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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[bigserve99]

The only question I have at the moment for the last question I asked, is the formulas you used to solve it.

Solving for t
Is this the standard equation used when you always need to find t?
x=V0x * t

Well, that's a tough question. I mean, this equation is the definition of velocity: V = delta x / delta t. However, it is not the only equation that can be used to find t in a problem.

Solving for the height of the building y
Is this the standard equation used to always find y?
y=0.5at^2

I used "y" in this case instead of "x" because these are displacements in two different directions given our standard 2d plane. Still, the equation displacement = 1/2 at^2 is just a part of the equation displacement = vo*t + 1/2 at^2, just like the equation before it (x = Vo * t, in that case there was no acceleration in the x direction so I left off the end)


You know, I think it would help if you were to focus more on the relationships between acceleration, velocity, and displacement. Really try and understand the graphs of each, and how each is related to the other. What I mean, there are only 2 definitive quatities in all the kinemeatics equations.....

displacement, and time.

Everything else (velocity, acceleration) is just ratio's of the two. The kinematic equations, help to simplfy the problems and relationships between these quantities given a constant acceleration.

Also, break up every kinematic problem, and really just about every problem in physics I've come accross that uses vectors in to cooridinates (usually just x and y, but be careful where cross products are present, the third dimension comes into play there)

I hope these suggestions help. Anyone else want to chime in with any advice.

Good luck,

~Bigserve99

 

[swatchgirl]

Hey guys,

Anyone have a link to how contact lenses work in terms of optics?

I know for nearsightedness you need a diverging lens to correct the problem (image in front of retina)... and for farsightedness (image behind retina), you need a converging lens to correct the problem.

For example, diverging lenses form an image on the same side as the source of light .. always.. how would this help with nearsightedness?

I'm having trouble picturing how the lenses would help.. thanks!

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/refrn/u14l6e.html

Scroll down and read the part on Correction for Nearsightedness. The diagrams there may also be helpful.

For farsightedness, picture the same diagrams but with a converging lens that helps bring the image forward from behind the retina by converging the incoming light before it reaches the cornea and the lens.

 

[DDS4tehwin]

.

 

[axp107]

I'm having trouble with circuits involving both a resistor and a circuit. Anyone have any link/help with what happens when a circuit containing a resistor and circuit is charged and discharged... namely the discharging part. Any links would help.

Not only voltage of the capacitor... but what happens to the current as a capacitor is discharging? What happens to the Voltage in the resistor as discharging occurs? I can never find websites that summarize all possibilities when it comes to a circuit containing ONE resistor and ONE capacitor.

Apparently, as you discharge, V capacitor goes to 0 (which makes sense) and V resistor also goes to 0... which makes no sense to me... how can the "voltage" just disappear? Shouldn't the resistors voltage go back up?

 

[LauraDO]

Any idea how to solve this one

A clock is set against an accurate standard and 30 days later is found to be in error by 12 seconds. What is the percentage of error in its time keeping?

 

[halekulani]

I'm having trouble with circuits involving both a resistor and a circuit. Anyone have any link/help with what happens when a circuit containing a resistor and circuit is charged and discharged... namely the discharging part. Any links would help.

Not only voltage of the capacitor... but what happens to the current as a capacitor is discharging? What happens to the Voltage in the resistor as discharging occurs? I can never find websites that summarize all possibilities when it comes to a circuit containing ONE resistor and ONE capacitor.

Apparently, as you discharge, V capacitor goes to 0 (which makes sense) and V resistor also goes to 0... which makes no sense to me... how can the "voltage" just disappear? Shouldn't the resistors voltage go back up?

when you discharge a capacitor, current flows.

voltage disappears because current stops flowing through the resistor over time...or at least that's how i understand it. someone correct me if i'm wrong.

 

[da8s0859q]

Any idea how to solve this one

A clock is set against an accurate standard and 30 days later is found to be in error by 12 seconds. What is the percentage of error in its time keeping?

What I would do:

30 days = 2,592,000 seconds

Clock in question counts either 2,591,088 or 2,592,012 seconds - makes no difference. A variance of 12 seconds.

