Physics Question Thread 2

[Pembleton]

Hi guys,

I don't have my physics textbook nearby and I was wondering if someone could remind me how to figure out the center of mass of two objects on a plank, for instance.

For example a 10 kg box is 1 m from the edge of a 10 m plank and a 20 kg box is 2 m from the other edge of the plank. Where is the center of mass on the plank?

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[heymanooh1]

Electric power for long distance transmission is stepped up to a very high voltage to:

a) produce currents of higher density
b) produce higher currents in transmission wires
c) to make less insulation necessary
d) to cut down the heat loss in transmission wires


Answer is D).. can anyone explain?

p = v squared divided by r.

resistance increases as the wire length gets longer (r = [rho* length]/area).
consequence of long wires is that there's more energy loss compared to shorter wire.

since r increases with length, v needs to also increase.

capice.

...I know what I'm doing...

sure ya are buddy

 

[axp107]

p = v squared divided by r.

resistance increases as the wire length gets longer (r = [rho* length]/area).
consequence of long wires is that there's more energy loss compared to shorter wire.

since r increases with length, v needs to also increase.

capice.

sure ya are buddy

you're wrong

explanation given:

power delivered is constant.. P = V^2/R. Power lost is I^2 R. Since R is fixed by the material and dimensions of the line, using a high voltage at lower current maximizes the power to the user and minimizes the heat loss in the transmission line.

 

[heymanooh1]

you're wrong

explanation given:

power delivered is constant.. P = V^2/R. Power lost is I^2 R. Since R is fixed by the material and dimensions of the line, using a high voltage at lower current maximizes the power to the user and minimizes the heat loss in the transmission line.

uh, hello the equation for power is generalized. p = v^2/r = i ^2*r, via v = i*r substitution.

for p delivered = v^2/r = constant

v has to increase since r is increasing with length (e.g. the total resistance in a wire 1m long is less than the total wire resistance of a 500m wire) because r = ([rho* length]/area). if it's a homogeneous wire, only length is changing from pt a to pt b.

just because there's an alternate explanation doesn't invalidate another reason that's both physically & mathematically consistent maybe you ought to ask your "reliable" kaplan instructor. better yet, a physics prof. that specializes in em.

 

[eltor07]

Yeah, but by increasing resistance you still are losing it to heat. You didn't answer the question as to upping Voltage will reduce heat loss. So your explanation is incomplete.

The question is saying that there is a long distance transmission. Ok, so length is fixed and thus resistance is fixed as well. Now there is power loss due to the heat. The question asks why is voltage stepped up in this case. The reason being since resistance is fixed, upping the voltage will decrease the current and thus decrease the power loss due to the heat of the resistor.

 

[tik-tik-clock]

A second object besides the balloon is suspended
motionless when released at depth d1. What can be
concluded?


I am confused what the suspended motionless means...does it mean that it did not submerge to the depth d1 or does it mean it floats on the surface of the water?
The answer explanation says that the densities of the object and the water have to be equal for the object to be suspended motionless. How did they come up with this? Whats the logic?

Thanks

 

[eltor07]

A second object besides the balloon is suspended
motionless when released at depth d1. What can be
concluded?


I am confused what the suspended motionless means...does it mean that it did not submerge to the depth d1 or does it mean it floats on the surface of the water?
The answer explanation says that the densities of the object and the water have to be equal for the object to be suspended motionless. How did they come up with this? Whats the logic?

Thanks

I think what is happening is the object is in the fluid, but is not moving. If you put the object a little higher than that point, it would sink, if you lowered it beyond that point it would rise. No matter what it would return to that exact same spot.

At the moment, I can't explain how the densities have to be equal, but if you think about if the object has a greater density than the liquid at the surface it will sink, but if the object has less density it will float.

It has something to do with when pressure the fluid exhibits on the object, which pushes it up equals the weight of the object (or the weight of the fluid displaced) then the object will remain stationary.

 

[eltor07]

I guess if the densities were the same it wouldn't matter where you put the object because it will remain motionless.

The force of buoyancy would be equal to the force the object exerts on the water. row(fluid)V(fluid)g = row(object)V(object)g

 

[tik-tik-clock]

I think what is happening is the object is in the fluid, but is not moving. If you put the object a little higher than that point, it would sink, if you lowered it beyond that point it would rise. No matter what it would return to that exact same spot.

