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
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• what you need to know about physics for the MCAT
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• how best to tackle the MCAT physical sciences section

Unacceptable topics:

• actual MCAT questions or passages, or close paraphrasings thereof
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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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[amestramgram]

but old wire/ new wire though?

 

[awk]

More Diameter = More Current = More Power!

 

[BrokenGlass]

but old wire/ new wire though?

R 1= rho * (L/A1) = rho * (L/(pi * (d1/2)^2))
R 2= rho * (L/A2) = rho * (L/(pi * ((2*d1)/2)^2))

R1/R2 = 1/(1/4) = 4 (R here is resistance, not radius)

P1 = V^2/R1
P2 = V^2/R2

P2/P1 = R1/R2 = 1/(1/4)) = 4

Hopefully no algebra mistakes here...

 

[killinsound]

R 1= rho * (L/A1) = rho * (L/(pi * (d1/2)^2))
R 2= rho * (L/A2) = rho * (L/(pi * ((2*d1)/2)^2))

R1/R2 = 1/(1/4) = 4 (R here is resistance, not radius)

P1 = V^2/R1
P2 = V^2/R2

P2/P1 = R1/R2 = 1/(1/4)) = 4

Hopefully no algebra mistakes here...

if you look at my explanation, it matches yours if P2 = new wire

Old / New = 1/4

Or like yours, New / Old = 4/1

However, in the book they have it as Old / new = 4 / 1

They say something along the lines of

P = IV and V = IR

so P = I^2 R

If r goes down, in order to maintain the same current, the power must decrease accordingly.

Is this the correct logic to have?

I'm really confused.

 

[BrokenGlass]

if you look at my explanation, it matches yours if P2 = new wire

Old / New = 1/4

Or like yours, New / Old = 4/1

However, in the book they have it as Old / new = 4 / 1

They say something along the lines of

P = IV and V = IR

so P = I^2 R

If r goes down, in order to maintain the same current, the power must decrease accordingly.

Is this the correct logic to have?

I'm really confused.

OK, we all agree that resistance goes down by a factor of 4 when we double the diameter of the wire.

V = I * R => I = V/R. If we decrease resistance by a factor of 4, we increase current by a factor of 4,
since V (potential difference) depends only on the battery and therefore stays the same.

P = Energy/time = (q*V)/t = (q/t)*V = I * V. So P = I*V (this equation applies to power dissipated by the
wire). I believe P = V^2/R and P = I^2*R apply only when we are dealing with resistors, so lets use
P = I * V.

From P = I * V, if current goes up by a factor of 4 (as shown above) and voltage tays the same, power must
go up by a factor of 4. So I am still getting the same answer.

Please post the exact problem statement and the solution provided by the answer key.

 

[BrokenGlass]

test

 

[gridiron]

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!?!?!?

Hey! There are multiple equations for power. First is the equation for the rate of electric energy transfer: P=iV. The other equations give the rate of energy dissipation: P=i^2R and P=V^2/R.

You need to calculate the resistance for the two sets of wires. The equation for this is:

R = rho * (L/A)
where A is the cross sectional area of the wire.

The problem states that the diameter of the new wire is twice that of the old wire. I will denote the old wire as 1 and the new wire as 1. This means:

r1 = d1/2
r2 = d2
because d2 = 2d1.

This means the resistance for wire 1 is:

rho * (L/(pi*d^2)/4)
which equals:

4*rho*L/pi*d^2

For wire 2:

rho*L/pi*d^2

Now, you need to solve for the power. The problems does not say what electric potential this wire is connected to so you can use the equation P=i^2R. If you plug in the values obtained from above, you will get a ratio of 4 to 1for the old wire to the new wire. You may ask yourself whether this makes sense. You might say that since the resistance has decreased for the new wire, more current should be able to flow through. However, you have to understand how energy is dissipated in a wire with resistance. The energy transfer is due to the collisions between the electrons in the wire and the molecules of the resistor.

You can solve the problem another way using the idea of current density--the current per unit area through the element. However, since the problem stated that the wires have the same length but different cross sectional area, I believe it wants you to use the principles of resistance and power.

Hope this helps and good .

 

[killinsound]

Hey! There are multiple equations for power. First is the equation for the rate of electric energy transfer: P=iV. The other equations give the rate of energy dissipation: P=i^2R and P=V^2/R.

You need to calculate the resistance for the two sets of wires. The equation for this is:

R = rho * (L/A)
where A is the cross sectional area of the wire.

The problem states that the diameter of the new wire is twice that of the old wire. I will denote the old wire as 1 and the new wire as 1. This means:

r1 = d1/2
r2 = d2
because d2 = 2d1.

This means the resistance for wire 1 is:

rho * (L/(pi*d^2)/4)
which equals:

4*rho*L/pi*d^2

For wire 2:

rho*L/pi*d^2

Now, you need to solve for the power. The problems does not say what electric potential this wire is connected to so you can use the equation P=i^2R. If you plug in the values obtained from above, you will get a ratio of 4 to 1for the old wire to the new wire. You may ask yourself whether this makes sense. You might say that since the resistance has decreased for the new wire, more current should be able to flow through. However, you have to understand how energy is dissipated in a wire with resistance. The energy transfer is due to the collisions between the electrons in the wire and the molecules of the resistor.