12 seconds out of 2,592,000 = 12/2592000 ~= 4.6296e-6 % (0.0000046296%).

 

[peternohibi]

Hi.
Please help me in this project.
We were told to do a project in making structures made out of 25 paper folders that could hold up to 25 lbs. of weight in a height of 25 inches.
What structure would be the most advisable to use.
Please, I really need help we are allowed to make it in a week.
I would like to thank those who would help in advance.

 

[axp107]

edit: nvm

 

[RPedigo]

Hi.
Please help me in this project.
We were told to do a project in making structures made out of 25 paper folders that could hold up to 25 lbs. of weight in a height of 25 inches.
What structure would be the most advisable to use.
Please, I really need help we are allowed to make it in a week.
I would like to thank those who would help in advance.

This is not really the place for homework help-- this is the MCAT question forum.

However, keep in mind that triangles are the most durable and strong shape. Lots of triangles in a row would be quite strong.

 

[ChubbyChaser]

Hi.
Please help me in this project.
We were told to do a project in making structures made out of 25 paper folders that could hold up to 25 lbs. of weight in a height of 25 inches.
What structure would be the most advisable to use.
Please, I really need help we are allowed to make it in a week.
I would like to thank those who would help in advance.

I think your best bet would be to make the 25 folders into columns, i think columns would work better than triangles. I did it in 9th grade, and 6 pieces of paper held up something ridiculous like 30 textbooks or something.

 

[mrmandrake]

Hey guys,

Quick question. I know when you raise something of mass m to height h the work you need to do is mgh right? The work required is the force required to counter the force of gravity over the distance h.

Now we also know that the work done on something can change it's kinetic energy so that its velocity changes, hence W = KE.

So if I was raising something not at a constant v but accelerating it to a height h, the work done would be W = mgh + 1/2mv^2 right?

If this is true, is this formula stated anywhere formally? Specifically, does it have a name? (i.e. W = KE is the Work-Energy theorem).

I was working on a problem that used the above formula combining PE and KE but couldnt find it in my TPR books. Thanks guys.

 

[cheezer]

Hey guys,

Quick question. I know when you raise something of mass m to height h the work you need to do is mgh right? The work required is the force required to counter the force of gravity over the distance h.

Now we also know that the work done on something can change it's kinetic energy so that its velocity changes, hence W = KE.

So if I was raising something not at a constant v but accelerating it to a height h, the work done would be W = mgh + 1/2mv^2 right?

If this is true, is this formula stated anywhere formally? Specifically, does it have a name? (i.e. W = KE is the Work-Energy theorem).

I was working on a problem that used the above formula combining PE and KE but couldnt find it in my TPR books. Thanks guys.

Conservation of Energy

Ui + Ki = Uf + Kf

 

[BlackSails]

Hey guys,

Quick question. I know when you raise something of mass m to height h the work you need to do is mgh right? The work required is the force required to counter the force of gravity over the distance h.

Now we also know that the work done on something can change it's kinetic energy so that its velocity changes, hence W = KE.

So if I was raising something not at a constant v but accelerating it to a height h, the work done would be W = mgh + 1/2mv^2 right?

If this is true, is this formula stated anywhere formally? Specifically, does it have a name? (i.e. W = KE is the Work-Energy theorem).

I was working on a problem that used the above formula combining PE and KE but couldnt find it in my TPR books. Thanks guys.

Work = F*dX
In lifting, F = mg.
The velocity v does not figure into the work. Work (in a conservative system, which basically just means no friction) is a state function. That means it depends ONLY on the endpoints.

 

[spoudaios]

Hi guys...

I found this in the reference section on SDN:

If the density of the fluid above the surface cannot be neglected (i.e., it's a liquid or a very dense gas), then the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface. In practice, subtract the density of the lighter fluid from the density of the the heavier fluid, use the principle above for solving problems neglecting the upper fluid to get a density of the object, and then add the density of the lighter fluid back in to get the virtual density of the object. (If you're finding how much is below the surface of the heavier fluid, subtract the density of the lighter one from both the heavier fluid and the object, and find the ratio as above.) An example shows it better:

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.

