Physics question thread

[Shrike]

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.

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[gridiron]

A car is driving on a level road at a contant speed 8m/s when it attempts to execute a turn about curve of effective radius 10m. For the following questions, we will assume the turn is successful, that is, the car performs the turns as the driver intends. The static coeff of friction between the tires and the road is .9, the kinetic coeff of friction is .7.

What force provides the centripetal force?
A. Gravity
B. The normal force
C. Static friction
D. Kinetic friction

Answer is C. I thought that it would be kinetic friction since the car is moving. The explanation for static friction is: Since the tires are not slipping on the road, the appropriate friction is static.

When a force which acts on a body causes it to slide along a surface, a frictional force from the surface acts on the body. If the body does not slide across the surface, that the frictional force is a static force. For a level curve, the frictional force provides the centripetal force due to the frictional force between the tires of the car and the road. If you draw the free body diagram for the car on the level curve, you will have the following forces:

1. The weight of the car
2. Normal force due to surface interactions
3. Frictional force, pointing inward, due to tire/road interactions

Why? Think of when you are driving a car. When you guide the steering wheel of the car at a sharp left turn, you slide across your seat toward the right--kind of like the seat slid beneath you. What is happening? The car actually wants to stay in a straight path. But, due to the wheel of the car changing direction due to the curve, the wheel will have a natural tendency to move in a direction away from the center. In response, the road applies a frictional force, which is directed inward and is perpendicular to the direction of the car--the force keeps the car going in a circle.

 

[rcd]

A car is driving on a level road at a contant speed 8m/s when it attempts to execute a turn about curve of effective radius 10m. For the following questions, we will assume the turn is successful, that is, the car performs the turns as the driver intends. The static coeff of friction between the tires and the road is .9, the kinetic coeff of friction is .7.

What force provides the centripetal force?
A. Gravity
B. The normal force
C. Static friction
D. Kinetic friction

Answer is C. I thought that it would be kinetic friction since the car is moving. The explanation for static friction is: Since the tires are not slipping on the road, the appropriate friction is static.

deja vu, I was just thinking about this yesterday or sometime today while driving.

Anyway, the reasoning behind 0 velocity between the surface and tire is this:

The net velocity of any point at the edge of a tire can be broken down into 2 smaller velocities:
1. The overall velocity of the tire as a whole.
2. The rotational velocity of the point.

So, you just add the two, remembering that velocity includes direction. For a point at the surface, v1 + v2 = 0, so you're talking no movement relative to the surface. So you're talking static friction.

 

[midn]

When a force which acts on a body causes it to slide along a surface, a frictional force from the surface acts on the body. If the body does not slide across the surface, that the frictional force is a static force. For a level curve, the frictional force provides the centripetal force due to the frictional force between the tires of the car and the road. If you draw the free body diagram for the car on the level curve, you will have the following forces:

1. The weight of the car
2. Normal force due to surface interactions
3. Frictional force, pointing inward, due to tire/road interactions

Why? Think of when you are driving a car. When you guide the steering wheel of the car at a sharp left turn, you slide across your seat toward the right--kind of like the seat slid beneath you. What is happening? The car actually wants to stay in a straight path. But, due to the wheel of the car changing direction due to the curve, the wheel will have a natural tendency to move in a direction away from the center. In response, the road applies a frictional force, which is directed inward and is perpendicular to the direction of the car--the force keeps the car going in a circle.

Centripetal force really confuses me. Why would the frictional force point inward if centripetal force is in the same direction? I thought friction would point away since it is trying to oppose motion.

I'm probably wrong because if we consider a satellite orbiting the earth, the gravitational force would be pulling it into the earth in the same direction as the centripetal force and centripetal force would be equal to the force of gravity. So what force stops the satellite from just getting sucked into the earth?

Bah.

 

[gridiron]

Centripetal force really confuses me. Why would the frictional force point inward if centripetal force is in the same direction? I thought friction would point away since it is trying to oppose motion.

I'm probably wrong because if we consider a satellite orbiting the earth, the gravitational force would be pulling it into the earth in the same direction as the centripetal force and centripetal force would be equal to the force of gravity. So what force stops the satellite from just getting sucked into the earth?

Bah.

Don't worry too much as to what force keeps the satellite from getting sucked into the earth--the idea is beyond the scope of the MCAT.

Imagine you are driving your car in a circular arc. When you encounter a turn in the road, after you have been going in a straight path for a while, the car turns sharply jamming your body against the car. The force keeping you going in the circle is the centripetal force. By Newton's second law, F=ma, your acceleration is also directed toward the center of the circular arc. A centripetal force accelerates a body by changing the direction of the body's velocity. Naturally, as you encounter the curve in your car, the car wants to keep going straight--Newton's first law. However, since the you steer the car otherwise, there must be force acting on the car go accelerate it toward the center of the circular arc. The friction between the wheels of the car and the road is what provides this force. Why not kinetic friction? The centripetal force does not change the speed of the body, only the velocity. The velocity of a body moving in a circle is always tangent to the path. The centripetal force always points inward--thus the centripetal force is always perpendicular to the velocity of the body. This means the work the centripetal force does on the body, W = fdcostheta, is zero since theta is always 90 degrees--the centripetal force is always perpendicular to the displacement of the object. So, as the object moves in a circle, the centripetal force does not alter the mechanical energy of the object since it does not do work on the displacement of the object. This means the kinetic energy remains constant and the speed, not velocity, of the object is constant as well. That is why static friction is taken into consideration and not kinetic friction--the traction between the wheels of the car and the surface of the road.

 

[midn]

Don't worry too much as to what force keeps the satellite from getting sucked into the earth--the idea is beyond the scope of the MCAT.

