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TheMuffinMan

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Hi, it's me again with another physics problem that I need help with. ^_^ lol


A car is driving at 44.7 mi/hr. First, I just did the m/s conversion and got 19.983 m/s. Okay, the coefficient of friction between the tires and the road is .13. It asks what the minimum distance is in which the car will stop.

Okay, so I wanted to calculate force of friction, so I thought of Ff = mu*Fn, but I can't calculate Fn without the mass of the car. Also, even if I had force of friction, I wouldn't really know what to do next. I know the distance formulas x = ViT + 1/2at^2 and Vf^2 = Vi^2 + 2ax, but I'm not really sure of how to apply those to actual forces, since I don't the horizontal deceleration of the car.

For example, I tried (19.983)^2 = 0 + 2ax, but again, no a. =[

So, I could try to find a, but even with a = (Vf - Vi)/t, it won't help because I don't know time either. D:

Please help. I really appreciate it, and thank you in advance.

I know this is an old question and obviously you don't need the answer, but I'm wondering why your class is having you do kinematics approaches to problems involving Friction, when you could just do Work/Energy formulas

and I know you got the answer to your most recent question, but just for future reference: kinetic friction is usually negative, as it is working 180 degree away from the displacement
 

Marly

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Okay, I need ideas for a cause and effect paper. For some odd reason when I look into topics/ideas for one I get a bit confused. I could really use some specific ideas. I keep thinking that the format should be something like: ________ causes __________ and here's why. Is that remotely correct? Gah, I seriously don't know what it is about cause and effect that confuses me, but it does. So would anyone be willing to elaborate on this? Or give me some ideas?

[edit] Okay nevermind on this. Ironically, I was stressing about how this is due tomorrow and how I shouldn't've put it off until the last minute when the idea of the causes and effects of procrastination popped into my head. So yea, awesome topic; procrastination.
 
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Solar

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Just a quick question since my memory has gone blank

We only use numerical prefixes for naming molecular compounds, right?

Bit rusty on this stuff, but I'm pretty sure I'm right
 

The Fishman

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Just a quick question since my memory has gone blank

We only use numerical prefixes for naming molecular compounds, right?

Bit rusty on this stuff, but I'm pretty sure I'm right

You mean organic compounds like 2-methyl propane or 1,2,3-tribromo butane? Or just like SiO_2_ or H_2_O
 

Dogenzaka

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I'm having a lot of trouble trying to comprehend orbitals and energy levels and this s,p,d,f shit in chemistry. This is the first time I've ever really been exposed to this :(

Please let me know if I understand this correctly...

So each energy level, n=1, n=2, n=3, etc. has orbital shapes it can take upon?

n=1 has an S orbital (called 1s) which is shaped like a sphere....described as l=0, and ml = 0.

So... n=2 has two shapes it can take on? A second S orbital (called 2s) shaped like a double sphere (a sphere within a sphere)...described as l=0, and a P orbital (called 2p), shaped like a dumbbell, which itself takes on three different orientations (ml = -1, 0, 1)...and whose shape is described as l=1?

Then the n=3 energy level has three different main shapes or whatever it can take upon....The S orbital (called 3s...it's basically just like the 1s and 2s orbitals...but maybe looks a little different, I don't know?)...described as l=0, the P orbital (called 3p), shaped like a dumbbell, which takes on three different orientations (ml = -1, 0, 1)...and whose shape is described as l=1, and the D orbital (called 3d), kind of shaped like a clover, which takes on five different orientations (ml = -2, -1, 0, 1, 2).

These orbitals stack upon each other in space.....and the electrons occupying that energy level (say, n=2) can choose to make their electron clouds take the shape of ANY of the orbitals in that energy level, EXCEPT that only TWO of them can occupy the same orbital at once (spinning in opposite directions, + and -).

Therefore, if you have an atom with 10 electrons.....two of them will occupy the lower 1s orbital (part of the n=1 energy level). Then two more will occupy the 2s orbital, and six will occupy the three different 2p orbitals (ml = -1, 0, 1), all in the n=2 energy level.

So in total, your 10 electrons are in two energy levels, spontaneously jumping up and down energy levels whenever possible, whenever exposed to photons, etc. They are in two S orbitals, one in the n=1, one in the n=2, and in three P orbitals, all in the n=2 level.

Now, there are some weird things to note such as the S orbital occupying the n=4 level (4s) and the five D orbitals occupying the n=3 level (3d) are degenerate orbitals, meaning they have the same energy. And if you have enough electrons to fill up all the orbitals up to 4s, 4s will then jump down down underneath the 3d orbitals?

