (the forces involved).

In another example problem, "an object wieghing 150N, whose density is 1.5kg/L, and it is completely submerged in a flask containing 2 immiscible fluids. If the densities of the fluids are 1 kg/L and 2 kg/L respectively, what volume of the object will be submerged in the denser fluid?" To answer this question, I was supposed to assume that the object is floating, so Fb=mg.

So to summarize, my questions are:
1) When can I assume the object is floating, and I can use Fb=mg? When can I not assume it is floating? Why is it in the first example, I cannot assume it is floating, but in second example I can?

2)What is the free energy diagram for the first example? Why is there a Wapparent? If there is a Wapparent, why is the object not sinking to the bottom of the container?

I'm so confused

What other factors contribute to rate of effusion aside from molecular mass? Do molecular structure or atomic radii play any part in rate of effusion? would an cation effuse slower? THanks.

googlinggoogler said:

Hi, I've had a lot of trouble with questions dealing with this very important concept that keeps coming up in problems. The book says that if an object is floating, then bouyant force (Fb) is equal to the weight of the object.

One example problem I had difficulty with involves a cube with volume of 1000cc, weighing 16N in air. When half of it is submerged in an unknown liquid, it weighs only 10 N. The question asks for the density of the liquid. However, I would like to know the free energy diagram and the forces that are acting on the cube. In my opinion, since the density of the object is 1.6g/1000cc (density is mass/volume, the object has mass of 1.6g, and volume of 1000cc), while the answer says the liquid density is 1.2 g/cc, then the object is less dense than the liquid, and it should float. And for a floating object, Fb=weight of the object. Therefore, if the diagram consists of Fb acting up, and mg acting down, then there should be no net force on the object, right? So what does it mean when it says it has an apparent weight of 10N? Since Wactual-Fb=Wapparent, and Fb=actual weight of the object, then why isn't Wapparent zero? And if there is a Wapparent, why is the object not sinking? I am having trouble visualizing what is going on in this problem (the forces involved).

In another example problem, "an object wieghing 150N, whose density is 1.5kg/L, and it is completely submerged in a flask containing 2 immiscible fluids. If the densities of the fluids are 1 kg/L and 2 kg/L respectively, what volume of the object will be submerged in the denser fluid?" To answer this question, I was supposed to assume that the object is floating, so Fb=mg.

So to summarize, my questions are:
1) When can I assume the object is floating, and I can use Fb=mg? When can I not assume it is floating? Why is it in the first example, I cannot assume it is floating, but in second example I can?

2)What is the free energy diagram for the first example? Why is there a Wapparent? If there is a Wapparent, why is the object not sinking to the bottom of the container?

I'm so confused

Click to expand...

Since in the first problem, the object has an apparent weight, it means that there is a force equal to this weight holding the object up along with the bouyant force, thus, the object is not simply "floating."

To solve this first problem: 1/2 volume of block= 500cc. This volume has an equivalent force of 6N. Density of fluid = 6N/10(N/kg)/500cc = 0.0012 kg/cc or 1.2g/cc.

The second problem is similar but without the extra force holding the object up other than the bouyant force (Fb=mg).

Edit: See BorkenGlass's post below for the second question.

MEG@COOL said:

What other factors contribute to rate of effusion aside from molecular mass? Do molecular structure or atomic radii play any part in rate of effusion? would an cation effuse slower? THanks.

Click to expand...

Ideal gases consist of identical particles with negligible volume. These particles undergo perfectly elastic collisions and exhibit no intermolecular forces.
For ideal gases, KEavg = 3/2 RT. In other words, KEavg ~ T. If you have 2 gases at the same temperature, you can set their kinetic energies equal and
solve for the ratios of their velocities.

Rate of effusion (and diffusion) of an ideal gas depens only on how fast the the partciles are moving and how massive they are. For an ideal gas, molecular
structure and atomic radius would play no role because ideal gas particles themselves occupy no volume. Ideal gas cannot consist of cations because cations
would exibit repulsive forces. I believe all the effusion and diffusion problems deal with gases that are assumed to behave ideally.

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/graham.htm

googlinggoogler said:

Hi, I've had a lot of trouble with questions dealing with this very important concept that keeps coming up in problems. The book says that if an object is floating, then bouyant force (Fb) is equal to the weight of the object.