Imagine you are driving your car in a circular arc. When you encounter a turn in the road, after you have been going in a straight path for a while, the car turns sharply jamming your body against the car. The force keeping you going in the circle is the centripetal force. By Newton's second law, F=ma, your acceleration is also directed toward the center of the circular arc. A centripetal force accelerates a body by changing the direction of the body's velocity. Naturally, as you encounter the curve in your car, the car wants to keep going straight--Newton's first law. However, since the you steer the car otherwise, there must be force acting on the car go accelerate it toward the center of the circular arc. The friction between the wheels of the car and the road is what provides this force. Why not kinetic friction? The centripetal force does not change the speed of the body, only the velocity. The velocity of a body moving in a circle is always tangent to the path. The centripetal force always points inward--thus the centripetal force is always perpendicular to the velocity of the body. This means the work the centripetal force does on the body, W = fdcostheta, is zero since theta is always 90 degrees--the centripetal force is always perpendicular to the displacement of the object. So, as the object moves in a circle, the centripetal force does not alter the mechanical energy of the object since it does not do work on the displacement of the object. This means the kinetic energy remains constant and the speed, not velocity, of the object is constant as well. That is why static friction is taken into consideration and not kinetic friction--the traction between the wheels of the car and the surface of the road.

So is it safe to say that centripetal force always has to be caused by something that points in the same direction of the resultant centripetal force?

I understood everything you said except the part that says because kinetic energy remains constant, kinetic friction does not apply. If there was a box sliding down a ramp at constant velocity (and speed), isn't there kinetic friction there with constant kinetic energy?

The only difference I see between the two examples is that mechanical energy is changing here with gravitational potential.

Could you please explain the kinetic energy part more and how that is relevant to kinetic vs. static friction?

 

[gridiron]

So is it safe to say that centripetal force always has to be caused by something that points in the same direction of the resultant centripetal force?

Yes. The direction of the net force, which should direct inward, is the direction of the centripetal force. This happens during uniform circular motion. That is the only way you can have an acceleration--according to Newton's second law Fnet = ma.

I understood everything you said except the part that says because kinetic energy remains constant, kinetic friction does not apply. If there was a box sliding down a ramp at constant velocity (and speed), isn't there kinetic friction there with constant kinetic energy?

The only difference I see between the two examples is that mechanical energy is changing here with gravitational potential.

Could you please explain the kinetic energy part more and how that is relevant to kinetic vs. static friction?

Kinetic friction is the frictional force which opposes the motion of a sliding body. Although the velocity of a body changes during uniform circular motion, its speed remains constant--velocity is a vector and speed is a scalar. Because the speed remains constant, the kinetic energy of the vehicle remains constant--no work is done on the body. In the case of a body sliding down a ramp, the kinetic energy of the body increases due to work done by the gravitational force of the body--provided the ramp is frictionless. In terms of a car moving in a circle, the force which accelerates the body by changing its velocity and keeping it from traveling in a tangential path is the static frictional force. As the tires move along the road, there are interactions between the tire and the road--this interaction is what makes the turn of a car around a curve in the road possible. Otherwise, the car would fly tangentially off the path.

 

[midn]

Yes. The direction of the net force, which should direct inward, is the direction of the centripetal force. This happens during uniform circular motion. That is the only way you can have an acceleration--according to Newton's second law Fnet = ma.



Kinetic friction is the frictional force which opposes the motion of a sliding body. Although the velocity of a body changes during uniform circular motion, its speed remains constant--velocity is a vector and speed is a scalar. Because the speed remains constant, the kinetic energy of the vehicle remains constant--no work is done on the body. In the case of a body sliding down a ramp, the kinetic energy of the body increases due to work done by the gravitational force of the body--provided the ramp is frictionless. In terms of a car moving in a circle, the force which accelerates the body by changing its velocity and keeping it from traveling in a tangential path is the static frictional force. As the tires move along the road, there are interactions between the tire and the road--this interaction is what makes the turn of a car around a curve in the road possible. Otherwise, the car would fly tangentially off the path.

Thanks for all your help, but it's just not clicking for me. I'll stick to another method of imagining static versus kinetic friction.

I'll just imagine that some point on the tires while the car is in motion never actually slides relative to the surface since it only makes contact with the road at a single point. Therefore, it is static friction.

Now, another nit-picky question that has been bothering me: what is an eV exactly? I am not actually to my electricity lectures yet, but it came up in one of the lecture exams in my EK book. I had to find the energy in a certain amount eV. The passage only gave me the charge of an electron 1.6 x 10^-19 C. According to Wikipedia, an eV "is the amount of kinetic energy gained by a single unbound electron when it passes through an electrostatic potential difference of one volt." How was I supposed to take the charge of the electron and translate it into energy? Is there 1.6 x 10^-19 C per electron volt?

Stupid question, I know.

Thanks for your time.

 

[Anastasis]

Well, eV is a measure of energy so to find the amount of energy in J in a certain number of eV you only need to convert the energy from eV to J

1 eV = 1.602 x 10^-19 C

I think eV exists because it's an easier unit for use when dealing with the very low levels of energy on an atomic level.

To covert the charge on the electron to energy just use

W=qV

W = energy in Joules
q = charge
V = potential difference in Volts

I don't know; I'm very tired. Maybe that didn't answer your question at all but I hope it helped.

 

[gridiron]

Thanks for all your help, but it's just not clicking for me. I'll stick to another method of imagining static versus kinetic friction.

I'll just imagine that some point on the tires while the car is in motion never actually slides relative to the surface since it only makes contact with the road at a single point. Therefore, it is static friction.

Now, another nit-picky question that has been bothering me: what is an eV exactly? I am not actually to my electricity lectures yet, but it came up in one of the lecture exams in my EK book. I had to find the energy in a certain amount eV. The passage only gave me the charge of an electron 1.6 x 10^-19 C. According to Wikipedia, an eV "is the amount of kinetic energy gained by a single unbound electron when it passes through an electrostatic potential difference of one volt." How was I supposed to take the charge of the electron and translate it into energy? Is there 1.6 x 10^-19 C per electron volt?

Stupid question, I know.

Thanks for your time.

Actually, don't worry too much about why static friction is the cause of the centripetal force for the MCAT. Just know the important facets to uniform circular motion: how the velocity vector is drawn, what is the centripetal force, direction of the centripetal force and acceleration, why centripetal force doesn't do work on the body, the equations to solve for force and acceleration, general method to solving problems, and applications such as roller coaster problems and satellites. If you know that, then you should be fine for the level necessary for the MCAT.

 

[gridiron]

Well, eV is a measure of energy so to find the amount of energy in J in a certain number of eV you only need to convert the energy from eV to J

1 eV = 1.602 x 10^-19 C

I think eV exists because it's an easier unit for use when dealing with the very low levels of energy on an atomic level.