You can solve the problem another way using the idea of current density--the current per unit area through the element. However, since the problem stated that the wires have the same length but different cross sectional area, I believe it wants you to use the principles of resistance and power.

Hope this helps and good .

Thanks! that makes a lot more sense.

I think what I was struggling with was whether or not electric potential stayed constant.

 

[BrokenGlass]

Hey! There are multiple equations for power. First is the equation for the rate of electric energy transfer: P=iV. The other equations give the rate of energy dissipation: P=i^2R and P=V^2/R.

You need to calculate the resistance for the two sets of wires. The equation for this is:

R = rho * (L/A)
where A is the cross sectional area of the wire.

The problem states that the diameter of the new wire is twice that of the old wire. I will denote the old wire as 1 and the new wire as 1. This means:

r1 = d1/2
r2 = d2
because d2 = 2d1.

This means the resistance for wire 1 is:

rho * (L/(pi*d^2)/4)
which equals:

4*rho*L/pi*d^2

For wire 2:

rho*L/pi*d^2

Now, you need to solve for the power. The problems does not say what electric potential this wire is connected to so you can use the equation P=i^2R. If you plug in the values obtained from above, you will get a ratio of 4 to 1for the old wire to the new wire. You may ask yourself whether this makes sense. You might say that since the resistance has decreased for the new wire, more current should be able to flow through. However, you have to understand how energy is dissipated in a wire with resistance. The energy transfer is due to the collisions between the electrons in the wire and the molecules of the resistor.

You can solve the problem another way using the idea of current density--the current per unit area through the element. However, since the problem stated that the wires have the same length but different cross sectional area, I believe it wants you to use the principles of resistance and power.

Hope this helps and good .

Why would current stay the same when a thin wire is replaced with a thick wire? If the rest of the circuit is unchanged, current
would not stay the same, right? Without knowing what the credited answer is, why would we assume that current stays the same?

 

[killinsound]

Why would current stay the same when a thin wire is replaced with a thick wire? If the rest of the circuit is unchanged, current
would not stay the same, right? Without knowing what the credited answer is, why would we assume that current stays the same?

I think gridiron was just trying to show another way of doing it given that current is constant.


I feel really really dumb for asking this, but the EK question of the day today has really stumped me...

Bob and Joe pull in opposite directions on the same rope. Bob has a mass of 100 kg. Joe has a mass of 50 kg. Bob pulls with a force of 200 N accelerating Joe at 4 m/s/s. If the rope is massless, with what force does Joe pull on the rope?

 

[BrokenGlass]

I think gridiron was just trying to show another way of doing it given that current is constant.


I feel really really dumb for asking this, but the EK question of the day today has really stumped me...

Bob and Joe pull in opposite directions on the same rope. Bob has a mass of 100 kg. Joe has a mass of 50 kg. Bob pulls with a force of 200 N accelerating Joe at 4 m/s/s. If the rope is massless, with what force does Joe pull on the rope?

A massless rope simply transfers force between Bob and Joe. From Newton's 3rd law, if Bob pulls with a force of 200N on Joe (through the rope), Joe pulls with an equal and opposite force on Bob. So Joe pulls on Bob (through the rope) with 200N of force.

 

[dochoov]

A massless rope simply transfers force between Bob and Joe. From Newton's 3rd law, if Bob pulls with a force of 200N on Joe (through the rope), Joe pulls with an equal and opposite force on Bob. So Joe pulls on Bob (through the rope) with 200N of force.

are you sure? I'm thinking Newton's 2nd law: sum of forces=ma

200N (bob force)- X (joe force)=m(joe)a(joe)

solve for X ---> F(joe)=110N

Am I confused here too?

 

[dochoov]

I have a physics homework problem with a circuit that has three resistors. One is in series, another is in series and parallel, and the other is in parallel. (I can't think of how to explain this better). How can I begin to solve this problem. This question comes from my physics homework. Is it out of the scope of the MCAT?

 

[gridiron]

I have a physics homework problem with a circuit that has three resistors. One is in series, another is in series and parallel, and the other is in parallel. (I can't think of how to explain this better). How can I begin to solve this problem. This question comes from my physics homework. Is it out of the scope of the MCAT?

No, this is not out of the scope of the MCAT. Could you perhaps provide a picture or drawing? I can help you better if I see the actual question. The MCAT loves circuit problems!

 

[BrokenGlass]

are you sure? I'm thinking Newton's 2nd law: sum of forces=ma

200N (bob force)- X (joe force)=m(joe)a(joe)

solve for X ---> F(joe)=110N

Am I confused here too?

Joe ---200N------>   <---?N---- Bob                      
50kg                            100kg

F = ma, a = F/m = 200/50 = 4 m/s/s (so as you can see, it is redundant for the problem statement to mention Joe's acceleration
because we can figure it out from other given information).

You cannot use F = ma to figure out force on Bob (applied by Joe via the rope) because you don't know Bob's acceleration.

Notice that the 200N force and the force we are asked to find are NOT acting on the same object, which is why it is incorrect to
solve this problem the way you did. In many physics problems you need to choose a system and consider all the forces acting on that system.

If we choose Joe as our system, we can solve for Joe's acceleration (which is already given by the problem statement).