One example problem I had difficulty with involves a cube with volume of 1000cc, weighing 16N in air. When half of it is submerged in an unknown liquid, it weighs only 10 N. The question asks for the density of the liquid. However, I would like to know the free energy diagram and the forces that are acting on the cube. In my opinion, since the density of the object is 1.6g/1000cc (density is mass/volume, the object has mass of 1.6g, and volume of 1000cc), while the answer says the liquid density is 1.2 g/cc, then the object is less dense than the liquid, and it should float. And for a floating object, Fb=weight of the object. Therefore, if the diagram consists of Fb acting up, and mg acting down, then there should be no net force on the object, right? So what does it mean when it says it has an apparent weight of 10N? Since Wactual-Fb=Wapparent, and Fb=actual weight of the object, then why isn't Wapparent zero? And if there is a Wapparent, why is the object not sinking? I am having trouble visualizing what is going on in this problem (the forces involved).

In another example problem, "an object wieghing 150N, whose density is 1.5kg/L, and it is completely submerged in a flask containing 2 immiscible fluids. If the densities of the fluids are 1 kg/L and 2 kg/L respectively, what volume of the object will be submerged in the denser fluid?" To answer this question, I was supposed to assume that the object is floating, so Fb=mg.

So to summarize, my questions are:
1) When can I assume the object is floating, and I can use Fb=mg? When can I not assume it is floating? Why is it in the first example, I cannot assume it is floating, but in second example I can?

2)What is the free energy diagram for the first example? Why is there a Wapparent? If there is a Wapparent, why is the object not sinking to the bottom of the container?

I'm so confused

Click to expand...

1) A floating object is in equilibrium, so the net force on it is zero. The only forces acting on a floating object are the buoyant
force and force due to Earth's gravity. When an object floats, these forces are equal and opposite.

When half of the object is submerged in an unknown liquid, it weighs only 10 N. The problem doesn't state or imply that the object would not eventually sink. As it sinks, more and more of it gets submerged in the fluid until the entire object is submerged. Since
buoyant force is equal to the weight of the fluid displaced, as more and more of the object gets submerged, more and more of
the fluid gets dispalced, and the buoyant force increases until the object can displace no more fluid, at which point Fb equal its
max value. The greater the Fb, the less the apparent weight.

Since our object weighs 16N in air, its mass is 1.6kg.

Object density is 1.6kg/1000cc = 1600 g/1000 cc = 1.6g/cc and liquid density is 1.2 g/cc (see below), so the object is MORE dense then the liquid, so the object would sink. As it's sinking, there will be point in time where half of it is submerged. At this point, the apparent weight is 10N.

1 cubic meter = 1,000,000 cc
500 cc = 0.0005 cubic meters
Fb = 16N - 10N = 6N
Fb = weight of the fluid displaced = mass of the fluid dispalced * g = density of fluid dispaced * volume of fluid dispalced * g
Fb = rho * 0.0005 * 10 = 6N
=> rho = 1200 kg/ cubic meter = 1200000g/1000000 cc = 1.2 g/cc

2) density = mass/volume

If an object weighs 150 N in air, its mass is 15 kg. Since its density = 1.5kg/L, its volume must be mass/density = 15/1.5 = 10 L.

Note that the denser of the 2 fluids will be on the bottom of the container, while the less dense of the 2 fluids will be sitting on top
of the fluid that is more dense. Since the object (whose density is 1.5kg/L) is more dense than the top fluid (whose density is 1
kg/L) it will sink in the top fluid. But the object will not sink all the way to the bottom of the container because the bottom fluid (whose density is 2 kg/L) is more dense than the object. So the object is submerged in the top fluid and is partially submerged in the bottom fluid.

Remember that Fb = weight of the fluid displaced. Since the object is floating, Fb = mg = 150N.

In what follows subscript 1 is referring to fluid 1 and subscript 2 is referring to fluid 2.

Fb = Fb1 + Fb2 since each of 2 fluids is exertinig a buoyant force on the object.

Fb1 + Fb2 = 150N.

The object doesn't displace the amount of first (i.e. lighter) fluid equal to its volume because part of the object is out of the first fluid, but is submerged in second (i.e. heavier fluid). In other words, the object goes through the top (lighter fluid), dips into the bottom (heavier fluid) and floats (i.e. ends up in equilibrium).

So we need to solve a system of 2 equations in 2 unknowns to see how much of the object is in fluid 1 and how much of it is in fluid 2.