To covert the charge on the electron to energy just use

W=qV

W = energy in Joules
q = charge
V = potential difference in Volts

I don't know; I'm very tired. Maybe that didn't answer your question at all but I hope it helped.

Correct

 

[endofevangelion]

Why is the speed of sound (longitudinal wave) higher in glass than air while the speed of light (transverse wave) lower in glass than in air?

 

[Anastasis]

Why is the speed of sound (longitudinal wave) higher in glass than air while the speed of light (transverse wave) lower in glass than in air?

Because sound waves actually displace the material they travel through - so a substance where the molecules are closer together (ie glass) will allow the wave to move faster than in a substance where the molecules are further apart (ie air). That's breaking it down to its basics. From Wikipedia:
"In general, the speed of sound is proportional to the square root of the ratio of the stiffness of the medium and its density"

But, light is an electromagnetic wave. It is an entity unto itself and not the result of displacing matter. In fact, it can impart momentum and energy to matter so it makes sense that it would travel fastest in an environment where there is less matter blocking the way (ie air or a vacuum).

 

[aevea]

Why is the speed of sound (longitudinal wave) higher in glass than air while the speed of light (transverse wave) lower in glass than in air?

The difference isn't about transverse vs. longitudinal, but rather about matter wave vs. field wave.

The waves that we experience as sound are propagated through the motion of particles. Particles in solids and liquids such as glass and water resist compression better than gases such as air. Therefore, when a matter compression wave propagates through the solids the particles move less and "snap back" faster so the wave front travels faster.

Electromagnetic waves don't rely on the motion of particles but rather on the fluctuations of electric and magnetic fields. Denser (clear) substances such as glass and water interfere with and slow the propagation of the fields and therefore slow the wave front.

 

[midn]

I just started the wave stuff and realized that I never learned this properly back in physics.

I do not understand standing waves at all. First, in order to have a standing wave with all of its harmonics, my EK book states you need to have both ends tied down to create nodes or both ends loose to create antinodes. In order to get the first harmonic, you must have two nodes and the middle of the string would be the antinode. How do you get the first harmonic when you have one end of a string tied to the wall and the other end loose? We are told that only odd harmonics can happen in this case (including the first), but it was just stated that you must have two nodes to achieve the first harmonic.

Ok, next we are given an equation L=(number of harmonic)(wavelength of harmonic)/2. What good is this equation? Does it allow me to solve for the wavelength of certain harmonic numbers so I can figure out the resonant frequencies?

Lastly, why do you get only the odd harmonics when you have one end tied down and the other end loose? Do the even nodes become antinodes by meeting constructively?

Thanks in advance.

 

[BrokenGlass]

I remember reading somewhere that in those nuclear reactions in which a radioactive particle is converted into 2 gamma rays,
(E = mc^2), the reason the 2 gamma rays move away from each other
in a straight line is because of conservation of linear momentum
(in other words, the linear momentum has to be zero both before
and after the reverse collison). But since gamma rays are just
high-energy photons and photons are massless, how can they
have momentum? Shouldn't their momentum ALWAYS be zero? So how
can the conservation of momentum explain the straight line motion
of gamma rays away from each other? Thank you!

 

[spicedmanna]

I remember reading somewhere that in those nuclear reactions in which a radioactive particle is converted into 2 gamma rays,
(E = mc^2), the reason the 2 gamma rays move away from each other
in a straight line is because of conservation of linear momentum
(in other words, the linear momentum has to be zero both before
and after the reverse collison). But since gamma rays are just
high-energy photons and photons are massless, how can they
have momentum? Shouldn't their momentum ALWAYS be zero? So how
can the conservation of momentum explain the straight line motion
of gamma rays away from each other? Thank you!

Ah, yes, it's the age old question, "how can something have momentum, if it doesn't have mass?" Well, first, you need to dissociate yourself from the traditional Newtonian way of thinking of momentum. Let's look at things from a relativistic point of view. If you think in terms of energy, you can relate energy with momentum, wavelength and Planck's constant. I think we can agree that photons have energy. Indeed the energy of a quanta of photons is expressed thusly:

E = hf​

We also know that frequency (f) can be related to the speed of light and the wavelength:

f = c/lambda​

By substitution, we now have:

E = h(c/lambda)​

If we rearrange the above equation, solving for wavelength (lambda), we get:

lambda = h(c/E)​

Let's call that equation A. We'll come back to that shortly.

Now, let's consider energy-mass equivalency: E = (gamma)mc^2. Let's call that equation B. Also, consider that momentum can be expressed as P = (gamma)mv; let's call that equation C. In both cases, gamma is the relativity factor which is equal to 1/(sqrt(1-((v^2)/c^2)).

Let's now equate equation B and equation C:

P/E = v/(c^2)​

If we consider what happens to the above equation as mass approaches zero, we know that v approaches c. Hence, under the stated condition and upon rearrangement we have:

P = E/c​

Let's come back now to equation A. Upon inspection, we find that c/E is really just 1/P, right? So by substitution, we get:

lambda = h/P​

Another way of stating this is:

P = h/lambda​

So, based on all of the above, we can say that photons have momenta.

 

[gridiron]

I just started the wave stuff and realized that I never learned this properly back in physics.

I do not understand standing waves at all. First, in order to have a standing wave with all of its harmonics, my EK book states you need to have both ends tied down to create nodes or both ends loose to create antinodes. In order to get the first harmonic, you must have two nodes and the middle of the string would be the antinode. How do you get the first harmonic when you have one end of a string tied to the wall and the other end loose? We are told that only odd harmonics can happen in this case (including the first), but it was just stated that you must have two nodes to achieve the first harmonic.

Ok, next we are given an equation L=(number of harmonic)(wavelength of harmonic)/2. What good is this equation? Does it allow me to solve for the wavelength of certain harmonic numbers so I can figure out the resonant frequencies?

Lastly, why do you get only the odd harmonics when you have one end tied down and the other end loose? Do the even nodes become antinodes by meeting constructively?

Thanks in advance.