If we choose Bob as our system, we can solve for Bob's acceleration (but only after applying Newton's 3rd law and concluding that Joe pulls on Bob with the force of 200N).

The problem statement is not even asking for Bob's acceleration, so all we have to do is apply Newton's 3rd law here.

Finally, notice that forces in the vertical direction on Joe cancel out; forces in the vertical direction on Bob cancel out. So the net
force on Joe is the force with which Bob is pulling and the net force on Bob is the force with which Joe is pulling. The net force on Joe
and the net force on Bob are an action-reaction pair of forces (from Newton's 3rd law). Therefore these forces are equal in magnitude
and opposite in direction.

 

[y0ter]

Pressure Volume question. I came across the following question on a Kaplan practice exam and am having a hard time understanding how it relates to U=P(Vf-Vi).

Power = blood pressure * volume of blood pumped/time over which blood is pumped

I don't understand how volume has chaged while pumped. I'm trying to envision a piston system but i can't relate it to the circulatory system.

 

[BrokenGlass]

Pressure Volume question. I came across the following question on a Kaplan practice exam and am having a hard time understanding how it relates to U=P(Vf-Vi).

Power = blood pressure * volume of blood pumped/time over which blood is pumped

I don't understand how volume has chaged while pumped. I'm trying to envision a piston system but i can't relate it to the circulatory system.

Please post the problem statement and the answer key in their entirety

 

[BrokenGlass]

Pressure Volume question. I came across the following question on a Kaplan practice exam and am having a hard time understanding how it relates to U=P(Vf-Vi).

Power = blood pressure * volume of blood pumped/time over which blood is pumped

I don't understand how volume has chaged while pumped. I'm trying to envision a piston system but i can't relate it to the circulatory system.

Please post the problem statement and the answer key in their entirety

 

[minsoox1983]

This question is from the physical sciences of Kaplan Full length test#10.

I did exactly same as the solution til where I highlighted with red.

From what I understand, Current (I) splits in parallel circuit inversely proportional to amount of R. I=V/R

Voltage (V) splits in series, direct proportional to R. V=IR

But, I just don't understand their reasoning of I(total) = 3I(of 5m ohm) * I(of 5m ohm)

Shouldn't be I(total) = I(5m ohm) + I(2.5m ohm)
Since 5m R is twice greater than the 2.5m R,

I(x) will be I(total)*(2/3) which is about 1.6A . Isn't it ?

Thanks for your help

 

[BrokenGlass]

This question is from the physical sciences of Kaplan Full length test#10.

I did exactly same as the solution til where I highlighted with red.

From what I understand, Current (I) splits in parallel circuit inversely proportional to amount of R. I=V/R

Voltage (V) splits in series, direct proportional to R. V=IR

But, I just don't understand their reasoning of I(total) = 3I(of 5m ohm) * I(of 5m ohm)

Shouldn't be I(total) = I(5m ohm) + I(2.5m ohm)
Since 5m R is twice greater than the 2.5m R,

I(x) will be I(total)*(2/3) which is about 1.6A . Isn't it ?

Thanks for your help

test

 

[BrokenGlass]

This question is from the physical sciences of Kaplan Full length test#10.

I did exactly same as the solution til where I highlighted with red.

From what I understand, Current (I) splits in parallel circuit inversely proportional to amount of R. I=V/R

Voltage (V) splits in series, direct proportional to R. V=IR

But, I just don't understand their reasoning of I(total) = 3I(of 5m ohm) * I(of 5m ohm)

Shouldn't be I(total) = I(5m ohm) + I(2.5m ohm)
Since 5m R is twice greater than the 2.5m R,

I(x) will be I(total)*(2/3) which is about 1.6A . Isn't it ?

Thanks for your help

I = 2.57 Amperes. This is the (total) current at point d (which is also the current coming out of the battery).

Since the resistance of R2 is half the resistance of the 5 mili-Ohm resistor, twice as much current will flow through R2 than
through the 5 mili-Ohm resistor. Let the amount of current at point X be x. Then the currect that flows through R2 is 2x.

x + 2x = 2.57
3x = 2.57
x = 0.857 A

In the sample solution what you probably think is a multiplication sign is actually a period (the part you underlined in red).

In your solution, I think it should be "I(x) will be I(total)*(1/3) which is about 0.57A." I think that's why you weren't getting
the right answer.

But I don't understand where 10^-5 and 10^-6 come from in the answer choices.

 

[y0ter]

Please post the problem statement and the answer key in their entirety

BrokenGlass, the entire problem is as follows:

During intense exercise, the volume of blood pumped per second by an athlete's heart increases by a factor of 7, and his blood pressure increases by 80%. By what factor does the power output of the heart increase during exercise?
A. 1.2
B. 3.5
C. 7
D. 8.4

Thanks.

 

[minsoox1983]

My question is,, how do I tell which one is first/second/third harmonic wave from the figure 1a ?


Passage IV

The timbre, or quality, of a musical tone depends on the number and relative strengths of the harmonics including the fundamental frequency of the note. Figure 1a illustrates the first three harmonics of a tone. The addition of the first two harmonics is pictured in Figure 1b, and the addition of the first 3 harmonics is shown in Figure 1c.