Let V1 represent part of the object (by volume) that's in fluid 1 and V2 represent part of the object (by volume) that's in fluid 2. Remember that Fb = rho * V * g

Fb1 + Fb2 = Fb = mg
V = V1 + V2


(1 * V1 * 10) + (2 * V2 * 10) = 150N
V1 + V2 = 10L

10*V1 + 20*V2 = 150
V1 = 10 - V2

10 * (10 - V2) + 20 * V2 = 150
100 + 10 * V2 = 150
10 * V2 = 50
v2 = 5L
V1 = 5L

So 5L of the object will be submerged in the denser fluid. This solution assumes that the top fluid is deep enough to completely cover the object.

BrokenGlass said:

Ideal gases consist of identical particles with negligible volume. These particles undergo perfectly elastic collisions and exhibit no intermolecular forces.
For ideal gases, KEavg = 3/2 RT. In other words, KEavg ~ T. If you have 2 gases at the same temperature, you can set their kinetic energies equal and
solve for the ratios of their velocities.

Rate of effusion (and diffusion) of an ideal gas depens only on how fast the the partciles are moving and how massive they are. For an ideal gas, molecular
structure and atomic radius would play no role because ideal gas particles themselves occupy no volume. Ideal gas cannot consist of cations because cations
would exibit repulsive forces. I believe all the effusion and diffusion problems deal with gases that are assumed to behave ideally.

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/graham.htm

Click to expand...

What about for real gasses? Just wondering although i know this is outside the scope of the mcat. THanks

Message deleted.

Hey guys, I don't have a specific question, but I came across a problem the other day, which had a box on an incline (30 degrees), and it was being pushed up the ramp with x ( i think 3m/s2) acceleration. I know you have to transfer force into the x and y directions, but for the y direction the solutions said to: =F(sin30)+N-mg=0, thus N=mg-F(sin30). Where did they get zero and why did they add N-mg to the y component? I thought F was equal to mg in the y direction....actually I'm going to find this problem so it'll make more sense lol


Q: A 50-kg box is pulled by a force at an angle of 30 to the horizontal. If the box accelerates at a rate of 3 m/s2 , and coefficient of sliding friction is 0.4, what is the magnitude of force pulling the box?

HristosKaran said:

Hey guys, I don't have a specific question, but I came across a problem the other day, which had a box on an incline (30 degrees), and it was being pushed up the ramp with x ( i think 3m/s2) acceleration. I know you have to transfer force into the x and y directions, but for the y direction the solutions said to: =F(sin30)+N-mg=0, thus N=mg-F(sin30). Where did they get zero and why did they add N-mg to the y component? I thought F was equal to mg in the y direction....actually I'm going to find this problem so it'll make more sense lol


Q: A 50-kg box is pulled by a force at an angle of 30 to the horizontal. If the box accelerates at a rate of 3 m/s2 , and coefficient of sliding friction is 0.4, what is the magnitude of force pulling the box?

Click to expand...

1) There is no ramp in this problem. The box is being pulled along a horizontal surface by a force at an angle of 30 degrees to the horizontal.

2) Let Fp denote F pulling.
Let fk denote the force of kinetic friction.
The coefficient of kinetic friction is 0.4 (given) and is denoted by mu_k.
Let N denote the normal force.

3) Let Y direction be perpendicular to the horizontal surface. Let positive Y direction be up.

4) Let X direction be parallel to the horizontal surface. Let positie X direction be to the right.

5) Resolve the pulling force into X and Y components and do force analysis in the X and Y directions separately.

6) Y direction: Fnet = ma = Fp * sin 30 + N - mg = 0.

The solution seems to assume that the box is not lifted by the pulling force, i.e. the box is always in contact with the surface on which
it's sliding, so the acceleration in the y direciton is zero. The only effect of the pulling force in the Y direction is to decrease the normal
force. In other words, normal force would have been greater if the pulling force was applied parallel to the surface on which box is sliding.

7) X direction: Fnet = ma = Fp * Cos 30 - fk, where fk = mu_k * N

8) Fp * Sin 30 + N - mg = 0 (equation in the X direction)
Fp * cos 30 - mu_k * N = ma (equation in the Y direction)

We have 2 equations and 2 unknowns (Fp and N)

N = mg - Fp * sin 30

Fp * cos 30 - mu_k (mg - Fp * sin 30) = ma

Fp * 0.9 - 0.4 (50 * 10 - Fp * 0.5) = 50 * 3

0.9Fp - 200 + 0.2 Fp = 150

1.1Fp = 350

Fp = 320N

HristosKaran said:

Hey guys, I don't have a specific question, but I came across a problem the other day, which had a box on an incline (30 degrees), and it was being pushed up the ramp with x ( i think 3m/s2) acceleration. I know you have to transfer force into the x and y directions, but for the y direction the solutions said to: =F(sin30)+N-mg=0, thus N=mg-F(sin30). Where did they get zero and why did they add N-mg to the y component? I thought F was equal to mg in the y direction....actually I'm going to find this problem so it'll make more sense lol


Q: A 50-kg box is pulled by a force at an angle of 30 to the horizontal. If the box accelerates at a rate of 3 m/s2 , and coefficient of sliding friction is 0.4, what is the magnitude of force pulling the box?