Hey! The idea of one end of the rope loose and the other end tied to the wall is used to develop the idea of standing waves. If you wiggle the rope with your hand, you will create waves of a frequency f that will travel down the rope. As the waves hit the wall, they will be reflected. Now, you will have two waves on the rope--the wave you generate with your hand and the reflected wave. Because two waves exist on the rope, interference will occur. If you wiggle the rope with just the "right" frequency, you will create standing waves where the wave you create and the reflected wave add up. This is called a standing wave. If you were to take frameshots of this standing wave you will notice two distinct points. Some points will not vibrate at all and these are nodes. The point where the amplitude is maximized, halfway between two nodes, is the antinode. There are only certain conditions where standing waves will be produced. If you have a rope of length L, consider the simplest standing wave that can be formed. This is when there are nodes at both ends of the rope. Now a complete wavelength is defined as the distance between two crests or two troughs, however in the case of only two nodes, you have only half a wave. Therefore, L = 0.5lambda. Now consider when you have a complete wave. A complete wave has three nodes. This would be the second harmonic. Now L = lambda. A pattern will emerge where you will get:

lambda [where n is some whole number] = 2L/n

n is very important here. It is the harmonic number. The first harmonic is the fundamental harmonic and is lambda(1)... Also, lambda multiplied by frequency is the velocity for a wave. You can substitute variables in the above equation and obtain:

f = nv/2L

These equations can be used to find the harmonic wavlength, the velocity given a frequency, harmonic and length or other harmonic frequencies given the fundamental frequency.

 

[tp709]

There was something regarding this on my august exam that caught me off guard, I'm not quite clear on this concept,

but what happens to the density of a solid when heated?
I know that if it is heated it expands...does that mean it becomes less dense?

 

[deleted103644]

There was something regarding this on my august exam that caught me off guard, I'm not quite clear on this concept,

but what happens to the density of a solid when heated?
I know that if it is heated it expands...does that mean it becomes less dense?

Just think about it for a second. Density is defined as mass/volume, right? Now, assuming you're not heating the gold enough to start shedding off atoms or subatomic particles, the mass will stay the same when you heat it. So if d1 = m1/v1 and when you heat the metal d2 = m1/v2 where v2 > v1, logically d2 < d1.

 

[deleted103644]

I remember reading somewhere that in those nuclear reactions in which a radioactive particle is converted into 2 gamma rays,
(E = mc^2), the reason the 2 gamma rays move away from each other
in a straight line is because of conservation of linear momentum
(in other words, the linear momentum has to be zero both before
and after the reverse collison). But since gamma rays are just
high-energy photons and photons are massless, how can they
have momentum? Shouldn't their momentum ALWAYS be zero? So how
can the conservation of momentum explain the straight line motion
of gamma rays away from each other? Thank you!

Anything with kinetic energy has momentum, even if it doesn't have mass!

 

[midn]

Hey! The idea of one end of the rope loose and the other end tied to the wall is used to develop the idea of standing waves. If you wiggle the rope with your hand, you will create waves of a frequency f that will travel down the rope. As the waves hit the wall, they will be reflected. Now, you will have two waves on the rope--the wave you generate with your hand and the reflected wave. Because two waves exist on the rope, interference will occur. If you wiggle the rope with just the "right" frequency, you will create standing waves where the wave you create and the reflected wave add up. This is called a standing wave. If you were to take frameshots of this standing wave you will notice two distinct points. Some points will not vibrate at all and these are nodes. The point where the amplitude is maximized, halfway between two nodes, is the antinode. There are only certain conditions where standing waves will be produced. If you have a rope of length L, consider the simplest standing wave that can be formed. This is when there are nodes at both ends of the rope. Now a complete wavelength is defined as the distance between two crests or two troughs, however in the case of only two nodes, you have only half a wave. Therefore, L = 0.5lambda. Now consider when you have a complete wave. A complete wave has three nodes. This would be the second harmonic. Now L = lambda. A pattern will emerge where you will get:

lambda [where n is some whole number] = 2L/n

n is very important here. It is the harmonic number. The first harmonic is the fundamental harmonic and is lambda(1)... Also, lambda multiplied by frequency is the velocity for a wave. You can substitute variables in the above equation and obtain:

f = nv/2L

These equations can be used to find the harmonic wavlength, the velocity given a frequency, harmonic and length or other harmonic frequencies given the fundamental frequency.

Thanks a lot, that helps quite a bit. If you have the time, could you explain conceptually why only the odd harmonics (and thus, I guess only complete wavelengths) occur when you have one end loose and the other tied down?

 

[evoviiigsr]

Traction force is referring to the force exerted on the femur.

 

[gridiron]

Thanks a lot, that helps quite a bit. If you have the time, could you explain conceptually why only the odd harmonics (and thus, I guess only complete wavelengths) occur when you have one end loose and the other tied down?

Let me pm you this answer--it will require a drawing.

 

[gridiron]

Traction force is referring to the force exerted on the femur.

Just of curiosity, does this question concern the MCAT?

 

[evoviiigsr]

lol its mcat practice, seemed harder than any pulley problem ive had in physics

 

[byeh2004]

Hey guys, I was reading on fluid dynamics the other day and I had a few questions about the concepts of archimedes principle.

How does a boat or ship carrying hundreds of lbs worth of material float while that same stuff would sink to the bottom of the ocean if dumped overboard?
- totally clueless on this one; is it because the density of the ship is lighter than the density of the water?

Will the buoyancy felt by your body in a pool be more if you stretch your body out flat or if you curl up into a ball?
- Wouldn't it not matter because your volume is always going to be the same?

Thanks!

 

[gridiron]

Hey guys, I was reading on fluid dynamics the other day and I had a few questions about the concepts of archimedes principle.

How does a boat or ship carrying hundreds of lbs worth of material float while that same stuff would sink to the bottom of the ocean if dumped overboard?
- totally clueless on this one; is it because the density of the ship is lighter than the density of the water?

Will the buoyancy felt by your body in a pool be more if you stretch your body out flat or if you curl up into a ball?
- Wouldn't it not matter because your volume is always going to be the same?

Thanks!

Hi. The reason the ship will float and not sink is because it will displace a large amount of water. The buoyant force by definition is equal to the weight of the fluid displaced. In equation form this is equal to:

m(f)*g

Where m(f) is the mass of the fluid displaced. In fluid dynamics, the density of the substance is often used, so m(f) = rho*V. (rho is equal to density). From this equation. if the volume of the object is made larger, more water will be displaced leading to a greater buoyant force created. As a result, the ship will not sink. Also, and you will not need to know this for the MCAT, part of the bottom of a ship is filled with air in order to make it lighter than water. Best of .