**I am trying to understand the concept so i don't want to follow the advice of substracting the density of the 2 fluids.
Can someone help me doing this in another way?

How do you draw the picture of this problem? The same as if it were water and air (instead, now we have very heavy gas instead of air?)

Thanks​

 

[halekulani]

the way i visualized it was thinking of two liquids, the denser one is obviously on the bottom (eg. water and oil)

the buoyant force is the sum of the weights of the respective fluids displaced by the portions of the object above and below the surface
Weight = density * volume
for volume, just use fractions

2/3 * 1.6 = 1.066666 (buoyant force from dense medium)
1/3 * 1.0 = 0.333333 (buoyant force from light medium) [ you might think this doesn't make sense because you think the whole object displaced all of the light medium so it should be 1 * 1.0, but you're calculating it in terms of which fluid is acting on the block. only the top 1/3 of the block is being acted on by the light medium and therefore only 1/3 of it experiences the buoyant force. this way you calculate the sum of all the forces acting on the block. the dense medium acts on 2/3 of the block and the light medium acts on 1/3
add them together to get ~1.4 ]

the point of subtracting densities is to put things in terms of relativity. if you subtract the lighter density from the heavy, it's as if you have an object with air above it and it is floating in a medium with density of 0.6g/cm^3. since the object had to sink to the bottom to reach the second medium, obviously the displaced V = 100%, and the buoyant force is 1.0g/cm^3 * 1. that's why you add 1 to 0.4 to get 1.4.

i really recommend trying to understanding the concept using subtraction of densities.

 

[DShalo]

Just wondering if anyone could help me solve this problem. I have a general idea, however, my answer doesn't match with the correct answer.

Specifically deals with Newtons Second Law

Question:
A jet catapult on an aircraft carrier accelerates a 2000-kg plane uniformly from rest to a launch speed of 320 km/h in 2.0s. What is the magnitude of the net force on the plane?

Answer Provided:
8.9x10^4 N

I appreciate your help very much.

Convert 320km/h to m/s. This should be 88.9m/s.
Divide 88.9 m/s by 2 seconds so u obtain the planes acceleration. this should be 44.4 m/s^2.
Now multiply acceleration by the mass of the plane. you get - 88888.88N

 

[BlackSails]

Just wondering if anyone could help me solve this problem. I have a general idea, however, my answer doesn't match with the correct answer.

Specifically deals with Newtons Second Law

Question:
A jet catapult on an aircraft carrier accelerates a 2000-kg plane uniformly from rest to a launch speed of 320 km/h in 2.0s. What is the magnitude of the net force on the plane?

Answer Provided:
8.9x10^4 N

I appreciate your help very much.

Convert 320km/h to m/s. This should be 88.9m/s.
Divide 88.9 m/s by 2 seconds so u obtain the planes acceleration. this should be 44.4 m/s^2.
Now multiply acceleration by the mass of the plane. you get - 88888.88N

What do you need help with? You have the right answer?

 

[MundaneMD]

Maybe he is confused about the scientific notation part?

88888.88 = 8.888888 x 10^4

8.888888 rounds to 8.9

So 88888.88 = 8.9x10^4

?

 

[axp107]

What's the best way to do "work done by gravity" problems....?

Signs are important... +ive and -ive work, can someone explain the difference

a) say a ball is falling from height 10m, what is the work done by gravity?

b) a ball is thrown up, what is the work done by gravity?

c) what would be a situation when no work is done... (0 work) ?

 

[MundaneMD]

Work = Force x Distance x cos(&#952

where θ = the angle between the direction of the force (gravity), and the direction of the displacement.

a) If the ball is falling, the ball is moving downwards, and gravity is pulling the ball downwards. θ = 0, so cos(&#952 = 1. This means that the work done by gravity is positive.

b) If the ball is thrown up, the ball is moving upwards, and gravity is pulling the ball downwards. θ = 180, so cos(&#952 = -1. This means that the work done by gravity is negative.

c) If the cos(&#952 = 0, no work will be done. Some examples of this would be if the ball were not moving at all (just sitting on a table), or moving perpendicular to the force of gravity (rolling on the floor). You can even consider the exact moment that the ball has no velocity in the middle of a trajectory, because if the ball has no velocity vector, there can't be any θ with respect to the force of gravity.