At the moment, I can't explain how the densities have to be equal, but if you think about if the object has a greater density than the liquid at the surface it will sink, but if the object has less density it will float.

It has something to do with when pressure the fluid exhibits on the object, which pushes it up equals the weight of the object (or the weight of the fluid displaced) then the object will remain stationary.

That totally makes sense.. I was just confused what motionless meant.. with respect to what..
now its clear..thanks!

 

[eltor07]

Ok, whew. I was getting confused and trying to figure it out, too.

 

[wes431]

A 3kg object is released from rest at a height of 5 m on a curved frictionless ramp. At the foot of the ramp is a spring of force constant k= 100 N/m. The object slides down the ramp and into the spring, compressing it a distance of x before coming to a rest.
a) What is x?
b) Does the object continue to move after it comes to a rest? If yes, how high will it go up the slope before it comes to rest?

 

[bigman225]

Assuming it started at rest and energy is conserved (note the frictionless statement), that's an application of mechanical conservation. The variables are a big clue:
a)
k = 100n/m
h=5
m=3kg
1/2kx^2 + mgh = constant
mg(initial height) = 1/2k("compression distance")^2
(3kg)(10m/s^2)(5m) = 1/2(100n/m)x^2
x = sqrt(150/50)
= sqrt (3)
b)
Because energy conserved (1/2kx^2 + mgh = c), it should return to the same height.

One thing that is interesting to note is that it is completely meaningless that the ramp is curved.

 

[KnuxNole]

This Physics question is rattling my brain and I was wondering if anyone can help me understand it.

The question states: From ground level, a person at Point A throws a snowball horizontally to the right at 25 m/s. Where does the snowball land?

The diagram that accompanies the passage shows that Point A and Point B are seperated by 50 m(Point B is to the right of Point A).

The answer that I put was that the snowball will land 12.5 m to the right of Point B. However, the correct answer is that the snowball will land 50 m to the right of Point B. Does anyone know how to arrive at this answer? Thanks.

 

[eltor07]

I understand he is throwing the ball horizontally and ground level, but the ball has to be some height off the ground in order to answer this question.

 

[drellis8]

"When a car moves under its own power at constant velocity, the frictional force between the road and the tires that propel the car is:

A) static and in the direction of the motion of the car - CORRECT
B) static and in the opposite direction to the motion of the car
C) kinetic and in the direction of the motion of the car
D) kinetic and in the direction opposite to the motion of the car

I understand that static friction is the mode of friction here because it is non-sliding friction; however, I don't see how it can be in the same direction as the car. Isn't friction supposed to oppose the object's motion??

Thanks!

 

[cheezer]

"When a car moves under its own power at constant velocity, the frictional force between the road and the tires that propel the car is:

A) static and in the direction of the motion of the car - CORRECT
B) static and in the opposite direction to the motion of the car
C) kinetic and in the direction of the motion of the car
D) kinetic and in the direction opposite to the motion of the car

I understand that static friction is the mode of friction here because it is non-sliding friction; however, I don't see how it can be in the same direction as the car. Isn't friction supposed to oppose the object's motion??

Thanks!

what direction are the tires moving (i.e. the direction of the force)? imagine this same car going up an incline. hint hint

 

[cheezer]

This Physics question is rattling my brain and I was wondering if anyone can help me understand it.

The question states: From ground level, a person at Point A throws a snowball horizontally to the right at 25 m/s. Where does the snowball land?

The diagram that accompanies the passage shows that Point A and Point B are seperated by 50 m(Point B is to the right of Point A).

The answer that I put was that the snowball will land 12.5 m to the right of Point B. However, the correct answer is that the snowball will land 50 m to the right of Point B. Does anyone know how to arrive at this answer? Thanks.

you do know horizontal and vertical motion are independant right? to answer this problem, you need to find how high it is off the ground. then use a kinematics equation to figure out how much time it takes to hit the ground. then you use this time and plug it into an equation for the (x) kinematics equation. you should arrive at the correct answer.

 

[BananaMonkey]

Q:An electron is ejected from the cathode by a photon with an energy slightly greater than the work function of the cathode. How will the final kinetic energy of the electron upon reaching the anode compare to its initial potential energy immediately after it has been ejected?

A) It will be 2 times as large.
B) It will be approximately equal.
C) It will be 1/4 as large.
D) It will be 0.