Figure 1 Elements of a complex tone





22. Which of the waveforms shown in Figure 1 has the shortest period?

A) First harmonic

B) Second harmonic

C) Third harmonic **
The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer choice C is the best answer.


D) The waveform in Figure 1c


Thank you!

-Minsoo

 

[BrokenGlass]

BrokenGlass, the entire problem is as follows:

During intense exercise, the volume of blood pumped per second by an athlete's heart increases by a factor of 7, and his blood pressure increases by 80%. By what factor does the power output of the heart increase during exercise?
A. 1.2
B. 3.5
C. 7
D. 8.4

Thanks.

Work = P * (Vf-Vi) describes pressure-volume work done at constant pressure.

Power = Energy/Time (i.e. power is rate of energy transfer).

Fluid pressure can be thought of random translational kinetic energy of the fluid particles per unit volume.

Pressure = Energy/Volume => Energy = Pressure * Volume.

Power = Energy/Time = (Pressure * Volume) / Time = (blood pressure * volume of blood pumped ) / time over which blood is pumped

When you exercise, your heart rate increases and consequently your heart pumps more blood and therefore delivers more oxygen to
your tissues per unit time. Your breathing rate also goes up because your cells need more oxygen to make ATP at a faster rate.

In this problem we need to focus on the normal heart rate and heart volume (i.e. for an individual at rest) vs. increased but still
constant heart rate and heart volume during exercise. So we need to take the ratio of "new power" to "old power".

New Power / Old Power = ((New Pressure * New Volume)/Time ) / ((Old Pressure * Old Volume)/Time ) =

= (New Pressure * New Volume) / (Old Pressure * Old Volume) =

= ((1.8 * Old Pressure) * (7 * Old Volume)) / (Old Pressure * Old Volume) = 12.6


Of course, 12.6 is not among the answer choices. What is the credited answer choice? Maybe it all comes down to correctly
interpreting what "heart increases by a factor of 7" and "blood pressure increases by 80%" mean. What is Kaplan's sample solution?

 

[BrokenGlass]

My question is,, how do I tell which one is first/second/third harmonic wave from the figure 1a ?


Passage IV

The timbre, or quality, of a musical tone depends on the number and relative strengths of the harmonics including the fundamental frequency of the note. Figure 1a illustrates the first three harmonics of a tone. The addition of the first two harmonics is pictured in Figure 1b, and the addition of the first 3 harmonics is shown in Figure 1c.

Figure 1 Elements of a complex tone





22. Which of the waveforms shown in Figure 1 has the shortest period?

A) First harmonic

B) Second harmonic

C) Third harmonic **
The tone with the shortest period has the shortest wavelength. In Figure 1a, the period of the third harmonic (the curve with the smaller dashes) is seen to be shorter than the other two harmonics. Thus, answer choice C is the best answer.


D) The waveform in Figure 1c


Thank you!

-Minsoo

speed = frequency * wavelength

For standing waves, when both ends define a node or both ends define an anti-node (this is known as an "open pipe" for sound waves),
you need to use the following formulas:

L = (n * wavelength )/2, where n = 1,2,3,....

wavelength = 2L/n

frequency = speed/wavelength = nc/2L, where c is speed


when one end defines a node and the other end defines an anti-node (this is known as a "closed pipe" for sound waves), you need to use
the following formula:

L = (n * wavelength )/4, where n = 1,3,5,....

wavelength = 4L/n

frequency = speed/wavelength = nc/4L, where c is speed


Since in the problem you posted both ends are nodes, the equation that applies here is

wavelength = 2L/n.

When n=1, wavelength = 2L and we have the first harmonic.
When n=2, wavelength = L and we have the second harmonic.
When n=3, wavelength = 2L/3 and we have the third harmonic.

2L > L > 2L/3

As you can see, the first harmonic (whose corresponding frequency is also called the fundamental frequency) has the longest wavelength, and so on. Also notice that stringlength L is half the wavelength of the first harmonic.

 

[BrokenGlass]

speed = frequency * wavelength

For standing waves, when both ends define a node or both ends define an anti-node (this is known as an "open pipe" for sound waves),
you need to use the following formulas:

L = (n * wavelength )/2, where n = 1,2,3,....

wavelength = 2L/n

frequency = speed/wavelength = nc/2L, where c is speed


when one end defines a node and the other end defines an anti-node (this is known as a "closed pipe" for sound waves), you need to use
the following formula:

L = (n * wavelength )/4, where n = 1,3,5,....

wavelength = 4L/n

frequency = speed/wavelength = nc/4L, where c is speed


Since in the problem you posted both ends are nodes, the equation that applies here is

wavelength = 2L/n.

When n=1, wavelength = 2L and we have the first harmonic.
When n=2, wavelength = L and we have the second harmonic.
When n=3, wavelength = 2L/3 and we have the third harmonic.

2L > L > 2L/3

As you can see, the first harmonic (whose corresponding frequency is also called the fundamental frequency) has the longest wavelength, and so on. Also notice that stringlength L is half the wavelength of the first harmonic.

Period = 1/frequency
Wavelength = c/frequency

So longest period corresponds to longest wavelength and shortest period corresponds to the shortest wavelength.

 

[minsoox1983]

wow! Even though I knew all the formulas of ㅅ=2L/n and 4L/n and c=fㅅ, I didn't know how to use them.