In short, if θ = 90 (moving perpendicular to gravity), or if θ = n/a (ball is stationary), gravity will have a work of 0.

θ is what determines whether the work done by a force is positive, negative, or 0.

 

[realmadridstorm]

d

 

[wes431]

A boy standing on a 15 m ledge throws a 0.2 kg rock horizontally at 10 m/s. How far does the rock travel horizontally before it strikes the ground?

a) 10 square root 1.5
b) 10 square root 3 (answer)
c) 15 m
d) 30 m

 

[halekulani]

what's the question?

if you want to know how to solve the problem, find out how long it takes for the rock to reach the ground, then just multiply that time by the horizontal speed.

 

[Isoprop]

A boy standing on a 15 m ledge throws a 0.2 kg rock horizontally at 10 m/s. How far does the rock travel horizontally before it strikes the ground?

a) 10 square root 1.5
b) 10 square root 3 (answer)
c) 15 m
d) 30 m

to find the time the rock hits the bottom (y direction):
H = v0t + 0.5gt^2
15 = 0 + 5t^2
t = sqrt(3)

to find how far the rock goes, consider the x direction at constant velocity:
d = vt
d = 10*sqrt(3)

 

[Phanster]

Hey can someone help me with this problem?

In a popular lecture demonstration, a projectile is fired at a falling target. The projectile leaves the gun at the same instant that the target is dropped from rest. Assuming that the gun is initially aimed at the target, show that the projectile will hit the target, (One restriction of this experiment is that the projectile must reach the target before the target strikes the floor.)

 

[ChubbyChaser]

All objects fall vertically at the same rate 9.8 m/s^2, so if the projectile is fired fast enough in the horizontal direction and aimed directly at the falling target, they should collide before striking the ground.

Idk, im sure someone else might be able to explain it better, but i hope it helps.

 

[SBK]

I'm having trouble with this problem from TPR:

" A uniform meter stick weighing 20 N has a 50N attached to its left end and a 20N weight on its right end. The entire bar apparatus is hung from a rope. What is the tension in the rope? And, where should the rope be placed with respect to the left end of the bar so that the bar remains level?


I understand that F(T) is 100N for the first part...

And I know that we have to balance the torques.... so if you assume the rope is at point 'X', the left side has torque = F* L = 50X...

Now, how do you account for the bar weight and right side and bar weight since no other dimensions are given?

(Correct answer = 40cm)

 

[halekulani]

is it possible the answer is 33cm? i can't get 40cm :\

if it's a uniform meter stick, i think you assume the 20N is in the middle (center of mass)
here's how i would set it up...may be wrong though.

50x = 20 ( .5 - x ) + 20 ( 1 -x )

 

[MundaneMD]

A uniform meter stick weighing 20 N has a 50N attached to its left end and a 20N weight on its right end. The entire bar apparatus is hung from a rope. What is the tension in the rope? And, where should the rope be placed with respect to the left end of the bar so that the bar remains level?


I understand that F(T) is 100N for the first part...

It's 90N...how did you get 100N?

I agree with halekulani, I also get 33.3cm instead of 40cm.

 

[BerkReviewTeach]

It's 90N...how did you get 100N?

I agree with halekulani, I also get 33.3cm instead of 40cm.

is it possible the answer is 33cm? i can't get 40cm :\

if it's a uniform meter stick, i think you assume the 20N is in the middle (center of mass)
here's how i would set it up...may be wrong though.

50x = 20 ( .5 - x ) + 20 ( 1 -x )

I too get 90N and 0.33 cm for the numbers presented here. But I think there is an error.

If the answers are supposed to be 100N and 0.40cm, there is a typo in either the original question or what the original posted entered. If the masses and the ends of the bar were 50N and 30N and the bar was 20N, then those two answers would be correct. I think they meant 50N, 30N, and 20N.