Click to expand...

Hey! A good way to approach this question is to first draw a free body diagram from the verbal statement of the problem. From your question, are you saying the box is being pulled up a ramp? I ask this because when you actually write the question, it seems as if the box isn't being pulled up a ramp. Regardless, I will outline a method to solve this problem for a box being pulled up a ramp. If it is the other type of problem, just correct me and I will solve it the other way.

When you draw the free body diagram, you want to assign a coordinate system for the forces. Since the problem is dealing with an angle, it is a good idea to translate the forces so that the x axis is parallel to the surface of the ramp and the y axis is perpendicular to the surface of the ramp. The first force you should label is the weight of the block. Gravity is always acting downward, so the force should be drawn straight down from the center of the box. The force should be broken down into components because our axis is at an angle. Using trigonometry, the x component of the force would be mgsin(theta) and y component would be mgcos(theta). Another force is the force by which the box is being pulled with. The last two forces are the force of friction and the normal force. Now, how to assign sign to these forces? When a box is placed on a ramp, I think that it naturally will want to fall down--with the force (mgsin(theta)) in the direction of the acceleration. However, a force is pulling this block up. So, I would assign the pushing force as positive and the weight of the block in the x direction as negative since that force is acting opposite the displacement of the block. Friction always acts to impede motion so I would also assign it as a negative force. Since the displacement of the block occurs in the x direction, all you have to do is to sum up all the forces and set the net force equal to ma--by Newton's second law. The only unknown is F, the pulling force, and you have one equation to solve for it. I hope this helps and good .

Thanks for your help you guys, the other website I took this problem from had a diagram underneath where the guy drew a ramp, but now that I read the problem, there definitely isn't a ramp present. And thanks for your help solving it, b/c either way I would have had difficulty!

gridiron said:

Hey! A good way to approach this question is to first draw a free body diagram from the verbal statement of the problem. From your question, are you saying the box is being pulled up a ramp? I ask this because when you actually write the question, it seems as if the box isn't being pulled up a ramp. Regardless, I will outline a method to solve this problem for a box being pulled up a ramp. If it is the other type of problem, just correct me and I will solve it the other way.

When you draw the free body diagram, you want to assign a coordinate system for the forces. Since the problem is dealing with an angle, it is a good idea to translate the forces so that the x axis is parallel to the surface of the ramp and the y axis is perpendicular to the surface of the ramp. The first force you should label is the weight of the block. Gravity is always acting downward, so the force should be drawn straight down from the center of the box. The force should be broken down into components because our axis is at an angle. Using trigonometry, the x component of the force would be mgsin(theta) and y component would be mgcos(theta). Another force is the force by which the box is being pulled with. The last two forces are the force of friction and the normal force. Now, how to assign sign to these forces? When a box is placed on a ramp, I think that it naturally will want to fall down--with the force (mgsin(theta)) in the direction of the acceleration. However, a force is pulling this block up. So, I would assign the pushing force as positive and the weight of the block in the x direction as negative since that force is acting opposite the displacement of the block. Friction always acts to impede motion so I would also assign it as a negative force. Since the displacement of the block occurs in the x direction, all you have to do is to sum up all the forces and set the net force equal to ma--by Newton's second law. The only unknown is F, the pulling force, and you have one equation to solve for it. I hope this helps and good .

Click to expand...

I frequently mis-use equations for this type of simple kinematic questions.

For the question below, I struggled with what kind of equation i should use. And i chose x=Vot + (1/2)at^2 since I know total distance it needs to travel, initial v, and acceleration (=g) and looking for t.

Then, i got stuck as i tried to rewrite the equation into t=... so that i can test the relationship b/t the gravity and time, particularly for this situation.

t(Vo+.5at) = x
... can't solve for t !!? spent my precious 3minutes for this thing and have no idea..