Solution: The near equality of the photon energy and the work function means that little initial kinetic energy will be left for the electron. This initial kinetic energy is small compared to the 50 eV it will gain from the potential difference between electrodes.

I happened to get this question correct, but I believe it was through faulty reasoning. I figured that this question was driving at the relationship between KE, PE and Total Energy. I would have said they were equal regardless of the energy of the incoming photon. I am confused.

 

[drellis8]

Thanks for the response cheezer! I have another question folks...

A 100kg man dangles a 50kg mass from the end of a rope. If he stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be: (Note: ignore any frictional forces)

Answer: 333 N

in the explanation it says "In effect, gravity is only allowed to work on the mass, but it is accelerating both the mass and the man. Since the mass is 1/2 the man's, the acceleration is 1/3 of g."

I just needed some clarification on how they got 1/3 of g? Thanks!

 

[KnuxNole]

Thanks for the response cheezer! I have another question folks...

A 100kg man dangles a 50kg mass from the end of a rope. If he stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be: (Note: ignore any frictional forces)

Answer: 333 N

in the explanation it says "In effect, gravity is only allowed to work on the mass, but it is accelerating both the mass and the man. Since the mass is 1/2 the man's, the acceleration is 1/3 of g."

I just needed some clarification on how they got 1/3 of g? Thanks!

First, identify your system: The man and mass. Now, set the force due to gravity equal to the force of the system:

Fg= m*g
Fs= (m1+m2)*a

Fg = Fs

10*50 = (100+50)a

a will equal 3.33, which is 1/3 of g.

 

[gymgirl]

A 600 kg hovercraft traveling across a frozen lake
accelerates from rest to a speed of 60 kilometers per
hour in 4 seconds. What is the average force produced
by the thrust fan?​

A. ​

​

4.2 N​

B. ​

​

145 N​

C. ​

​

2500 N​

D. 9000 N

The answer is C. My question is, why is it that when they are finding the acceleration, they do not average the velocity (initial + final)/2. Because I know that EK averages it. Makes a whole lot of difference. Or am I just not seeing what should be seen?

Thanks!

 

[MundaneMD]

A 600 kg hovercraft traveling across a frozen lake​

accelerates from rest to a speed of 60 kilometers per
hour in 4 seconds. What is the average force produced
by the thrust fan?​

A.

4.2 N

B. ​

145 N

C. ​

2500 N
D. 9000 N​

The answer is C. My question is, why is it that when they are finding the acceleration, they do not average the velocity (initial + final)/2. Because I know that EK averages it. Makes a whole lot of difference. Or am I just not seeing what should be seen? ​

Thanks!​

Note: The problem says average force. To find the average force, you need the average acceleration. The equation you have, (Vi + Vf)/2, is average velocity. Average acceleration is the (change in velocity) / (change in time).​

0 km/hr --> 60 km/hr​

Average acceleration = (change in velocity) / (change in time)​

a(avg) = 60 / 4 = 15 km/hr^2​

but now, I'll convert the units into standard units (meters, Kilograms, seconds)​

a(avg) = 15 km/hr^2 = 15000 m/hr^2 = 15000/3600 m/s^2 = (I'll leave it as a fraction to make life easier in the next step)​

F = ma
F(avg) = m x a(avg)
F(avg) = 600 kg x (15000/3600) m/s^2 = 15000/600 N = 2500 N​

 

[gymgirl]

Note: The problem says average force. To find the average force, you need the average acceleration. The equation you have, (Vi + Vf)/2, is average velocity. Average acceleration is the (change in velocity) / (change in time).​

0 km/hr --> 60 km/hr​

Average acceleration = (change in velocity) / (change in time)​

a(avg) = 60 / 4 = 15 km/hr^2​

but now, I'll convert the units into standard units (meters, Kilograms, seconds)​

a(avg) = 15 km/hr^2 = 15000 m/hr^2 = 15000/3600 m/s^2 = (I'll leave it as a fraction to make life easier in the next step)​

F = ma
F(avg) = m x a(avg)

F(avg) = 600 kg x (15000/3600) m/s^2 = 15000/600 N = 2500 N​

Ofcourse I understand that
Average acceleration = (change in velocity) / (change in time)
But my real questions is, in the above equation, wouldn't the change in velocity be the average velocity/ change in time, therefore (v0+vf)/2 ? I understand the rest of the problem. But my concern is , when it is that you would average the velocity in such situations and when not to.
Thanks