 

[PoorSanDiegan]

Thanks for all your help, but it's just not clicking for me. I'll stick to another method of imagining static versus kinetic friction.

I'll just imagine that some point on the tires while the car is in motion never actually slides relative to the surface since it only makes contact with the road at a single point. Therefore, it is static friction.

Now, another nit-picky question that has been bothering me: what is an eV exactly? I am not actually to my electricity lectures yet, but it came up in one of the lecture exams in my EK book. I had to find the energy in a certain amount eV. The passage only gave me the charge of an electron 1.6 x 10^-19 C. According to Wikipedia, an eV "is the amount of kinetic energy gained by a single unbound electron when it passes through an electrostatic potential difference of one volt." How was I supposed to take the charge of the electron and translate it into energy? Is there 1.6 x 10^-19 C per electron volt?

Stupid question, I know.

Thanks for your time.

Hi Midn - I know this might be a little late - but I had the same question (I have EK too).

the response: 1 eV = 1.602 x 10^-19 C
I think was meant to be: 1 eV = 1.602 x 10^-19 C * V
which is equal to: 1 ev = 1.602 x 10^-19 J

For the problem I think you're referring to, since 1 electron carries 1.6e-19 Couloubs, think of it as:

1 electron-volt * (1.6e-19 Couloub) / (1 electron) = 1.6e-19 C*V

Sorry if this is repetitive!

 

[WestSideSunDanc]

How long does it take for a 33 rpm record to rotate once, at constant speed?

 

[deleted36260]

How long does it take for a 33 rpm record to rotate once, at constant speed?

There's two things about problems like this to remember (someone may be able to give a smoother answer).

1> Think of 33 rpm as "chopping a minute into 33 parts". How long is each part? The issue so many people have is the immediate want to say, "look how easy, just 33rpm = 33 PER 60 seconds = 33/60 = .55"
Say to yourself, when measuring rotation, "each rotation is a chopped up part of the whole time unit in question".


2> The other thing is to simply check yourself when you get confused with some easy math. 60 rpm would be one revolution per second. Right? Okay, no problem there. 60 rpm is much faster than 33 rpm, so OBVIOUSLY 33rpm has to take longer to get around that record than 1 second.

So, flip your division 60/33 =1.8

If we were going 30rpm, we'd be taking twice as long as 60 rpm to get around. So it'd be 2 seconds. 33 is close to 30. So our 1.8, if we know nothing else, is pretty damn close to all the values expected that we can easily work out in our head

 

[deleted36260]

When reading the physics Q&A, I got to this:

Question: An object is floating at the surface of two fluids; two thirds of its volume is below the surface. If the densities of the two fluids are 1.0 g/cm^3 and 1.6g/cm^3, what is the density of the object?

Answer: Subtract the density of the lighter fluid from that of the heavier, and we get 0.6 g/cm^3. Two thirds of the object is below the surface, so that yields a density of 2/3 * 0.6 = 0.4 g/cm^3. Then we need to add the density of the first fluid back in: 0.4 + 1.0 = 1.4 g/cm^3.

Perhaps I don't understand the setup of the problem (what the experiment looks like), but I don't get why you add back in the first fluid, or while it's taken out. Can someone better explain, or sketch me something in mspaint?

 

[lisichka]

hello,
i have an exam this monday, and this question keeps on bugging me. please help.

if we have two floating blocks, each weighing 5 kg. one made of wood, another-->steel, and both of them float in the same fluid. Steel block is 60 % submerged, and wood is 40 percent submerged. kaplan say that both of them MUST have same Fbuyoant, b/c they have the same mass and they both FLOAT(if one would sink, it would be diff) But....,how can they have the same Fb if they have different VOLUMES submerged in the fluid? As far as i know, Fb=fluid density*Vsub*g.

Can someone help me with this. thank you very much.

 

[lisichka]

oops sorry
double message

 

[BrokenGlass]

hello,
i have an exam this monday, and this question keeps on bugging me. please help.

if we have two floating blocks, each weighing 5 kg. one made of wood, another-->steel, and both of them float in the same fluid. Steel block is 60 % submerged, and wood is 40 percent submerged. kaplan say that both of them MUST have same Fbuyoant, b/c they have the same mass and they both FLOAT(if one would sink, it would be diff) But....,how can they have the same Fb if they have different VOLUMES submerged in the fluid? As far as i know, Fb=fluid density*Vsub*g.

Can someone help me with this. thank you very much.

Since Fb is equal to the weight of the fluid displaced, both the wood block and the steel block can displace the same volume (and therefore the same weight) of fluid and hence experience identical Fb. The volume of the steel block is less than the volume of the wooden block (let sb represent steel block and wb represent wooden block. density = mass/volume => volume = mass/density;
Volume of sb = mass of sb/ density of sb; Volume of wb = mass of wb / density of wb; since masses of two
blocks are equal, and density of sb > density of wb, volume of sb < volume of wb) so 60% of steel block could be equal to 40% of the wood block by volume.

Another way to look at it is as follows: both the steel block and the wood block have the same mass and therefore the same weight. Since they both float, they are in equilibrium, so there must be a force up to balance out their weight. Since their weights are the same, their Fb must also be the same. Anyhow, that's my understanding, so if it's incorrect, I hope someone will correct me.

 

[BrokenGlass]

For a symmetrical (i.e. constant radius of coils, constant spacing between coils) solenoinds, why is the magnetic field (inside the solenoid) strongest in the middle (and weakest at the right and left ends)?

Also, for assymetrical solenoids (suppose coils get bigger from left to right, but the spacing between the coils stays the same), why is the magnetic field inside the solenoid strongest at the end where coils are smaller?

I think the relevant equation is B = mew * i * n, but I don't understand how do derive these results from this equation.

I sort of understand that the magnetic fields between adjacent coils cancel each other out, and I think that's part of the reason magnetic field outside the solenoid is very weak, but not zero because there is a right hand rule which says that if you curl your fingers in the direction the current is running through the coils of a solenoid, your thumb points in the direction of the magnetic field coming out of the solenoid, and since magnetic fields have to form loops, there must be at least one loop on the outside (from north pole to south pole). In all the pictures I have seen, the magnetic field lines are very dense on the inside of a solenoid and they are the most dense right in the middle, but I don't really understand why.