1. Can anyone ellaborate the answer's explanation ?
2. and why isn't my approach not working ?
3. How can we so easily assume that time is inversely proportional to the gravity (that is if g/6 and you get 6t) , considering the gravity changes its direction as the thrown ball begins to fall down. Less gravity would cause less time taken to reach the highest pt (less pushing down force) but more time to reach to the ground from the highest pt (less pushing down force) ?



Question. Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time t. If the value of g were reduced to g/6 (as on the moon), then t would:

**A) increase by a factor of 6.
The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6. Choice A states this fact.

B) increase by a factor of 6^(1/2).

C) decrease by a factor of 6.

D) decrease by a factor of 6^(1/2).


Thank you !

minsoox1983 said:

I frequently mis-use equations for this type of simple kinematic questions.

For the question below, I struggled with what kind of equation i should use. And i chose x=Vot + (1/2)at^2 since I know total distance it needs to travel, initial v, and acceleration (=g) and looking for t.

Then, i got stuck as i tried to rewrite the equation into t=... so that i can test the relationship b/t the gravity and time, particularly for this situation.

t(Vo+.5at) = x
... can't solve for t !!? spent my precious 3minutes for this thing and have no idea..

1. Can anyone ellaborate the answer's explanation ?
2. and why isn't my approach not working ?
3. How can we so easily assume that time is inversely proportional to the gravity (that is if g/6 and you get 6t) , considering the gravity changes its direction as the thrown ball begins to fall down. Less gravity would cause less time taken to reach the highest pt (less pushing down force) but more time to reach to the ground from the highest pt (less pushing down force) ?



Question. Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time t. If the value of g were reduced to g/6 (as on the moon), then t would:

**A) increase by a factor of 6.
The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6. Choice A states this fact.

B) increase by a factor of 6^(1/2).

C) decrease by a factor of 6.

D) decrease by a factor of 6^(1/2).


Thank you !

Click to expand...

For a problem like this, you might want to use a kinematic formula that isolates t more simply. You chose a displacement formula, which forces you into a quadratic form, and so you run into trouble.

I would use a simpler one. How about solving for average velocity:

V=Vo + at;

Plug in g:

V=Vo + gt;

then solve for t:

t = (V-Vo)/g

Now, what ever you do on one side, do on the other and see what happens.

Since g is in the denominator, dividing g by 6 is the same as multiplying the whole term on the right by 6. What you do on the right, you do on the left. So it takes 6 times the time. Now, more logically, if you have one sixth of the gravitational acceleration, force due to gravity is MUCH weaker... so wouldn't it take a lot LONGER for the projectile motion to occur?

minsoox1983 said:

I frequently mis-use equations for this type of simple kinematic questions.

For the question below, I struggled with what kind of equation i should use. And i chose x=Vot + (1/2)at^2 since I know total distance it needs to travel, initial v, and acceleration (=g) and looking for t.

Then, i got stuck as i tried to rewrite the equation into t=... so that i can test the relationship b/t the gravity and time, particularly for this situation.

t(Vo+.5at) = x
... can't solve for t !!? spent my precious 3minutes for this thing and have no idea..

1. Can anyone ellaborate the answer's explanation ?
2. and why isn't my approach not working ?
3. How can we so easily assume that time is inversely proportional to the gravity (that is if g/6 and you get 6t) , considering the gravity changes its direction as the thrown ball begins to fall down. Less gravity would cause less time taken to reach the highest pt (less pushing down force) but more time to reach to the ground from the highest pt (less pushing down force) ?



Question. Suppose that a ball is thrown vertically upward from earth with velocity v, and returns to its original height in a time t. If the value of g were reduced to g/6 (as on the moon), then t would:

**A) increase by a factor of 6.
The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6. Choice A states this fact.

B) increase by a factor of 6^(1/2).

C) decrease by a factor of 6.

D) decrease by a factor of 6^(1/2).


Thank you !

Click to expand...

Lets find the time to go up to max height. Note that at max height, V = 0. Define positive direction up. So v0 vector points up and g vector points down.

V=Vo + at
0 = V0 - gt
v0 = gt
t = V0/g

The time for the round trip is twice as long (assuming there is no air resistance). t for the round trip = 2V0/g. If g is reduced by a factor of 6, t is increased by a factor of 6 because t and g are inversely proportional.

Thank you for your help! now I understood!

Hi, I had a question about how spring constants add up.

For example, if a single object is bound by springs on either side with the same spring constant, how does one go about finding the effective spring constant of the entire system?

A mathematical derivation would help me the most.