 

[Begaster]

I too get 90N and 0.33 cm for the numbers presented here. But I think there is an error.

If the answers are supposed to be 100N and 0.40cm, there is a typo in either the original question or what the original posted entered. If the masses and the ends of the bar were 50N and 30N and the bar was 20N, then those two answers would be correct. I think they meant 50N, 30N, and 20N.

Wow, good catch.

 

[halekulani]

dang

all that fuss for a freakin error
good catch.

 

[bozz]

Can this concept be applied only when you have a plain wire in parallel along with a resistor... the current wouldn't pass through the resistor at all, causing a short circuit

What about if you have a capacitor in parallel with a resistor? Does no current pass through the resistor?

Here's a pic to illustrate

 

[smeagol]

Current will pass through both the resistor and the capacitor. The current in the capacitor is 90 degrees out of phase with the current in the resistor. I highly doubt you need to know this for the MCAT, or how to calculate the currents flowing in each unless it is explained in the passage.

 

[bozz]

Thanks for the reply ^ ... appreciated

A circuit topic that constantly confuses me is "heat lost" from resistors

Electric power for transmission is stepped up to a high voltage in order to :
1) produce currents of higher density
2) to produce higher currents in the transmission wires
3) to make less insulation necessary
4) to cut down the heat loss in the transmission wires

I thought since R is fixed, increasing Voltage would increase I, the current.

the answer is 4)

 

[MundaneMD]

I thought since R is fixed, increasing Voltage would increase I, the current.

I think you're mixing up the concept of Ohm's Law.

First, some definitions:

Ohm's Law
V=iR
This equation refers to a voltage drop across resistance.
Voltage drop = (current in the wire) x (resistance in the wire)
V(lost) = i(wire) x R(wire)

P=iV
This equation can have two meanings:
Power transmitted = (current in the wire) x (voltage across the wire)
P(wire) = i(wire) x V(wire)

or

Power lost = (current in the wire) x (voltage lost due to resistance)
P(lost) = i(wire) x V(lost)

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

Now, I'll deal with this problem in two parts:
1) Show that the current drops due to the increased voltage
2) Show that a decreased current will lead to a decrease in the power lost

1) P(wire) = i(wire) x V(wire)
The power in a wire is always constant. This means that there is an inverse relationship between the current in the wire, and the voltage being transmitted across the wire. Increasing one will decrease the other. This means that an increase in voltage will cause a decrease in the current flowing through the wire.

2) P(lost) = i(wire) x V(lost)
Now we're talking about a loss in power due to a voltage drop (due to resistance). This is where Ohms Law comes in.
V(lost) = i(wire) x R(wire).
P(lost) = i(wire) x i(wire) x R(wire)
Since the current in the wire is lower, less power will be lost in the wire.


Hope that helps.

 

[NickMB]

Any idea how to solve this one

A clock is set against an accurate standard and 30 days later is found to be in error by 12 seconds. What is the percentage of error in its time keeping?

The rate of error is 12 seconds per 30 days. Units are not consistent, however. Convert 30 days to seconds. Don't forget to multiply by 100 to get the percentage.

For quick time saving estimation on the MCAT, consider that the amount of seconds in 30 of days is in the millions ( 30 x 60 x 60 x 24 = somewhere in the 2 millions), which means that 12 divided by a number that high will yield a figure that's to the power of -6. When you multiply by 100, you'll get a figure to the power of -4. Any answer that's larger than that is automatically incorrect.

 

[spoudaios]

Hi guys...

More electricity topics:

1>. Is work state function? In my chem textbook, it says that work and heat are not state functions. In physics, work is a state function. The work to move charge from A to B only depends on potential diff. and not the path.

2>. Graph between Power (P) vs Resistance (R) where:
P=I^2xR. Also, P= V^2/R
Does this mean there are 2 possible graphs on this?