Thanks for your explanation and they really make sense!

 

[ThreeJ]

Hello I just have a few questions for projectiles. I was wondering when asked for the maximum height a projectile attains after being struck at an angle on 65 to the horzontial with an initial velocity of 85m/s. It takes 7.85sec to reach this max height. The answer is 302m.
So it is basically s=Vit+1/2at2 (wehere the last 2 is for squared). basicallt just then use 1/2 at2 and omit the Vi treating it as 0 even though they give you a value in the question for Vi?

Does this seem right or does it look like a dodgy answer.

 

[castlmere]

That answer is totally dodgy.

You have the right kinematic equation. What you need to make sure to do is use the proper initial velocity by making a proper vector triangle and then noting that the initial vertical velocity is the total initial velocity times the sine of the angle. This, coupled with the acceleration should result in the proper answer.

Alternatively you could note that at maximum height all the vertical kinetic energy is transformed into graviational potential energy. Either method works.

 

[gridiron]

Hello I just have a few questions for projectiles. I was wondering when asked for the maximum height a projectile attains after being struck at an angle on 65 to the horzontial with an initial velocity of 85m/s. It takes 7.85sec to reach this max height. The answer is 302m.
So it is basically s=Vit+1/2at2 (wehere the last 2 is for squared). basicallt just then use 1/2 at2 and omit the Vi treating it as 0 even though they give you a value in the question for Vi?

Does this seem right or does it look like a dodgy answer.

The answer seems right. Because they gave you an angle that you do not know the sine or cosine of, you need to do some manipulation. To find the time it takes the projectile to reach the top, you need to do the following:

V final = V initial - gt
V final is equal to zero at the very top.

So, v initial is equal to gt. However, v initial is equal to (Initial Velocity multiplied by the sine of 65). You don't know the sine of 65 but you don't need to worry about it. V initial is simply the time multiplied by approximately 10. So the v initial quantity is equal to 78.5--according to the 7.85 seconds provided in the question. So, to find the max height, you use:

y = (v initial)t - 0.5gt^2.
I will round 7.85 to 8 and 78.5 to 80 to simply things, and you get approximately:

640 - 320 which is approximately 320.

Since I rounded up, the actual answer is a little lower.

However, if you use a calculator, you will find the time it takes to reach the very top to be 7.7 seconds. I wonder why the question would say 7.85 seconds?

 

[BrokenGlass]

Hello I just have a few questions for projectiles. I was wondering when asked for the maximum height a projectile attains after being struck at an angle on 65 to the horzontial with an initial velocity of 85m/s. It takes 7.85sec to reach this max height. The answer is 302m.
So it is basically s=Vit+1/2at2 (wehere the last 2 is for squared). basicallt just then use 1/2 at2 and omit the Vi treating it as 0 even though they give you a value in the question for Vi?

Does this seem right or does it look like a dodgy answer.

Here is an alternative solution:

Vavg = (Vi + Vf)/2 (when acceleration is constant)

distance = Vavg * time

distance = (0 + (85*Sin65))/2 * 7.85

Sin 60 = 0.87. Sin 65 is a little bigger than sin 60 (think of a unit circle).
So lets say sin 65 = 0.9

distance = (0 + (85*0.9))/2 * 7.85 = (85*0.9)/2 * 8 = 306 m, which
is probably close enough

 

[gridiron]

Here is an alternative solution:

Vavg = (Vi + Vf)/2 (when acceleration is constant)

distance = Vavg * time

distance = (0 + (85*Sin65))/2 * 7.85

Sin 60 = 0.87. Sin 65 is a little bigger than sin 60 (think of a unit circle).
So lets say sin 65 = 0.9

distance = (0 + (85*0.9))/2 * 7.85 = (85*0.9)/2 * 8 = 306 m, which
is probably close enough

You actually raise a very important point--to be aware of how the sine of an angle changes from 0 to 90 degrees. I have seen questions where you are not provided, for example, the sine of 63 degrees--you have to estimate based on the sine of 60 degrees. You should know the sine and cosine of 30, 60 and 45 degrees but also how to estimate the sine and cosine of other angles as well.

 

[Shrike]

Another point arising from BG's post: you do not need to be careful with your math. It may be otherwise in your practice material, but on the MCAT being within ten percent is plenty good enough.

So: If we know it takes 7.85 seconds to reach the top, that's about 8. Using the equation that v = v(0) +at, we get v(0) (in the y direction) is about 80.

Using d = (average of v(0) and v)x(time), we get d = 40 x 8 = 320.

Close enough.

 

[HristosKaran]

A ball of mass 2kg is thrown horizontally at a speed of 5.4 m/s off a 10 m high roof. A second ball of mass 4kg is thrown downward at an angle of 45 degrees below the horizontal at the same speed. The speeds of the 2 and 4 kg balls when they hit at the bottom are, respectively:

-Answer is 15 m/s, 15 m/s

Can someone explain to me why this is and why they are hitting the ground at the same speed? Thanks

 

[BrokenGlass]

A ball of mass 2kg is thrown horizontally at a speed of 5.4 m/s off a 10 m high roof. A second ball of mass 4kg is thrown downward at an angle of 45 degrees below the horizontal at the same speed. The speeds of the 2 and 4 kg balls when they hit at the bottom are, respectively:

-Answer is 15 m/s, 15 m/s

Can someone explain to me why this is and why they are hitting the ground at the same speed? Thanks

Lets define the rooftop as our reference point and the downward direction as positive direction. Also, remember that
projectile motions in horizontal and vertical directions are independent of one another.

v^2 = Vo^2 + 2*a*delta d, where d is displacement.

v = sqrt(Vo^2 + 2*a*delta d), where sqrt means square root

For the 2kg ball we have: v = sqrt(0^2 + 2*10*10) = sqrt (200), which is a little less than 15 m/s (since 15^2 = 225).