 

[Foghorn]

Of course I understand that
Average acceleration = (change in velocity) / (change in time)
But my real questions is, in the above equation, wouldn't the change in velocity be the average velocity/ change in time, therefore (v0+vf)/2 ? I understand the rest of the problem. But my concern is , when it is that you would average the velocity in such situations and when not to.
Thanks

Either [change in velocity/change in time] or [(v0+vf)/2] can be used. If [ (v0+vf)/2 = v avg] is used, then [t avg] must be associated with it in order to get [a avg]; [change in time] isn't used for this equation because it would signify the difference between two quantities and not the average of two quantities. For this case [t avg] = [(t0 + tf)/2].

 

[gymgirl]

Either [change in velocity/change in time] or [(v0+vf)/2] can be used. If [ (v0+vf)/2 = v avg] is used, then [t avg] must be associated with it in order to get [a avg]; [change in time] isn't used for this equation because it would signify the difference between two quantities and not the average of two quantities. For this case [t avg] = [(t0 + tf)/2].

oops. i meant just time.
yeah, i just tried the two ways and it gives the same answer. Thats what you mean right? Just making sure.

 

[wes431]

A vibrating string has consecutive harmonics at wavelengths of 2.0 m and 4.0 m. What is the length of the string?

A) 1 m
B) 2 m (Answer)
C) 4 m
D) 8 m

 

[pyromatic]

Either [change in velocity/change in time] or [(v0+vf)/2] can be used. If [ (v0+vf)/2 = v avg] is used, then [t avg] must be associated with it in order to get [a avg]; [change in time] isn't used for this equation because it would signify the difference between two quantities and not the average of two quantities. For this case [t avg] = [(t0 + tf)/2].

This would only work in this problem and problems with a similar setup. I can simply redefine the time so that t0 = 10 and tf = 14, the same amount of time has elapsed; I've only changed when I started the clock. In this situation, your average time is 12, and you will get the wrong answer.

Similarly I can change coordinate systems so that v0 = 100 km/hr and vf = 160 km/hr. Again your average velocity becomes 130 km/hr, and you will get the wrong answer.

However you came to the correct answer by averaging the values is purely coincidence.

You must go back to the definitions:

v = dx/dt
a = dv/dt

F = dp/dt = d(mv)/dt = m*dv/dt for constant mass

The time average of any function is simply the time-integral of that function divided by the amount of time:

average F = [1/(tf - ti)] * integral(F*dt from ti to tf)
= [1/(tf-ti)]*integral((dp/dt)*dt from ti to tf) = [1/(tf-ti)]*integral(dp from p(ti) to p(tf))
= [1/(tf-ti)]*[p(tf) - p(ti)] = (m*vf - m*vi)/(tf - ti)
= mass * change in velocity / change in time

 

[KnuxNole]

I'm going over the EK Lecture question in the Physics Manual, and I am confused by their explanation of #60.

For those who do not have the book, the question states: A one meter board with uniform density, hangs in static equilibrium from a rope with tension T. A weight hangs from the left end of a board as shown. What is the mass of the board?

The diagram shows the board, on the far left side has a hanging mass of 3kg. Above that mass, it shows an arrow pointing to the right with 0.2m underneath it, with Tension pointing upward. Here is a diagram to make that last two sentences clearer:




^
0.2m |
____> | T
-------------------------------
|
|
|
[3kg]

The answer choices are

A) 1kg
B) 2kg(answer)
C) 3kg
D) 4kg

Their explanation says that the cc-torque is 3kg x 0.2m and the c-torque should be 0.3 m x 2kg. Several things I don't understand...

1) Where do they get the 0.3 from? Shouldn't it be 0.8 to add up to 1 meter
2) Isn't torque equal to the force times lever arm? Should the real equation be: 30 x 0.2 = 0.8m x F?

 

[axp107]

Pendulum question:

What if you start a pendulum off an a 90 degree angle... would it still go all the way up on the other side? and keep going? What about if you start it at the 12 of clock position?

 

[MundaneMD]

Pendulum question:

What if you start a pendulum off an a 90 degree angle... would it still go all the way up on the other side? and keep going? What about if you start it at the 12 of clock position?