3>. I am really confused about capacitance.
What would happen to
C; Q; V; E-field; Energy of capacitor when: (determine whether it will increase or decrease)

a>. capacitor is discharged from battery (after being fully charged) then pulled apart
b>. capacitor is charged then a dielectric is inserted

It seems like i have no problems determining the C and Q but i am not sure about V and E-field. Can someone clarify the concept?

For the reply by MundaneMD:
--> Power (wire) = V (wire) x I (wire) where power is a constant. Where does this definition come from? I probably will get this question wrong. I thought V = IR and if you increase V, I will increase providing R is constant.

 

[BlackSails]

1>. Is work state function? In my chem textbook, it says that work and heat are not state functions. In physics, work is a state function. The work to move charge from A to B only depends on potential diff. and not the path.

Work done by (or against) a conservative field is a state function. Work dones against anything else, is not a state function. Heat is a state function. In the thermodynamic identity Q=TdS-PdV, all the terms are state functions.

2>. Graph between Power (P) vs Resistance (R) where:
P=I^2xR. Also, P= V^2/R
Does this mean there are 2 possible graphs on this?

Yes, it does. P=IV=I^2*R=V^2/R

 

[MundaneMD]

For the reply by MundaneMD:
--> Power (wire) = V (wire) x I (wire) where power is a constant. Where does this definition come from? I probably will get this question wrong. I thought V = IR and if you increase V, I will increase providing R is constant.

Basically, P=iV is a generic formula. It can be applied to resistors, and to anything carrying a current.

current = charge/seconds
voltage = energy/charge

so power = energy/seconds

The energy/seconds can be used to measure the rate of energy being lost at a resistor (P lost), or to measure the amount of energy flowing through a wire (P wire).

 

[spoudaios]

Thanks for the answers guys...

Can someone help me with the capacitance please?

For MundaneMD: the reason why i am confused is that i never read in textbook which states P=VI is a constant. However, the explanation makes sense. Increasing the voltage is to minimize the energy lost as heat thus we have to reduce current. In some cases though, like using big appliances (washer and dryer), we need large current, correct? Then the resistance must be small (because according to V=IR... V is constant)

 

[MundaneMD]

the reason why i am confused is that i never read in textbook which states P=VI is a constant. However, the explanation makes sense. Increasing the voltage is to minimize the energy lost as heat thus we have to reduce current. In some cases though, like using big appliances (washer and dryer), we need large current, correct? Then the resistance must be small (because according to V=IR... V is constant)

Yea, I never knew about the P=iV being constant in a closed circuit until I did the same problem. My Kaplan teacher clarified it for me.

And yea, for a big appliance, you need a larger current. This means that the resistance of the wire must be lower. Have you ever held a 12-guage (20 amp) wire vs a 14-guage (15 amp) wire? The 12-guage is thicker. Thicker = more area = less resistance.

 

[MundaneMD]

3>. I am really confused about capacitance.
What would happen to
C; Q; V; E-field; Energy of capacitor when: (determine whether it will increase or decrease)

a>. capacitor is discharged from battery (after being fully charged) then pulled apart
b>. capacitor is charged then a dielectric is inserted

Are you saying that for (a), the capacitor is fully charged, and during discharge, the plates of the capacitor are pulled apart?

After seeing the explanation for (b), you can probably figure out (a) for yourself.

For (b), the capacitor is charged, and the dielectric is inserted, which causes K to increase.
C' = KC, so C' will increase.
Q will remain the same, since no charge was introduced.
V = Q/C', so V will decrease.
Energy = (1/2)C'V² = (1/2)C'(Q/C')² = (1/2)Q²/C', so Energy will decrease.
Electric Field (E) = V/d, so the magnitude of E will decrease, but the direction won't change.

 

[spoudaios]

Your answers for part b is exactly like mine. Thanks

I am more confused about part a. So, it is like what you said. Capacitor is charged fully. Then disconnected from the battery and then pulled apart.

Finding about what happen about C is easy but I am always not sure about V and E...

Do you think that Q will always be constant in ANY condition?

Are you saying that for (a), the capacitor is fully charged, and during discharge, the plates of the capacitor are pulled apart?