Alternatively, since the initial vertial velocity is zero, we can set its initial mechanical energy (all potential) equal to final mechanical energy (all kinetic) and solve for v.

m*g*h = (1/2) m*v^2. We would get v = sqrt(2*g*h) = sqrt(200).

As you can see from v = sqrt(2*g*h), mass doesn't affect the final velocity (assuming no air resistance). This is so
because F = ma, a = F/m = (mg)/m = g for all projectiles (assuming force due to gravity is the only force acting).

For the 4kg ball we have: v = sqrt((5.4*sin 45)^2 + 2*10*10) = = sqrt ((5.4 * 0.7)^2 + 200), which is about 15 m/s.

Note that 5.4*sin 45 is the initial velocity in the y direction.

 

[BrokenGlass]

Lets define the rooftop as our reference point and the downward direction as positive direction. Also, remember that
projectile motions in horizontal and vertical directions are independent of one another.

v^2 = Vo^2 + 2*a*delta d, where d is displacement.

v = sqrt(Vo^2 + 2*a*delta d), where sqrt means square root

For the 2kg ball we have: v = sqrt(0^2 + 2*10*10) = sqrt (200), which is a little less than 15 m/s (since 15^2 = 225).

Alternatively, since the initial vertial velocity is zero, we can set its initial mechanical energy (all potential) equal to final mechanical energy (all kinetic) and solve for v.

m*g*h = (1/2) m*v^2. We would get v = sqrt(2*g*h) = sqrt(200).

As you can see from v = sqrt(2*g*h), mass doesn't affect the final velocity (assuming no air resistance). This is so
because F = ma, a = F/m = (mg)/m = g for all projectiles (assuming force due to gravity is the only force acting).

For the 4kg ball we have: v = sqrt((5.4*sin 45)^2 + 2*10*10) = = sqrt ((5.4 * 0.7)^2 + 200), which is about 15 m/s.

Note that 5.4*sin 45 is the initial velocity in the y direction.

We can use V = Vo + a*t to see that time to reach the ground would be different for the 2kg and 4 kg balls.

 

[soonereng]

Hello I just have a few questions for projectiles. I was wondering when asked for the maximum height a projectile attains after being struck at an angle on 65 to the horzontial with an initial velocity of 85m/s. It takes 7.85sec to reach this max height. The answer is 302m.
So it is basically s=Vit+1/2at2 (wehere the last 2 is for squared). basicallt just then use 1/2 at2 and omit the Vi treating it as 0 even though they give you a value in the question for Vi?

Does this seem right or does it look like a dodgy answer.

Another way of doing it, energy balance between starting energy and top height energy:

Vertical energy => 1/2*m*(85*sin 65)^2 = m*9.81*h

The m cancels out => h = (85*sin 65)^2/(2*9.81) ~ 302 m

 

[Shrike]

A ball of mass 2kg is thrown horizontally at a speed of 5.4 m/s off a 10 m high roof. A second ball of mass 4kg is thrown downward at an angle of 45 degrees below the horizontal at the same speed. The speeds of the 2 and 4 kg balls when they hit at the bottom are, respectively:

-Answer is 15 m/s, 15 m/s

Can someone explain to me why this is and why they are hitting the ground at the same speed? Thanks

Much quicker than kinematics for figuring out that the speeds are the same:

At the top, each has kinetic energy and potential energy. Speeds are the same, so KE's are the same except for a factor of 2, because of the different masses. Heights are the same, so PE's are the same except for that same factor of two. Each then converts its KE to PE, by falling. Same addition to KE (per kilogram), so same final speed. In other words, each gains the same amount of KE per kilogram, and they start at the same speed, so they end at the same speed.

With numbers: each ball gains 100J/kg of KE. KE(0)/m = 0.5(5.4)^2 = approximately 15J/kg. KE(f)/m = 0.5v^2 = 115J/kg, so v = sqrt(230) = approximately 15 m/s.

 

[colorado_doc_2B]

Never mind

 

[wishuponastar]

So you have Power=I^2 (R) and Power= V^2 / R...I am not understanding how you can have a particular resistance that is large that will increase the power in the first equation and decrease the power in the second equation. I.e. why is it that R is in the numerator of Equation 1 and in the denominator of Equation 2? Thanks.

 

[gridiron]

So you have Power=I^2 (R) and Power= V^2 / R...I am not understanding how you can have a particular resistance that is large that will increase the power in the first equation and decrease the power in the second equation. I.e. why is it that R is in the numerator of Equation 1 and in the denominator of Equation 2? Thanks.