Furthermore, if we shoot photons from an electromagnetic wave with a high enough frequency to a certain metal plate, that energy can excite electrons to the point that they free themselves from the atom they are a part of, and are given off as radiation from the metal plate, and the atom now has one less electron, becoming positively ionized unless it gains it back?

So let's take Nitrogen for instance. Its atomic number is 7. Therefore, in its ideal state, it has 7 electrons.

These electrons occupy Nitrogen in various energy levels and orbitals, constantly jumping up and down the energy levels whenever possible.

Two electrons occupy Nitrogens first n=1 energy level, at the 1s orbital. Shaped like a sphere.
Then five electrons occupy the second n=2 energy level. Electrons occupy orbitals by themselves first before willing to let others spin with them. Two electrons occupy the 2s orbital, shaped like a double sphere. Then the last three electrons each populate one of the three orientations of the 2p orbital (ml = -1, 0, 1).

And all these orbitals and energy levels are just ways to describe the positions of electrons surrounding the atom.

Which, by the way, electrons, can exhibit the properties of both a wave and a particle. It's been stated that as a virtual particle, electrons can appear and disappear out of nothing in an electron cloud, and can simultaneously occupy more than one spot, if the time given is brief enough, and can even take all possible paths simultaneously and arrive at one spot. As a standing wave around the nucleus, an electron very quickly spawns in and out of existence at points in the electron cloud, but these points TEND to be where the wave crest of the standing wave of the electron is maximum. Places where the wave crest are at a minimum are places the electron generally avoids.

So it's best to think of an electron as not necessarily a single particle moving in a circle around a nucleus, but as a thick blanket taking on a certain shape, composed completely of one electron quickly appearing and disappearing in and out of existence in spots around the nucleus of the atom.

Did I get ANY of this right? :l

One pressing question, though:
But what is the "valence shell" of an atom? Is it the outermost orbital? That wouldn't make sense since I hear that the number is either 2, 6, or 8....and orbitals hold only two electrons at once, and if you mean the different orientations of an orbital available, such as ml=-2,-1,0,1,2 for the d orbital, that's 10 electrons...and still doesn't make sense to me. So what is it?
 

Vayne Mechanics

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n = the principal quantum number
l = orbital angular momentum number
ml (where "l" is actually subscript) = magnetic quantum number

"All orbitals with the same value of n are in the same principal electronic shell or principal level, and all orbitals with the same n and l values are in the same subshell, or sublevel."

The value of n represents principal electronic shells or the principal level.
- the first principal shell consists of orbitals where n=1; the second principal shell consists of orbitals where n=2
- n relates to the energy and most probable distance of an electron from the nucleus.
- higher value of n = greater electron energy and the farther (on average) the electron is from the nucleus
- quantum number l represents shape of an orbital, while ml represents orientation of the orbital

the number of subshells in a principal shell is the same as the number of allowed values of the orbital angular momentum number, l.
- when n=1, l=0 (one subshell); when n=2, l=0,1 (two subshells); when n=3, l=0,1,2 (three subshells); etc.
- in otherwords, the number of subshells is equal to the principal quantum number
- the subshells all have a given name
- l=0, s subshell; l=1, p subshell; l=2, d subshell; l=3, f subshell

the number of orbitals in a shubshell is the same as the number of allowed values of ml for that particular value of l
- the total number of orbitals in a subshell = 2l+1
- l=0, ml=0 (one s orbital); l=1, ml= 0, +/-1 (three p orbitals); l = 2, ml = 0, +/-1, +/-2 (5 d orbitals); etc
 

Nostalgia

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Hi. I'm back with another physics problem, lol.

An object is placed on a horizontal turntable with a radius of 21.5 cm for a radius. The object's mass is 67.5 g. The coefficient of static friction between the two is .32. It asks to find the maximum angular velocity the object can have while remaining stationary relative to the turntable. It wants it in revolutions per second.

The second part says that the turntable starts from rest at t = 0, and it has a constant angular acceleration of 3.82 rad/s^2. It asks at what time will the object begin to slip.

Please help!
 

Banishing Blade

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Forces acting on the object:
[Perpendicular to surface] Normal and Gravity
[Parallel to surface] Friction.