 

[BrokenGlass]

My understanding is that an ideal fluid has to meet 3 or 4 conditions, one of which is that it should not be viscous. Ideal fluids obey the continuity equation, i.e. the volume flow rate is constant over the entire pipe (but not necessarily constant over time). A1*v1 = A2*v2. The continuity equation is a consequence of conservation of mass (since rate of flow into the pipe must be equal to the rate of flow out of the pipe).

If any of these 4 conditions is violated, the fluid is a non-ideal (real) fluid. In particular, if viscosity (i.e. internal friction of the fluid) is not zero, Poiseuille's Law can be used to describe volume flow rate because it takes viscosity (i.e. resistance) into account.

Q = (pi*r4*(p2-p1)) / (8*n*L), where resistance = (8*n*L) / pi*r4

Here are my questions:

1) Since continuity equation doesn't apply to real fluids, the only conclusion I can draw is that for a real fluid, the rate of inflow is NOT equal to the rate of outflow (otherwise the continuity equation would apply). But wouldn't the fluid build up in the pipe? I don't understand how that would work.

2) Lets assume blood is viscous, so Poiseuille's Law applies. If we constrict a blood vessel, would the value of Q before constriction equal the value of Q after constriction? I guess what confuses me is I don't know if pressure would change. if pressure stays constant, the obviously Q goes down because Q is proportional to the fourth power of r. What equation do I need to use to see how pressure is affected?

3) The Reynolds number for the flow of a fluid of density

and viscosity eta through a pipe of inside diameter d is given by

RN =

*d *v / eta = 2*

* r* v / eta, where v is the velocity.

Q = (pi*r4*(p2-p1)) / (8*n*L) = A*v = pi*r2*v

How would RN be affected if the blood vessel is constricted? Would RN go up, down, or stay the same?

 

[BrokenGlass]

There's two things about problems like this to remember (someone may be able to give a smoother answer).

1> Think of 33 rpm as "chopping a minute into 33 parts". How long is each part? The issue so many people have is the immediate want to say, "look how easy, just 33rpm = 33 PER 60 seconds = 33/60 = .55"
Say to yourself, when measuring rotation, "each rotation is a chopped up part of the whole time unit in question".


2> The other thing is to simply check yourself when you get confused with some easy math. 60 rpm would be one revolution per second. Right? Okay, no problem there. 60 rpm is much faster than 33 rpm, so OBVIOUSLY 33rpm has to take longer to get around that record than 1 second.

So, flip your division 60/33 =1.8

If we were going 30rpm, we'd be taking twice as long as 60 rpm to get around. So it'd be 2 seconds. 33 is close to 30. So our 1.8, if we know nothing else, is pretty damn close to all the values expected that we can easily work out in our head

33 rpm means it takes 1 minute for 33 rotations. So we can think of this problem simply as a unit conversion problem. 1 min/ 33 rotations * 60 sec / 1 min = 60 sec/ 33 rotations = 1.8 sec / rotation

 

[killinsound]

Just say an element undergoes decay with a half-life of 3 hours by emitting alpha particles. Two moles of that element are isolated, and placed on a scale. 6 hours later it's mass is found to be:...

Okay, I think i am very confused about something. I thought a halflife means that after 3 hours, only half of it remains. Therefore to solve this question, after 6 hours only .25 of 2 moles remained, or 1/2 a mole remained of the original element. You just multiply this by the MW and you get the mass.

However, the study material I'm using said that the mass only decreases by the amount that are alpha particles. For instance 1.5 moles of alpha particles remain. 1.5 * 4(MW of alpha particle) = 6 grams. Did it only really reduce by 6 grams? I am having trouble seeing why they are using the alpha particles to judge and not the original element itself.

Thanks in advance.

 

[gridiron]

For a symmetrical (i.e. constant radius of coils, constant spacing between coils) solenoinds, why is the magnetic field (inside the solenoid) strongest in the middle (and weakest at the right and left ends)?

Also, for assymetrical solenoids (suppose coils get bigger from left to right, but the spacing between the coils stays the same), why is the magnetic field inside the solenoid strongest at the end where coils are smaller?

I think the relevant equation is B = mew * i * n, but I don't understand how do derive these results from this equation.

I sort of understand that the magnetic fields between adjacent coils cancel each other out, and I think that's part of the reason magnetic field outside the solenoid is very weak, but not zero because there is a right hand rule which says that if you curl your fingers in the direction the current is running through the coils of a solenoid, your thumb points in the direction of the magnetic field coming out of the solenoid, and since magnetic fields have to form loops, there must be at least one loop on the outside (from north pole to south pole). In all the pictures I have seen, the magnetic field lines are very dense on the inside of a solenoid and they are the most dense right in the middle, but I don't really understand why.

Hey! Solenoids are an application of Ampere's law. One assumption you need to make is that the length of the solenoid is much greater than its diameter. The magnetic field of the solenoid is the vector sum of the fields produced by the individual loops that make up the solenoid. At points inside the solenoid, the magnetic field is parallel to the center of the solenoid axis. To derive the equation that you have above, you need to use Ampere's law. Ampere's law is not tested on the MCAT. Ampere's law is actually an extension of the Biot-Savart Law and is given by:

The closed loop integral of the magetic field B multiplied by some differential distance ds is equal to mu multiplied by the enclosed current. Do not worry at all as to what the equation describes, but you need it to understand where equation for a solenoid comes from. You use Ampere's law to find the net magnetic field due to any distribution of currents. In most problems, B can be pulled out of the integral so ds becomes the distance around which you integrate. A classic example of the use of Ampere's law is to find the magnetic field outside a long straight wire with current. (Note, inside the closed loop integral, it is B dot ds, so you need to find the angle between the magnetic field and ds. Both vectors usually point in the same direction). If you choose an Amperian loop that is a concentric circle around the wire, ds is the summation of the line segments around the loop--thus it equals to 2piR which is the circumference of the loop. This product is equal to mu multiplied by the enclosed current, so you get the magnetic field to be:

B = mu multiplied by i/ 2piR

Now, this is where it can get much more complicated for the MCAT. If you apply Ampere's law to an solenoid, the enclosed current is equal to i multipied by n where n is the number of turns per unit length of the solenoid. If this is true, the loop encloses nh turns, where h is the length of the solenoid. But the magnetic field is encloses the same length, so h cancels and you obtain the equation you have above. This is why loops are used--they increase the magnetic field. The magnetic field at any point is tangent to the path; with multiple loops, you sum up the tangent vectors, and you net magnetic field strength increases.