When you have a battery connected to some conducting device like a resistor, energy is transferred to internal thermal energy as current passes through the terminals of the resistor--this actually increases the temperature of the resistor.

Current is the flow of electrons. As an electron moves through a resistor with some drift speed, it loses electric potential energy due to the collisions between the electron and the molecules of the resistor. This loss of electric potential energy is dissipated as heat in the resistor and the temperature increases. More current going through a resistor, provided the same resistance, will increase power because more electrons will collide with molecules of the resistor.

Both the above equations are derived from the power loss due to electric energy transfer. As current passes through some load such as a motor or electrical device, charge moves from a decrease in potential through the load. According to the principle of conservation of energy, this electric energy loss is accompanied by a transfer of energy is another form--power (P =iV). The reason R appears in the denominator is because if you substitute i=V/R into the P=iV equation, you get R in the denominator.

 

[ladida]

hi, i don't really get torques here are some questions

1) a person exerts a force of 45N on the end of a door 84cm wide. What is the magnitude of the torque if the force is exerted a) perpendicular to the door, and b) at a 60degree angle of the face of the door?

2) If the coefficient of static friction between tires and pavement is 0.75, caluclate the minimum torque that must be applied to the 66-cm-diametre tire of a 1080kg automobile in ordder to "lay rubber" (make the wheels spin, slipping as the car accelerates) Assume each wheel supports an equal share of the weight.

 

[gridiron]

Hey! To understand torque, consider the following example. If you were to tie a rope to a bucket and twirl it over your head, you can make it move in a circular path. However, what if you wanted to make the bucket spin. As you twirl the bucket over your head, the centripetal force keeps the bucket moving in a circular path. As the bucket moves in a circular path, its center of mass experiences an acceleration. Thus, in order to make the center of mass of an object accelerate, we need to exert a force--where the force in this example is the centripetal force. However, if we wanted to make the bucket spin, we need to exert a torque. Torque isn't a force rather its magnitude is a measure of the effectiveness of a force making an object spin. The equation for torque is usually derived using the classic example of using a wrench to tighten a bolt. This equation is as follows:

tau (t) = rFsin(theta)

Where r is the radius vector (the vector from the center of rotation, such as the center of mass of the object, to the point of application of the force), F is the force applied at some point on the object, and theta is the angle between the radius vector and force. If you were to plug in 90 degrees for theta, you would get the maximum torque (because the sine of 90 is 1) and if you plug in 0 degrees you get zero torque. This is important because it gives you two important concepts:

1.) The maximum torque will result when the force is applied in a perpendicular fashion from the center of rotation.
2.) You will get zero torque when the force is applied through the center of mass or through the pivot point.

This makes sense when using the wrench example. If you want to loosen a bolt, you place the wrench around the bolt and then push down with a force in order to displace the bolt. You won't be able to spin or rotate the bolt you don't push down--your force is applied through the pivot point.

You can alternatively use the lever arm method. This doesn't use an angle but rather you find the distance from the pivot point to the line along which the force is applied. The equation:

tau (t) = LF

Where L is the lever arm, the shortest distance between the pivot to the line along which the force is applied.

Also remember that when a force is applied at an angle to the pivot, only part of the force will cause torque. To answer your questions:

1.) If you want to rotate the door around its hinges, and you apply a perpendicular force to the end of the door, then you calculate the torque as follows:

t = (0.84)*(45N) ---> remember to convert 84 cm to m because of SI units

If you were to apply the force at 60 degrees then you have the theta =60 degrees.

2.) You want to find the torque necessary to rotate the wheels. The weight of the wheel is usually drawn through the center of mass. In order to rotate this wheel, the force of static friction needs to be overcome. The maximum force of static friction is equal to:

F stat friction = (0.75)(1080)(10)---> I rounded 9.8 to 10 for gravity.

This means, force > than the F stat. friction needs to be applied opposite the force of static friction in order to rotate the wheel around its center of mass. This force would be equal to the F stat force multiplied by radius of the wheel--assuming you want to rotate the wheel around its center of mass.

I hope this helps and good .

 

[ladida]

hi i understand that part now thank you
but.... i got lots more to ask xD

1) a 70kg adult sits at one end of a 10-m board , on the other end hwich sits his 30kg child. wehre shoudl teh pivot be palces so the board(ignore its mass) is balanced?
this one i get is 3m
but the continue part is
repeat problem taking into account the board's 15kg mass. the asnwer is 3.3m from the adult but i don get it.

hm..... i haf more but they have picutres....

hahah ^^ ill ask this for now
thank you

 

[BrokenGlass]

hi i understand that part now thank you
but.... i got lots more to ask xD

1) a 70kg adult sits at one end of a 10-m board , on the other end hwich sits his 30kg child. wehre shoudl teh pivot be palces so the board(ignore its mass) is balanced?
this one i get is 3m
but the continue part is
repeat problem taking into account the board's 15kg mass. the asnwer is 3.3m from the adult but i don get it.

hm..... i haf more but they have picutres....

hahah ^^ ill ask this for now
thank you

You need to include the weight of the board (150N) and the torque due to this force into your translational and rotational equilibrium equations. Assume the center of mass is at the center of the board (i.e. 5m from either end of the board).