After seeing the explanation for (b), you can probably figure out (a) for yourself.

For (b), the capacitor is charged, and the dielectric is inserted, which causes K to increase.
C' = KC, so C' will increase.
Q will remain the same, since no charge was introduced.
V = Q/C', so V will decrease.
Energy = (1/2)C'V² = (1/2)C'(Q/C')² = (1/2)Q²/C', so Energy will decrease.
Electric Field (E) = V/d, so the magnitude of E will decrease, but the direction won't change.

 

[armybound]

Here's a basic one, but it's bothering me.

There's a question about a fulcrum balancing weights.. the fulcrum's mass is 10kg, and the weights are 5kg and 40kg. La dee da, it asks for the normal force exerted by the fulcrum.

The answer they gave is ~550N. I figured it was ma for the 2 blocks its balancing, since it exerts an opposite force equal to their weight. but the answer being 550N means they also take into consideration the weight of the fulcrum itself.

My question is basically where does the normal force belong? does the normal force belong to the object with the weight, meaning my mass is 75kg but I'm also exerting a 750N normal force on myself? or is the earth exerting that normal force on me? it seems like the answer of 550N means the fulcrum is exerting a 550N normal force on itself against the ground.

 

[NickMB]

Here's a basic one, but it's bothering me.

There's a question about a fulcrum balancing weights.. the fulcrum's mass is 10kg, and the weights are 5kg and 40kg. La dee da, it asks for the normal force exerted by the fulcrum.

The answer they gave is ~550N. I figured it was ma for the 2 blocks its balancing, since it exerts an opposite force equal to their weight. but the answer being 550N means they also take into consideration the weight of the fulcrum itself.

My question is basically where does the normal force belong? does the normal force belong to the object with the weight, meaning my mass is 75kg but I'm also exerting a 750N normal force on myself? or is the earth exerting that normal force on me? it seems like the answer of 550N means the fulcrum is exerting a 550N normal force on itself against the ground.

Normal force is a contact force, so at any point where to objects are in contact with each other, there is a normal force. The normal force exerted BY the fulcrum is at the point of support, and it's directed upward, opposite the direction of the force of gravity. Say you have a seesaw, the point of support would be at the centre, with normal force pointing up.

Are you solving this problem using laws of rotational equilibrium and dynamics?

 

[armybound]

Normal force is a contact force, so at any point where to objects are in contact with each other, there is a normal force. The normal force exerted BY the fulcrum is at the point of support, and it's directed upward, opposite the direction of the force of gravity. Say you have a seesaw, the point of support would be at the centre, with normal force pointing up.

Are you solving this problem using laws of rotational equilibrium and dynamics?

yes.

so the normal force includes the weight of the fulcrum because the normal force comes from the center of the fulcrum?

 

[NickMB]

yes.

so the normal force includes the weight of the fulcrum because the normal force comes from the center of the fulcrum?

Yes

 

[MundaneMD]

I am more confused about part a. So, it is like what you said. Capacitor is charged fully. Then disconnected from the battery and then pulled apart.

Finding about what happen about C is easy but I am always not sure about V and E...

C = K x (Area / distance)
Since the distance between the plates will increase, the capacitance will decrease.
V = Q/C, so V will increase.
Energy = (1/2)C'V² = (1/2)C'(Q/C')² = (1/2)Q²/C', so Energy will increase.

Now this part is a bit tricky, because both the V and the d are changing.
Electric Field (E) = V/d
So lets plug in hypothetical numbers.
If the distance, d, was increased ten times the original distance, C would be 1/10 of the original value.
If C is 1/10 of the original value, V will be 10 times bigger.
If V is 10 times bigger, and the distance is 10 times bigger,
E = V/d = 10/10 = 1. So the electric field would not change.

Another way of determining the electric field:
E = V/d = Q/Cd = Qd/KAd = Q/KA
But Q, K, and A are constant in this case, so the electric field will not change.

Do you think that Q will always be constant in ANY condition?

Q only changes when the capacitor is being charged or discharged. Otherwise, assume that Q is constant.