In the case of an ideal solenoid, the reason the field weakens as you move away from the coils is because the fields created by the upper turns of the solenoid cancel the field created by the lower turns. In the case of an ideal solenoid, the magnetic field outside is very close to zero. To understand this, draw some loops on a piece of paper. Now, grasp the solenoid with you right hand and make your fingers follow the direction of the current in the windings. Where you thumb extends is the direction of the axial magnetic field. Do this for the top of the windings and the bottom, and what do you see? Since the coils follow circles, the fields cancel. The fields actually don't completely cancel, but the external magnetic field is relatively weak compared to the inside.

I don't know if I completely answered your question, but you will not need to know the details of the solenoid on the MCAT. Just know the basics like how to calculate the magnetic field due to current and how to use the right hand rule to find the direction of the magnetic field. Good luck .

 

[BrokenGlass]

Just say an element undergoes decay with a half-life of 3 hours by emitting alpha particles. Two moles of that element are isolated, and placed on a scale. 6 hours later it's mass is found to be:...

Okay, I think i am very confused about something. I thought a halflife means that after 3 hours, only half of it remains. Therefore to solve this question, after 6 hours only .25 of 2 moles remained, or 1/2 a mole remained of the original element. You just multiply this by the MW and you get the mass.

However, the study material I'm using said that the mass only decreases by the amount that are alpha particles. For instance 1.5 moles of alpha particles remain. 1.5 * 4(MW of alpha particle) = 6 grams. Did it only really reduce by 6 grams? I am having trouble seeing why they are using the alpha particles to judge and not the original element itself.

Thanks in advance.

There are 0.5 moles of undecayed element remaining after 2 half lives. That much is true. Notice that we assumed first order decay and hence constant half-life...as far as I know radioactive decay reactions are first order reactions. So if you multiply 0.5 moles by atomic weight
of your element, you get the mass of your UNDECAYED element remaining after 6 hours (a.k.a. 2 half lives).

This means that 2 - 0.5 = 1.5 moles of the substance have decayed away. Since one mole of element in question produces 1 mole of alpha
particles as it decays, 1.5 moles of alpha particles have decayed away. Alpha particles are helium nuclei. Atomic weight for helium is
4 grams/mole. 1.5 moles * 4 grams/ mole = 6 grams (we ignore mass of electrons since they are very light compared to protons
and neutrons, so a helium atom has the same mass as a helium nucleus for our purposes). Now subtract 6 grams from the pre-decay mass
of your element and that gives you your answer, which is TOTAL mass of your element remaning.

It appears that the solution key is giving the answer for the TOTAL remaining mass, while you are solving for the mass of UNDECAYED element. You are just answering a different question.

 

[googlinggoogler]

hello,
i have an exam this monday, and this question keeps on bugging me. please help.

if we have two floating blocks, each weighing 5 kg. one made of wood, another-->steel, and both of them float in the same fluid. Steel block is 60 % submerged, and wood is 40 percent submerged. kaplan say that both of them MUST have same Fbuyoant, b/c they have the same mass and they both FLOAT(if one would sink, it would be diff) But....,how can they have the same Fb if they have different VOLUMES submerged in the fluid? As far as i know, Fb=fluid density*Vsub*g.

Can someone help me with this. thank you very much.

I think perhaps your misunderstanding comes from the fact that they don't have to have different volumes submerged. Just because the steel block is 60% submerged, and the wood is 40% submerged, you don't know the original volumes of the steel and the wood blocks. Maybe the steel block is smaller than the wood block, which makes the 60% steel equal to the 40% wood. Therefore, Vsub would be equal in both cases. I think those values were just put in the problem as distractors. At least that's my take on it.

 

[gridiron]

My understanding is that an ideal fluid has to meet 3 or 4 conditions, one of which is that it should not be viscous. Ideal fluids obey the continuity equation, i.e. the volume flow rate is constant over the entire pipe (but not necessarily constant over time). A1*v1 = A2*v2. The continuity equation is a consequence of conservation of mass (since rate of flow into the pipe must be equal to the rate of flow out of the pipe).

If any of these 4 conditions is violated, the fluid is a non-ideal (real) fluid. In particular, if viscosity (i.e. internal friction of the fluid) is not zero, Poiseuille's Law can be used to describe volume flow rate because it takes viscosity (i.e. resistance) into account.

Q = (pi*r4*(p2-p1)) / (8*n*L), where resistance = (8*n*L) / pi*r4

Here are my questions:

1) Since continuity equation doesn't apply to real fluids, the only conclusion I can draw is that for a real fluid, the rate of inflow is NOT equal to the rate of outflow (otherwise the continuity equation would apply). But wouldn't the fluid build up in the pipe? I don't understand how that would work.

2) Lets assume blood is viscous, so Poiseuille's Law applies. If we constrict a blood vessel, would the value of Q before constriction equal the value of Q after constriction? I guess what confuses me is I don't know if pressure would change. if pressure stays constant, the obviously Q goes down because Q is proportional to the fourth power of r. What equation do I need to use to see how pressure is affected?

3) The Reynolds number for the flow of a fluid of density

and viscosity eta through a pipe of inside diameter d is given by

RN =

*d *v / eta = 2*

* r* v / eta, where v is the velocity.

Q = (pi*r4*(p2-p1)) / (8*n*L) = A*v = pi*r2*v

How would RN be affected if the blood vessel is constricted? Would RN go up, down, or stay the same?

Hey!

1.) I'm not dodging your question, but you do not need to worry about this for the MCAT. On the MCAT, you will be dealing with Newtonian or ideal fluids more than 95% of the time. If you are given a passage about blood, a non-Newtonian fluid but can be Newtonian at very high shear speeds, you will be given all the equations necessary. What you are dealing with in the question you asked is when the density varies from point to point--which is true for blood. However, you will not be asked that on the MCAT.