Thanks in advance.

midn said:

Hi, I had a question about how spring constants add up.

For example, if a single object is bound by springs on either side with the same spring constant, how does one go about finding the effective spring constant of the entire system?

A mathematical derivation would help me the most.

Thanks in advance.

Click to expand...

Assuming they individually have the same constant and on opposite sides of the object as stated, let's get the required force for a given delta x.

F=K*dx+K*dx=2K*dx ; equivalent spring constant Keq = 2K

If they were different constants:

F=K1*dx+K2*dx=(K1+K2)dx ; equivalent spring constant Keq = K1+ K2

So does it make a difference if the springs are attached to the same side or the opposite side?

midn said:

So does it make a difference if the springs are attached to the same side or the opposite side?

Click to expand...

It depends on how they are attached. If they are in line (spring-spring-object) it is different than the two springs side by side attached to the same side of the object which is more like your first question.

soonereng said:

It depends on how they are attached. If they are in line (spring-spring-object) it is different than the two springs side by side attached to the same side of the object which is more like your first question.

Click to expand...

Sorry, I meant 2X springs attached directly to the object on the same side. Is it the same result as two spring attached on opposite sides. If so, that baffles me.

I don't even want to think about spring-spring-object.

I wanted to post a question that someone posted in the regular MCAT forum about a physics problem. If anyone can help, that would be great!

I'm having a bit of difficulty solving this problem on my physics homework. I know this question isn't from an MCAT review book, but I figured maybe by posting this, it might help others understand some concepts a little better, so if anyone feels amibitious enough to help out, I'd greatly appreciate it.


A simple model of the membrane of a nerve cell is shown in the figure at the right (see link below). It consists of 2 batteries (ion pumps) with voltages V1=100mV and V2=50mV. The resistance to flow across the membrane is represented by two resistors with resistances R1=10K&#937; and R2=90K&#937;. The variability is represented by a switch, SW1. Four points are labeled by the letters a-d. Point b represents outside of membrane and point d inside the membrane.

Picture of circuit:
http://i107.photobucket.com/albums/m...icsCircuit.jpg

A) What is the voltage difference across the membrane when the switch is open?

B) What is the current flowing around the loop when the switch is closed?

C) What is the voltage drop across the resistor R1 when switch is open? closed?

D) What is the voltage drop across the resistor R2 when switch is open? closed?

E) What is the voltage difference across the membrane when switch is closed?

F) If the locations of resistance R1 and R2 were reversed, would the voltages across the cell membranes be different?
__________________

HippocratesX said:

I wanted to post a question that someone posted in the regular MCAT forum about a physics problem. If anyone can help, that would be great!

I'm having a bit of difficulty solving this problem on my physics homework. I know this question isn't from an MCAT review book, but I figured maybe by posting this, it might help others understand some concepts a little better, so if anyone feels amibitious enough to help out, I'd greatly appreciate it.


A simple model of the membrane of a nerve cell is shown in the figure at the right (see link below). It consists of 2 batteries (ion pumps) with voltages V1=100mV and V2=50mV. The resistance to flow across the membrane is represented by two resistors with resistances R1=10K&#937; and R2=90K&#937;. The variability is represented by a switch, SW1. Four points are labeled by the letters a-d. Point b represents outside of membrane and point d inside the membrane.

Picture of circuit:
http://i107.photobucket.com/albums/m...icsCircuit.jpg

A) What is the voltage difference across the membrane when the switch is open?

B) What is the current flowing around the loop when the switch is closed?

C) What is the voltage drop across the resistor R1 when switch is open? closed?

D) What is the voltage drop across the resistor R2 when switch is open? closed?

E) What is the voltage difference across the membrane when switch is closed?

F) If the locations of resistance R1 and R2 were reversed, would the voltages across the cell membranes be different?
__________________

Click to expand...

Hey! The link doesn't seem to work .

Can someone help with this question. I can only see that B is incorrect. Why why does D have to be the right answer? Doesn't A, C, and D work? For A, I considered a wire splitting to have R1 and R2 be parallel and then the wire comes back together and have R3 and R4 be parallel and then the wire comes back together. For C and D, I considered the circuit to have R1 and R2 in series and then the wire splits to have R3 and R4 parallel and then coming back together again. I don't know what I am missing here.

In the arrangement shown below, a current flows from P to Q. Which of the following could represent the current flowing through R1, R2, R3, and R4, respectively?