In theory, if there was no resistance of any kind (air, friction, etc.), the kinetic energy would convert 100% into potential energy, which would convert 100% back into kinetic, so yes. It would keep going.

If you start it at the 12 position, it would just fall straight down.

 

[MundaneMD]

I'm going over the EK Lecture question in the Physics Manual, and I am confused by their explanation of #60.

Their explanation says that the cc-torque is 3kg x 0.2m and the c-torque should be 0.3 m x 2kg. Several things I don't understand...

1) Where do they get the 0.3 from? Shouldn't it be 0.8 to add up to 1 meter
2) Isn't torque equal to the force times lever arm? Should the real equation be: 30 x 0.2 = 0.8m x F?

(Follow along with the picture, but keep in mind that it is not drawn to scale)

1) I think this is how you meant to draw your diagram, right?

2) I simplified the diagram, because a uniform board is assumed to have its center of balance in the middle of the board. (This is the reason the distance is .3, and not .8)

3) I got rid of the "Tension" point, and replaced it with a fulcrum (same thing, but maybe it will be easier to understand).

After these steps, this problem becomes a lot simpler. I think you should be able to solve the problem from this point on, but I'll write it out just in case:

The distance from the fulcrum on the left side is .2m, and the distance from the fulcrum on the right side is .3m

The counter clockwise force is 30N x .2m

The clockwise force is ? x .3m

? = 20N = 2kg

 

[MundaneMD]

A vibrating string has consecutive harmonics at wavelengths of 2.0 m and 4.0 m. What is the length of the string?

A) 1 m
B) 2 m (Answer)
C) 4 m
D) 8 m

First, the basic formula for a standing wave on a string:
λ = 2L/n
2L = n x λ

This is an inverse relationship; as n increases, λ decreases, because the L is constant in this case.

Thus, for 2.0 m, the harmonic is n.
For 4.0 m, the harmonic is lower, (n-1). (it is n, and n-1, because the harmonics are consecutive)

First,
2L = n x 2.0
L = n

Next,
2L = (n-1) x 4.0
L = (n-1) x 2.0

Substituting L = n from the first step,
L = (L-1) x 2.0
L = 2L - 2
-L = -2

L = 2 meters

 

[RedSox1982]

Ok... so here is an easy one that I'll flat out admit, I just don't have right yet.

If I have a string attached to some mass, the tension in the rope is (when it's not moving) mg.

Now say I start twirling the string in a vertical circle so it goes above my head and down to my toes in a circular path, around and around. Here's my question: When the mass is at its LOWEST POINT (closest to the ground), will the TENSION in the rope be MORE or LESS than mg... and why?

Let the debate begin!

Thanks.

 

[axp107]

Ok... so here is an easy one that I'll flat out admit, I just don't have right yet.

If I have a string attached to some mass, the tension in the rope is (when it's not moving) mg.

Now say I start twirling the string in a vertical circle so it goes above my head and down to my toes in a circular path, around and around. Here's my question: When the mass is at its LOWEST POINT (closest to the ground), will the TENSION in the rope be MORE or LESS than mg... and why?

Let the debate begin!

Thanks.

Tension up... gravity down... net force should be centripetal force (which points upwards towards the center) so..

T - mg = mv^2/r
T = mv^2/r + mg

So T > mg

 

[RedSox1982]

Tension up... gravity down... net force should be centripetal force (which points upwards towards the center) so..

T - mg = mv^2/r
T = mv^2/r + mg

So T > mg

I see how you arrived at your answer... but I'm having a problem with the conceptual aspect. mg is going down. Tension is going up. If Centripetal force is ALSO going up, shouldn't that "take" some of the weight off of the Tension?

Maybe I should put it like this: gravity is going down. Now you add the component of centripetal force which is going up. Add these two vectors, it is still going to point down, but less. The tension should be that AMOUNT same magnitude, but pointing up... shouldn't it?

See... I think the equation should be:

T + mv^2/r = mg

I know you gave me the right answer, and I'm thinking about this incorrectly, but could you help show me where I'm going wrong?

Thanks!

 

[MundaneMD]

I see how you arrived at your answer... but I'm having a problem with the conceptual aspect. mg is going down. Tension is going up. If Centripetal force is ALSO going up, shouldn't that "take" some of the weight off of the Tension?