2.) The pressure drop down the length of a tube is given by: the change in P divided by L. You can actually rearrange the equation you have above to solve for pressure. Pressure is equal to the force over the area. If the vessel constricts, the pressure decreases because the there is more force over the same area. As an approximation, you can use Bernoulli's law--it only applies to ideal fluids. The velocity increases when the cross sectional area decreases so the pressure will decrease.

3.) By definition, Reynolds number is the ratio of inertial forces to the viscous forces of the fluid. The reynolds number will decrease with a decrease in vessel diameter. This is because the transition from laminar flow to turbulent flow will be faster in a constricted vessel--inertial forces are dominant and eddies will be present.

 

[killinsound]

There are 0.5 moles of undecayed element remaining after 2 half lives. That much is true. Notice that we assumed first order decay and hence constant half-life...as far as I know radioactive decay reactions are first order reactions. So if you multiply 0.5 moles by atomic weight
of your element, you get the mass of your UNDECAYED element remaining after 6 hours (a.k.a. 2 half lives).

This means that 2 - 0.5 = 1.5 moles of the substance have decayed away. Since one mole of element in question produces 1 mole of alpha
particles as it decays, 1.5 moles of alpha particles have decayed away. Alpha particles are helium nuclei. Atomic weight for helium is
4 grams/mole. 1.5 moles * 4 grams/ mole = 6 grams (we ignore mass of electrons since they are very light compared to protons
and neutrons, so a helium atom has the same mass as a helium nucleus for our purposes). Now subtract 6 grams from the pre-decay mass
of your element and that gives you your answer, which is TOTAL mass of your element remaning.

It appears that the solution key is giving the answer for the TOTAL remaining mass, while you are solving for the mass of UNDECAYED element. You are just answering a different question.

Thanks a bunch! that makes perfect sense.

 

[awk]

Capacitors C1 = 4.0 µF and C2 = 2.0 µF are charged as a series combination across a 95 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor.

????

 

[gridiron]

Capacitors C1 = 4.0 &#181;F and C2 = 2.0 &#181;F are charged as a series combination across a 95 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate and negative plate to negative plate. Calculate the resulting charge on each capacitor.

????

.

 

[awk]

Right...those would be the charges if the initial setup was in Parallel but it was in series. I got as far to when the disconnection occured but I don't know what happens to charges that are connected positive to positive plate. How do they disperse?

 

[BrokenGlass]

A car is driving on a level road at a contant speed 8m/s when it attempts to execute a turn about curve of effective radius 10m. For the following questions, we will assume the turn is successful, that is, the car performs the turns as the driver intends. The static coeff of friction between the tires and the road is .9, the kinetic coeff of friction is .7.

What force provides the centripetal force?
A. Gravity
B. The normal force
C. Static friction
D. Kinetic friction

Answer is C. I thought that it would be kinetic friction since the car is moving. The explanation for static friction is: Since the tires are not slipping on the road, the appropriate friction is static.

I'll add my 2 cents.

When the roadbed is flat, N*sin theta component of the normal force is not contributing to the centripetal force. The entire centripetal
force is provided by static friction between the tires and the road.

In uniform circular motion, veloctiy vector and centripetal force vector (i.e. net force vector which points towards the center of the circle)
are always perpendicular to each other. The velocity vector at any given point is subjected to an acceleration that turns it, but does
not speed it up or slow it down. The acceleration vector points toward the center of the circle, so it's radial (or centripetal). Velocity
vector is always tangential (tangent to the circle at any given point).

Static friction exists before sliding occurs. Kinetic friction exists after sliding occurs. Since the car is not sliding in the radial direction
(there is no motion along the radial line), the frictional force is static.

Kinetic friction can never change DIRECTION of velocity because it's always colinear (acts along the same line) with velocity. Kinetic
friction can only change MAGNITUDE of velocity.

 

[killinsound]

Here's another problem I ran into.

A wire that delivers a current has a non negligible resistance. If a wire with the same length but double the diameter, what is the ratio of power output between the old wire and the new wire.

I got 1:4, but that is the wrong answer.

What I did is P = IV and I = V/r

so P = V^2/R

1/4 r1 = r2 (1 = old, 2 = new)

And P old = V^2/r1 and P new = V^2/r2 = 4V^2/r1

Pold / Pnew = 1:4

However, the answer is 4:1... why!?!?!?

 

[amestramgram]

Here's another problem I ran into.

A wire that delivers a current has a non negligible resistance. If a wire with the same length but double the diameter, what is the ratio of power output between the old wire and the new wire.

I got 1:4, but that is the wrong answer.

What I did is P = IV and I = V/r

so P = V^2/R

1/4 r1 = r2 (1 = old, 2 = new)

And P old = V^2/r1 and P new = V^2/r2 = 4V^2/r1

Pold / Pnew = 1:4

However, the answer is 4:1... why!?!?!?

R= rho * L/A

P1 = V^2/R1 (old)
P2 = V^2/R2 (new)

P1/P2 = R2/R1

rho the same

L the same

A1 = a
A2 = 4a^2

substitute

P1/P2 = R2/R1 = A1/A2 =1/4

I think your book may have a typing error I vote with you!

 

[BrokenGlass]

R= rho * L/A

P1 = V^2/R1 (old)
P2 = V^2/R2 (new)

P1/P2 = R2/R1

rho the same

L the same

A1 = a
A2 = 4a^2

substitute

P1/P2 = R2/R1 = A1/A2 =1/4

I think your book may have a typing error I vote with you!

I = V/R is Ohm's law. In Ohm's law, R is resistance, not radius of the wire.

Ohm's law can also be written as V = IR (voltage = current * resistance)

Resistance is given by R = p * (L/A), where p (rho) is resistivity, L is lenght of the wire and A is the cross sectional area of the wire.
When you increase the diameter of the wire, you are increasing the cross sectional area of the wire, so you decrease resistance
to current in the wire

Area = pi * radius ^ 2 = pi * (diameter/2)^2.

So if you double the diameter, you increase the cross-sectional area by a factor of 4, which means you decrease
resistance by a factor of 4.

Since P = V^2/R (P = power, V is voltage, R is resitance), you increase power by a factor of 4.

Intutively, if you have a fatter wire, current flows more easily through it because there is less resistance, so it's
easier to transfer energy. Since power is rate of energy transfer, power should go UP.