We have a mass, and it's obvious the object isn't moving perpendicular to the surface, so we can go ahead and calculate the normal force.
Normal Force = - Gravity Force
Normal Force = - (mass * -9.8 m/s^2)
Normal Force = mass * 9.8 m/s^2

Now we can calculate the frictional force
Frictional Force = u * Normal Force [that "u" is actually mu, the coefficient of friction]

Now let's look at this statement: "maximum angular velocity... while remaining stationary relative to the turntable"
"Angular velocity" is talking about moving in circles. We [should] know that circular movement means to be accelerating.
"While remaining stationary relative to the turntable" means that if you consider only the frame of reference that is the turntable, the object isn't moving. This means that the only motion the object has is the same motion as the turn table.
"Maximum" means that the magnitude of acceleration caused by the angular velocity is the same as the magnitude of acceleration caused by the frictional force. However, a frictional force is always in the opposite direction of motion.

Centripetal Acceleration = - Frictional force
Centripetal Acceleration = - ( u * Normal Force)

But there is another equation for the centripetal acceleration, and that is:
Centripetal Acceleration = v^2 / r
Let's rearrange to get:
v = sqrt(a*r)

How does this help? Well, angular velocity is w= v / r. We have an equation for v, and we know r. So,
w = sqrt(a * r) / r

Now just put it all together!
 

Dogenzaka

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I've got a short logarithmic equation....I've got it wrong twice and I need to know wtf to do.

log9(7x) + log9(x^3) - 4 = 0

That's all log-base-9

soo....

you add the 4

log9(7x) + log9(x^3) = 4

Combine the two logs

log9(7x^4) = 4

At which point...I had two strategies, and both gave me different, wrong answers. Math :(

9^4 = 7x^4

9^4/7 = x^4

(9^4/7)^1/4 = (x^4)^1/4

5.533309 = x (but I needed a fraction and my calculator won't give it to me so this is wrong anyway)

:(

the other way I thought was

4 log9(7x) = 4

log9(7x) = 1

9^1 = 7x

I tried 9/7 = x....which was wrong.

I'm confused :<
 
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Eyesore

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The second way you worked it is wrong, because you said that the ^4 was attached to all of 7x, not just x.

let me do a bit of work here:

9^4 = 7x^4

(9^4)/7 = x^4

((9^4)/7)^1/4 = (x^4)^1/4

(9/(7^(1/4)) = x

(9 * 7^(3/4))/7 = x

so then 7^3 = 343

(9 * 343^(1/4))/7 = x

and that's your answer. it looks a little bit nicer like this though.

dogenlogprob.png


you can't use a negative value for x, can you? I didn't think so, but there is also the possibility of a negative x value as the answer.

I hope that helps out a little bit.
 

Coffee Lover

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The class is College Algebra and the lessons are:
1.Exponential Equations
2. Log Equations

I need someone's help. I didn't understand the first lesson and the second lesson was today. I was absent today because of personal reasons.

I just need someone to tell me how to do it. You don't have to teach me, though I'd be eternally grateful, but please show me the steps and explain a little about the steps.

First lesson problems:

6^x-3/4 = square root of 6

9^x = 1/cubic root of 3

e^1-5x = 793

7^0.3x = 813

Second lesson problems:

Instructions: Solve, reject any value of x that is not the domain of original log expressions. Then, use a calculator to find approx.

log3x=4

lnx = 2

log3(x-4) = -3

log3(x-5) + log3 (x+1) = 3

2log3(x+4) = log3 9 + 2
 

Orion

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6^x-3/4 = square root of 6
One way to help is to rephrase the equation/s.
6^(x-3/4)=6^(1/2), because any number to the power of a half is also it's square root.
Because they are of the same base, we can then equate the powers.
Thus, x-3/4=1/2
Bring the pronumerals and numbers to either side of the equation through inverse operations.
Thus x=1/2 + 3/4=1

It's pretty much always the best idea to get each side to the same base.

9^x = 1/cubic root of 3
Rephrase; (3^2)^x=3^(-1/3)
One of the index laws. (x^m)^n = x^(m*n)
Doing away with the common base, 2x=-1/3
Thus x=-1/3

e^1-5x = 793
To find the value of (1-5x) you take the natural logarithm of 793.
So, ln(793)=1-5x
Thus (-1/5)*(ln(793)-1)=x
Numerically, x=-1.135...

7^0.3x = 813
Quickly, 0.3x=ln(813)/ln(7)
Then x = [10*ln(813)]/[3*ln(7)]
x=11.47831......
 

Coffee Lover

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Okay, on the first problem on the first lesson, what was your answer? The book has 5 as the answer. I think I wrote it wrong. x-3 is over 4.

I have some questions. Like on the third one, why did you move the 1? Could you go in depth a little on the third one?

Also, you lost me on the fourth one.