Maybe I should put it like this: gravity is going down. Now you add the component of centripetal force which is going up. Add these two vectors, it is still going to point down, but less. The tension should be that AMOUNT same magnitude, but pointing up... shouldn't it?

See... I think the equation should be:

T + mv^2/r = mg

I know you gave me the right answer, and I'm thinking about this incorrectly, but could you help show me where I'm going wrong?

Thanks!

The part I bolded is where you're going wrong (I think).

Look at it this way:

If you were to take a mass, and spin it in a circular motion, there would be centripetal acceleration towards the middle. This will create a certain amount of tension on the string.

Now, when the mass is at the top of the circle, gravity will pull the mass down. Since the centripetal acceleration will point down at the top of the circle, it will create an acceleration in the same direction as gravity. This is analogous to stepping on a scale in an elevator going downwards. The overall tension will be reduced.

When the mass is at the bottom of the circle, the tension in the string will continue pulling the mass as before. But, since gravity is now pulling the mass in the opposite direction as the string, it will add to the tension of the string. Thus, the tension of the string must counteract the force of gravity, and keep the centripetal force going. The overall tension will increase.

 

[RedSox1982]

Thanks guys... really appreciate it! I feel like an idiot right now for not seeing it before!

 

[axp107]

The key to getting this problem right is knowing that centripetal force is always the net force

You know how you can have a gazillion forces.. F1-F2+F3 + etc... = MA

Centripetal force is the MA.. its just rewritten!

 

[drellis8]

Hi Guys, sorry if this has been posted before, but pulleys are really bugging me in this EK 1001 book. Do problems on the real deal get this complicated? Anyway here's the problem (excuse the drawing lol):

Q: What is the tension T in the rope? (the two masses are 1kg each)
​

 

[mozdef]

Hey guys

I have a few questions from EK 1001 physics. I already checked their sites and couldn't find satisfactory answers and for some reason it won't let me create threads.

The first question goes like this:

In which direction will a compass point if it is held directly over a wire carrying a strong current? The answer is obviously that it will be point perpendicular to the wire (parallel to the magnetic field). Now in addition to this, the answer in the back states that the north end of the compass will point in the direction opposite the magnetic field (anti-parallel to the magnetic field). Now I thought that a compass in a magnetic field will always point in a way so that the north end points in the direction of the magnetic field (just like how a positive charge moves in the direction of the electric field). Can somebody clear this up for me?

The other question is question 891. There is a diagram for it so I guess you can only help me if you have the book. Now I thought the answer was A because first I held the top loop and determined that the magnetic field will come out of the page at the dot using the right hand rule. Then I held the bottom loop and determined that the magnetic field will also come out of the page at the dot. Thus the sum of these two vectors (both coming out of the page) would result in a stronger magnetic field. Why is this reasoning wrong - and also what effect does the radius of the loop have on the magnitude of the magnetic field at the dot?

Please I'd really appreciate some help. Thanks

 

[axp107]

A question about terminal velocity...

why is it that an object that takes longer to fall will have a smaller terminal velocity? if it takes longer, wouldn't it have a lot of time to reach a high velocity due to gravity?

 

[BlackSails]

A question about terminal velocity...

why is it that an object that takes longer to fall will have a smaller terminal velocity? if it takes longer, wouldn't it have a lot of time to reach a high velocity due to gravity?

If you drop two objects from the same height, and one takes longer to fall, which one had the greater terminal velocity?

 

[axp107]

If you drop two objects from the same height, and one takes longer to fall, which one had the greater terminal velocity?

I know my reasoning is wrong.. but I initially thought: the one that takes longer to fall since gravity has more time to act on it?
I'm assuming its... : the one that doesn't take as long to fall has a smaller drag force acting on it, resulting in a lower terminal velocity?

I guess my real question really is: does terminal velocity depend on WHEN the drag force occurs... for larger objects, does the drag force occur later.. while for lighter objects SOONER?

 

[BlackSails]

I know my reasoning is wrong.. but I initially thought: the one that takes longer to fall since gravity has more time to act on it?
I'm assuming its... : the one that doesn't take as long to fall has a smaller drag force acting on it, resulting in a lower terminal velocity?

I guess my real question really is: does terminal velocity depend on WHEN the drag force occurs... for larger objects, does the drag force occur later.. while for lighter objects SOONER?