A. 1.0 A, 2.0 A, 1.0 A, 2.0 A
B. 1.0 A, 1.0 A, 0.5 A, 1.0 A
C. 3.0 A, 3.0 A, 4.0 A, 2.0 A
D. 3.0 A, 3.0 A, 2.0 A, 4.0 A

hmcalley said:

Can someone help with this question. I can only see that B is incorrect. Why why does D have to be the right answer? Doesn't A, C, and D work? For A, I considered a wire splitting to have R1 and R2 be parallel and then the wire comes back together and have R3 and R4 be parallel and then the wire comes back together. For C and D, I considered the circuit to have R1 and R2 in series and then the wire splits to have R3 and R4 parallel and then coming back together again. I don't know what I am missing here.

In the arrangement shown below, a current flows from P to Q. Which of the following could represent the current flowing through R1, R2, R3, and R4, respectively?

A. 1.0 A, 2.0 A, 1.0 A, 2.0 A
B. 1.0 A, 1.0 A, 0.5 A, 1.0 A
C. 3.0 A, 3.0 A, 4.0 A, 2.0 A
D. 3.0 A, 3.0 A, 2.0 A, 4.0 A

Click to expand...

Hey? Where is the figure??

gridiron, sorry the diagram seems to have been taken down for the link i posted above. Someone else had posted this question in another forum, and I was curious as to the manner in which to solve it, so I re-posted it in this physics questions thread. Anyway, sorry about that!

I have a quesiton on grav potential energy:

Grav potential energy is Ug= - GMm/r^2 and is abbreviated as Ug = mgh. For some reasons, I thought g = GM/r^2. If so, the two equations above are not the same. Could somebody derive the second equation from the second and explain how the NEGATIVE change in the first equation disappeared in the second? It doesn't make sense to me that the negative sign can disappear from the first to second.

Thanks.

pezzang said:

I have a quesiton on grav potential energy:

Grav potential energy is Ug= - GMm/r^2 and is abbreviated as Ug = mgh. For some reasons, I thought g = GM/r^2. If so, the two equations above are not the same. Could somebody derive the second equation from the second and explain how the NEGATIVE change in the first equation disappeared in the second? It doesn't make sense to me that the negative sign can disappear from the first to second.

Thanks.

Click to expand...

http://www.phy.duke.edu/~rgb/Class/phy51/phy51/node7.html

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravPotEnergy.htm

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/GravPotential.html

http://physics.ucsc.edu/~josh/6A/book/gravity/node11.html

But there an easier way to derive U = -GMm/r. Start with F = GMm/r^2. Since work is expressed in units of energy, and
W = F*d, U = F*d should make sense in terms of units.

U = F*r = GMm/r (remember that r in GMm/r^2 represents distance between centers of mass). Now we need to add a
negative sign, resulting in U = -GMm/r. Why the negative sign? Because energy is REQUIRED to separate objects that attract
each other. This energy is stored as gravitational potential energy. When objects are infinitely far apart, graviational potential
energy is zero (i.e. the largest possible) when this forumula is used. Without the negative sign, the formula would say that
as we separate objects that attract each other, the gravitational potential energy decreases. That would make no sense
because we must put energy INTO the system (by doing work) to separate objects that attract each other. Where does the
energy that was added to the system go? It becomes gravitational potential energy!!!!

Remember that potential energy is always defined relative to some reference point. In this case, reference point is zero
gravitational potential energy.

Now lets look at U = mgh (gravitational potential energy near the surface of the earth). Again, energy is required to separate objects that attract each other. So to lift an object to a height h with respect to some artibraty point (usually with respect to
a point on the surface of the earth), we must apply a force mg (i.e. a force equal and opposite to force due to gravity). This
way the object is in equilibrium when it's being lifted because there is no net force acting on it. From definition of work (W = F*d* cos theta), work done would be mgh and this work is stored as gravitational potential energy.

Note: one major difference between U = -GMm/r and U = mgh is that in U = -GMm/r the distance is in the denominator while in
U = mgh the distance is in the numerator. That's why we need a negative sign in U = -GMm/r but we don't need a negative sign
in U = mgh

question: if a positive charge and a negative charge attract, do they move toward each other with constant velocity, constant acceleration, or increasing acceleration? same question for a charge in an electric field and two charges of the same charge. thanks.

cloosh said:

question: if a positive charge and a negative charge attract, do they move toward each other with constant velocity, constant acceleration, or increasing acceleration? same question for a charge in an electric field and two charges of the same charge. thanks.

Click to expand...