 

[googlinggoogler]

Hi, the equation I learned for doppler effect is:
fd=fs x ((v +or- vd)/(v +or- vs))
where fd is the frequency of the detector
fs is the frequency of the source
v is the speed in the medium
vd is the velocity of the detector
vs is the velocity of the source.

My question is, how does that equation convert into the other doppler effect equation:
(change of wavelength/wavelength)= v/c
where v is the relative velocity between source and detector
and c is the speed of the wave in the medium it propagates.

This question is related to a passage from AAMC CBT 4. I just don't recognize the second equation at all, but I needed to know it to solve two of the questions (equation wasn't supplied in the passage). Thanks in advance.

 

[Shrike]

My question is, how does [the usual Doppler] equation convert into the other Doppler effect equation:
(change of wavelength/wavelength)= v/c
where v is the relative velocity between source and detector
and c is the speed of the wave in the medium it propagates.

The second equation is the relativistic version; it works for EM waves. The difference there is that for purposes of EM Doppler shifts, only relative velocity matters (because both the detector and the source in effect "see" a wave speed of c relative to themselves); in the case of sound, or any other physical wave, it's possible for the detector/and or the source to moving relative to the medium, and therefore to see different wave speeds.

Essentially, the relativistic version works provided the detector is not moving (try it and see). [There's also an awkward version that works for a stationary detector.) With light, this is fine because detectors can be assumed to be fixed relative to EM waves; it doesn't work for sound, or any other mass-based wave. Using relative velocity for a doppler shift in sound et al is a good way to get a problem wrong.

 

[gridiron]

hi i understand that part now thank you
but.... i got lots more to ask xD

1) a 70kg adult sits at one end of a 10-m board , on the other end hwich sits his 30kg child. wehre shoudl teh pivot be palces so the board(ignore its mass) is balanced?
this one i get is 3m
but the continue part is
repeat problem taking into account the board's 15kg mass. the asnwer is 3.3m from the adult but i don get it.

hm..... i haf more but they have picutres....

hahah ^^ ill ask this for now
thank you

There are multiple ways to solve the problem, but I will demonstrate the long way in order to elucidate the concepts. You can start the problem in a similar fashion to the first part but now you have to take into consideration the mass of the board--this time, the mass of the board will cause some rotation around the pivot point. Now, you will have the torque due to the adult, torque due to the board and torque due to child. The only known distance is the center of mass of the board which is at the middle of the board--5m. Now, you have to assign a convention:

1.) I will derive all the torques with respect to the endpoint of the board. In other words, I will assume the adult and child are at the endpoints of the board. Using this convention, I will assign the x=0 position to the adult and x=10 position to the child. The x=5 position is assigned for the center of mass. The problem however does not ask for the these distances but rather for the pivot point.

2.) The weight of the adult, board and child all cause rotation around the pivot point. You want to assign the pivot point some distance r from the x=0 endpoint of the board. This way, the weight of the adult cause counterclockwise torque around the pivot point and the weight of the board and child cause clockwise torque around the pivot point. These torques must balance each other in a equilbrium system.

3.) Now calculate the torques:
For the adult: from the x=0 position, the weight of the adult exerts some torque from a distance of r from the pivot point. Thus the torque of the adult is=

Weight of the adult multipied by some distance r

For the board: From the x=5 position, the board exerts some torque from a distance of (5-r) from the pivot point. Thus the torque of the board is:

Weight of the board multipied by (5-r)

For the child: From the x=10 position, the child exerts some torque from a distance of (10-r) from the pivot point. Thus the torqe of the child is:

Weight of the child multipied by (10-r)

4.) If you plug in 700N, 150N and 300N respectively for the weights of the adult, board and child, you should obtain a r of approximately 3.3. Thus, from the x=0 position of the adult, you should place the pivot point 3.3 meters away. (*Note* I approximated gravity to 10 m/s^2 in obtaining the weights).

The postions used are arbritrary and you can assign any convention you wish. Try it another way and you should obtain the same answer. Good .

 

[V4viet]

A vibrating string has consecutive harmonics at wavelengths of 2 m and 4 m. What is the length of the string?

A. 1 m
B. 2 m
C. 4 m
D. 8 m

The answer is B..
I really don't understand what consecutive harmonics mean and how to go about solving this problem..can someone please help me?

 

[Shrike]

A vibrating string has consecutive harmonics at wavelengths of 2 m and 4 m. What is the length of the string?

"Consecutive harmonics" means harmonics with consecutive harmonic numbers, e.g., 1 and 2 (in this case), 2 and 3, 3 and 4, etc. The harmonic with n=1 is referred to as the first harmonic (also, the fundamental); n=2 is the second harmonic, and so forth.

The formula for wavelength of harmonics on a vibrating string is 2L/n, where L is the length of the string and n is the harmonic number, which can be any positive integer (1, 2, 3, etc.). Just try the answers, and see which one works.

A. L=1m can't work, because the longest harmonic has a wavelength of 2L = 2m, too short.

B. L=2m works: first harmonic (n=1) has a wavelength of 4m, second harmonic (n=2), has 2m.

C. L=4m has the right wavelengths, but they are n=2 (4m) and n=4 (2m). 2 and 4 aren't consecutive harmonic numbers.

D. Similar to C: L=8m has a 4th harmonic at 4m, and an 8th harmonic at 2m.