Thank you, though. I really appreciate it. :)
 

Banishing Blade

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One way to help is to rephrase the equation/s.
6^(x-3/4)=6^(1/2), because any number to the power of a half is also it's square root.
Because they are of the same base, we can then equate the powers.
Thus, x-3/4=1/2
Bring the pronumerals and numbers to either side of the equation through inverse operations.
Thus x=1/2 + 3/4=1
Check your math. 1/2 + 3/4 is not 1, rather 5/4

If the actual question was (x-3)/4, the process is the same, but the answer is indeed 5.
(x-3)/4 = 1/2. First multiply both sides by 4, then add 3 to both sides.

Rephrase; (3^2)^x=3^(-1/3)
One of the index laws. (x^m)^n = x^(m*n)
Doing away with the common base, 2x=-1/3
Thus x=-1/3
Again, your math is wrong. If 2x=-1/3, x = -1/6.

To find the value of (1-5x) you take the natural logarithm of 793.
So, ln(793)=1-5x
Thus (-1/5)*(ln(793)-1)=x
Numerically, x=-1.135...
e^(1-5x) = 793, take the ln of both sides[/quote]
1-5x = ln(793)
-5x = ln(793) - 1
x = (-ln(793) - 1) / 5

Quickly, 0.3x=ln(813)/ln(7)
Then x = [10*ln(813)]/[3*ln(7)]
x=11.47831......
7^0.3x = 813, take ln of both sides
(0.3x)(ln7) = ln(813)
0.3x = ln(813)/ln7
x = ln(813) / (0.3 * ln7)
 

Orion

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Okay, on the first problem on the first lesson, what was your answer? The book has 5 as the answer. I think I wrote it wrong. x-3 is over 4.
My apologies, I thought it was x minus three quarters.

6^[(x-3)/4]=6^(1/2)
Thus, (x-3)/4=1/2
So x-3=4*(1/2)=2
Therefore x = 2+3 = 5. Yeah

I have some questions. Like on the third one, why did you move the 1? Could you go in depth a little on the third one?
I used inverse operations so to simplify the expression, and leave x on one side of the equation, with it's actual value on the other.
I've got M=1-5x, applying the inverse operation to +1 on each side gives
M-1 = (+)1-5x-1
The plus one and minus one cancel
M-1 = 5x
Dividing M-1 by 5 gives us x
(M-1)/5 = x

Also, you lost me on the fourth one.
I worked backwards from the result I got for solving this on my graphics calculator.
Based on that previous log law I described, you can rephrase the equation as the following, taking each number to its natural log.
ln(7)*(0.3x)=ln(813)
So,
0.3x=ln(813)/ln(7)
Or, (3/10)x=ln(813)/ln(7
So, taking the inverse operation of three tenths to the side with the logs gives us
10*ln(813) = x
3*ln(7)
 

Coffee Lover

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Check your math. 1/2 + 3/4 is not 1, rather 5/4

If the actual question was (x-3)/4, the process is the same, but the answer is indeed 5.
(x-3)/4 = 1/2. First multiply both sides by 4, then add 3 to both sides.

Okay, you mutiply both sides by 4, that crosses out the fours, and then you isolate the x by adding 3s to each side. I think I got that.
 

Coffee Lover

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If 2x=-1/3, x = -1/6.


e^(1-5x) = 793, take the ln of both sides
1-5x = ln(793)
-5x = ln(793) - 1
x = (-ln(793) - 1) / 5


7^0.3x = 813, take ln of both sides
(0.3x)(ln7) = ln(813)
0.3x = ln(813)/ln7
x = ln(813) / (0.3 * ln7)

Okay, on the third problem after the -1/6 problem, I think I'm beginning to understand. ln and the e's cross out, leaving 1-5x=ln793. The next thing you do is isolate the x, by gettind rid of the 1. 5x=ln793-1. Then, divide by 5 to get the x alone some more. There's the answer. Is that right, is that how it's done?
 

Orion

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Check your math. 1/2 + 3/4 is not 1, rather 5/4

If the actual question was (x-3)/4, the process is the same, but the answer is indeed 5.
(x-3)/4 = 1/2. First multiply both sides by 4, then add 3 to both sides.
My bad.
Again, your math is wrong. If 2x=-1/3, x = -1/6.
We all make simple mistakes.
e^(1-5x) = 793, take the ln of both sides
1-5x = ln(793)
-5x = ln(793) - 1
x = (-ln(793) - 1) / 5

7^0.3x = 813, take ln of both sides
(0.3x)(ln7) = ln(813)
0.3x = ln(813)/ln7
x = ln(813) / (0.3 * ln7)
And both of these are exactly the same as my own, expressed in a slightly different form.
ln and the e's cross out,
Yeah, in e^(ln(x)) the exponential/natural log part simplifies to 1, leaving you only with the x value.
 
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