Less drag = HIGHER terminal velocity. Newton's law: F(net)=ma. The two forces are gravity, and drag, Since they act opposite to each other, the lower drag force = higher accleration and higher final velocity.

The drag force occurs as soon as an object is in the air. Its not constant, it increases with respect to velocity, which makes velocity approach the limit of whatever the terminal velocity is. F(drag)=-b*v. b is a constant depending on the shape of the object and the medium, while v is the current velocity. Through some calculus, you can derive an expression for the terminal velocity.

 

[BananaMonkey]

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

Answer: increase by a factor of 6.
The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6.


Ok so I know intuitively that the time would be longer. I think my brain is just fried at this moment, but how did they get t=2v/g? I know this is obvious, I just can't see it...

 

[MundaneMD]

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

Answer: increase by a factor of 6.
The round-trip time t for a ball thrown vertically is given by t = 2v/g. If g is replaced by g/6, then t is increased by a factor of 6.


Ok so I know intuitively that the time would be longer. I think my brain is just fried at this moment, but how did they get t=2v/g? I know this is obvious, I just can't see it...

Here is the formula for the displacement:

y = (V)(t) + (1/2)(a)(t^2)

First, set y=0, because total displacement is 0 (same height before and after).

0 = (V)(t) + (1/2)(a)(t^2)

Divide both sides by t to get rid of t=0 as a solution.

0 = V + (1/2)(a)(t)
-V = (1/2)(a)(t)
t = -2V/a

But, since a = -g,

t = 2v/g

 

[medstudent786]

hey can someone please explain to me the RIGHT HAND RULE!!! I hate it soo much, i always get confused and never know what direction the answer is. PLEASE SOMEONE EXPLAIN IT TO ME because I know its bound to show up on the real mcat for sure. How do you tell the direction for a force on a moving charge, direction for current carrying wire and any others?????

 

[gridiron]

hey can someone please explain to me the RIGHT HAND RULE!!! I hate it soo much, i always get confused and never know what direction the answer is. PLEASE SOMEONE EXPLAIN IT TO ME because I know its bound to show up on the real mcat for sure. How do you tell the direction for a force on a moving charge, direction for current carrying wire and any others?????

The first thing you need to understand is magnetic force. Magnetic force is the force which acts on a moving charge through a magnetic field. The formula is:

Fb = qvB (actually q(v cross B), but for the MCAT qvB is fine).

The direction of the pependicular force is ALWAYS perpendicular to both v and B. How do you find Fb? You use the right hand rule. Since the magnetic force deals with vectors in dimensions, you need a way to visualize what is coming in and out. In physics, the standard notation for "out of the page" is the circle or the circle with a dot and for "into the plane of the page" is the x or the x with circle. The way I remember it is using the concept of an arrowhead--when the arrow is coming at you the head is circular and when it is going away from you the back looks like a x. If the charge you are dealing with is positive, you use the right and rule and if the charge is negative you use the left hand rule. I always use the right hand--for negative charges everything is opposite of the positive charge because the right and left hand are mirror images. So an answer for the right hand is the opposite, a switch, for the negative charge. Whichever you choose, the thumb points in the direction of the velocity, the fingers point in the direction of the B field and the direction of Fb can be determined by which way the palm is facing--the direction of the force is perpendicular to the palm. That's it! Try this question as an example:

Say you have a positive charge that is moving north (velocity vector points up) in a magnetic field which points east (to the right). What is the direction of the magnetic force?

Answer: Break the problem into pieces first. You have a positive charge so you need to use the right hand. The fingers are pointing east and the thumb north. Your hand is flat, so that means the magnetic force is into the plane of the page.

Then, what about a negative charge. The force will point out of the plane of the page because everything for the right hand is the opposite for the left hand.

I hope this helps and good .

 

[BananaMonkey]

thanks!

I didn't realize you could divide by t...

 

[MundaneMD]

thanks!

I didn't realize you could divide by t...

Yup, its something that your math teacher would cringe at, but for our purposes, it makes sense.

We already know that at t=0, y=0, because the ball hasn't been thrown yet. So by dividing by t, we are eliminating that as a possible solution. This greatly simplifies the equation.

 

[ashley]

The answer that my book gives to this question doesn't really explain how to find the answer:

A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires is .1, how far does the car slide?

Answer choices are:

A. 50m
B. 100m
C. 200m
D. 400m