F = KQq/r^2 (Coulomb's Law)

Since there is a net force, there is acceleration (F = ma). Since force depends on the distance (i.e. inverse
of the square of the distance), as the charges that attract each other get closer to each other, the distance between them decreases and the force between them increases. Since force increases, acceleration cannot be constant. Acceleartion also increases.

All this assumes that there are no other forces acting on the charges.

BrokenGlass said:

F = KQq/r^2 (Coulomb's Law)

Since there is a net force, there is acceleration (F = ma). Since force depends on the distance (i.e. inverse
of the square of the distance), as the charges that attract each other get closer to each other, the distance between them decreases and the force between them increases. Since force increases, acceleration cannot be constant. Acceleartion also increases.

All this assumes that there are no other forces acting on the charges.

Click to expand...

BG's post raises an important concept tested on the MCAT--applications of Newton's second law. Remember, you should use the second law for the net force acting on an object. Also, keep in mind that the electrostatic force is an inverse square law. If the distance increases, the force will decrease and vice versa. Make sure you understand the concepts clearly as they come up on the test very frequently and understanding them can save you time.

There is a 24V battery connected to two 3 and 6 ohm resistors in parallel and another 4 ohm resistor in series after the two in parallel...sorry can't get image up...

How much current passes through the 3-ohm resistor?

2 amps
3 amps
8/3 amps
4 amps
4/3 amps

The answer is 8/3 amps...I can get the overall current, but I don't know how to break down the two parallel resistors to get individual currents by Kirchoff's rule...if someone could explain that, it would be extremely helpful!

HristosKaran said:

There is a 24V battery connected to two 3 and 6 ohm resistors in parallel and another 4 ohm resistor in series after the two in parallel...sorry can't get image up...

How much current passes through the 3-ohm resistor?

2 amps
3 amps
8/3 amps
4 amps
4/3 amps

The answer is 8/3 amps...I can get the overall current, but I don't know how to break down the two parallel resistors to get individual currents by Kirchoff's rule...if someone could explain that, it would be extremely helpful!

Click to expand...

Solution:

el aye said:

Solution:

Click to expand...

Thanks! But can you explain the math a little more on figure two, I know I'm being dense but I dont' really see how you got 8V..

I think he's try to explain voltage drop across the parallel resistor which will determine the current. Total resistance in the circuit is 6, so by using ohm current is 4 amp. Now you calculate the volatge drop across the parallel resistor
V = IR = 4 amp x 2 ohm (resistance across parallel cirucit) = 8

If the average velocity of an object is zero in some time interval what can you say about the displacement of the object during that interval?

dill_punjabi said:

If the average velocity of an object is zero in some time interval what can you say about the displacement of the object during that interval?

Click to expand...

Let me take a stab at this brain teaser. Average velocity can be defined as: avg V = delta(x)/delta(t). So you are saying that avg V=0. This means that delta(x) must be equal to 0 also (0=0). There is no change in displacement.

HristosKaran said:

Thanks! But can you explain the math a little more on figure two, I know I'm being dense but I dont' really see how you got 8V..

Click to expand...

In a series, the voltage drop on R (1 and 2) and R3 are different but the current is the same so, using V=IR3, we solve for the voltage drop at R3 by 4 amperes*2 ohms (R3) = 8V.

Remember: in a parallel circuit, the voltage drop for each resistor is the same while the current may be different and, in a series, the current is the same for resistors in series but the voltage drop may be different.

I am having a difficult time with Newton's Law of Universal Gravitation. Is there any one that can help me figure out this problem: I understand the problem but need to understand how to use the ratios that are involved. Can someone show me as it applies to these two problems.

1. Planet B has twice the mass of Planet A. Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10 m above the surface of Planet B and at the same time a 10 kg mass is dropped 10m above the surface of Planet A. If the mass on Planet B strikes the ground in 10 sec,the mass on Planet A strikes the ground in approx:
a.7, b. 10, c. 14, d.20 ans is a.
2. Planet A and B have the same mass . Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10m above the surface of Planet A and at the same a 5kg mass is dropped 10m above the surface of Planet B. If the mass on Planet B strikes the ground in 10 sec, the mass of Planet A strikes the ground in :
a. 2.5s b.5s c.10s d. 20s ans is b.

thanks

I have a brief question about a tricky momentum concept.

If you take any object at rest and it explodes, why is total momentum zero immediately after explosion? I know P=mv

My line of thinking is this: Each fragment has a mass and a (very high) velocity after the explosion, so how can it not have a momentum?