ベンゼン

The time it took to make this is equivalent to the average blog post I make.

電子配置

Can you spot the error? With all the positive feedback this book has received, I'm surprised they made such a simple mistake. Despite that, it's still a very good book (Organic Chemistry 6th Edition, Paula Y. Bruice)

有機化学モデルキット

Today I had to go pick up one of these:

A standard organic chemistry model kit. It's made in Japan; that means it's some legit stuff.

It also comes with a sheet that lists all the contents, as well as what each part is supposed to represent:

Light blue - Hydrogen
Black - Carbon
Blue - Nitrogen
Red - Oxygen
Grey - m(?); no idea what that is

There's some black tube within the box, but I have no idea what it's for, and the sheet gives no information about it.

What else to do but to play with it, right?

Guess what molecule I made: Ethane

The real bummer; the kit costs $40. Although I'll probably still use it next semester for Organic Chemistry II, most likely I won't be using it more than a few times. Is it worth $40. Not really, but I guess I could sell it to an underclassman for $30 later.

物理学もう一度: 式シートない

Really getting tired of no formula sheets. Oh well, last midterm anyways:

Physics Review Day 2

Circular Loop

A piece of wire is formed into a circular loop of radius 23 cm. The loop has a resistance of 141 Ω. A magnetic field of 0.9T is applied perpendicular to the plane of the loop and then increased at a constant rate by a factor of 2.2 in 17 s. Calculate the magnitude of the induced emf in the loop during that time.

We are informed that there is a change in magnetic field; this indicates flux, and we apply Faraday's Law:

|ε| = ΔΦ / ΔT where Φ = BAcosθ and cos0 = 1

We don't know B exactly, but we are given a constant factor of increase; therefore we can treat B as 1.

|ε| = (2.2B - B)[π(23.0 cm * 1 m / 100 m)²] / 17 s
|ε| = (1.2 T)[π(0.23 m)²] / 17 s
|ε| = 1.06×10^-2 V 

Calculate the current induced in the loop during that time.

Simple application of Ohm's Law:

I = ε / R
I = (1.06×10^-2 V ) / 141 Ω
I = 7.49×10^-5 A 

Calculate the average induced emf when the magnetic field is constant at 1.98 T while the loop is pulled horizontally out of the magnetic field region in 6.5 s.


Back to Faraday's Law. This time, the area changes as opposed to the magnetic field, but the same concept applies:

|ε| = ΔΦ / ΔT where Φ = BAcosθ and cos90 = 1
|ε| = (1.98 T)[0 -  π(23.0 cm * 1 m / 100 m)²] / 6.5 s
|ε| = 5.06×10^-2 V

Rotating Square Coil

What is the peak emf produced by a 74-turn square coil (of side l = 26.0 cm, as shown in the diagram below,) rotating on an axis with a frequency of 37.0 Hz in a uniform magnetic field of 0.695 T perpendicular to the coil's axis of rotation? 

Use the equation for a rotating coil (AKA an electric generator), considering the fact that peak emf is when the magnetic field is perpendicular to the coil:

ε = NABωsin(ωt) where ω = 2πf
ε = (74)(26.0 cm * 1 m / 100 m)² (0.695 T) * (2π*37.0) sin90
ε = 8.08×10² V 

Coiled Wire

You have a piece of thin wire that is 15.5m long, a constant uniform magnetic field of 0.155T, and a device that can rotate a coil at a fixed frequency of 87.5Hz. What is the radius of a circular coil made from this length of wire that will produce an AC e.m.f of maximum voltage 116V? (Neglect the amount of wire used in the connections.)

Once again, we use the rotating coil equation:

ε = NABωsin(ωt) where ω = 2πf

However, we don't know N, the number of turns. However, we do know the length of the wire L;
we can equate and find N:


L = 2πrN (Since the length is equal to the circumference times the number of turns)
N = L / 2πr


We can now plug in the second equation into the first, and solve:

ε = NABωsin(ωt) where ω = 2πf
ε = (L / 2πr)ABωsin(ωt)
116 V = (15.5 m / 2πr)(πr²)(0.155T)(2π*87.5 Hz)
r = 116 V / (15.5 m * π * 0.155 T * 87.5 Hz)
r =  1.76×10^-1 m 

Wavetrains

If light is emitted from an atom in little wavetrains, each lasting up to 2.90 × 10^-8 s, how long, at most, is such a disturbance in space? 

Wavelength is equal to the speed of light multiplied by the time:

L = ct = (3.0E+08 ms^-1)(2.90E-8 s)
L = 8.70 m 

If we approximate the wavelength as 510 nm, roughly how many waves long is the train? 


The number of waves is simply the length divided by the wavelength:

# Waves = L / λ = (8.70 m) / (510 nm * 1.0E-9 m / 1 nm)
# Waves = 1.71×10⁷ 

Laser Pulses

A laser that emits pulses of UV lasting 2.25 ns has a beam diameter of 2.25 mm. If each burst contains an energy of 2.70 J, what is the length in space of each pulse?

The length in space is simply the speed multiplied by the time (Kinematics):

l = ct = (3.0E+08) * (2.25 ns * 1.0E-9 s / 1 ns)
l = 6.75×10^-1 m

What is the average energy per unit volume (the energy density) in one of these pulses?  
 

A laser is a cylindrical beam, so the volume of that beam is the length in space:

V = πr² l 

V = π[(2.25 mm * 1.0 m / 1000 mm) / 2]² * 6.75×10^-1 m
V = 4.77E-3 m³

We can now find energy density by dividing the energy by the volume:


= 2.70 J / 4.77E-3 m³
= 1.01×10⁶ J/m^3 

Irradiance of a Candle


The irradiance 1 m from a candle flame is just about 4.65 × 10^-3 W/m². How much energy will arrive in 1.25 s on a disk having a 3.95- cm² area held as close to perpendicular as possible 1 m from the flame? 

Energy is related to irradiance by the following formula:

E = IAt = (4.65 × 10^-3 W/m²)(3.95 cm² * 1.0 m² / 1.0E+4 cm²)(1.25 s)
E = 2.30×10^-6 J 

Physics Review Day 1

Magnetic Force on an Electron

At SFU the magnetic field due to the earth is at 16.2^° to the vertical and has a magnitude of 6.34E-5T. An electron moves straight down at 2.98E+5ms⁻¹. Find the ratio of the magnitude of the magnetic force on the electron to its weight, mg.

Simple question; just plug in numbers

|Fm| / mg = qvBsinθ / mg

q = -1.6E-19 C
m = 9.1E-31 kg
g = 9.81 ms⁻²

= | (-1.6E-19 C)(2.98E+5ms⁻¹)(6.34E-5T)sin(16.2) | / (9.1E-31 kg)(9.81 m·s⁻²)
= 9.46×10¹⁰ (No units)

Charged Cork Ball

A cork ball carrying charge q has a mass of 2.10 g and is set in motion perpendicular to a uniform magnetic field of 0.90 T. What is the magnitude of q if its direction of motion changes by 5.0° in 1.0 s? 


Since the ball is set in motion perpendicular to a uniform magnetic field, it will undergo centripetal motion. In other words, centripetal motion is achieved when the centripetal force equals the magnetic force.

Fc = Fm
mv² / r = qvBsinθ
q = mv / rBsinθ

To find v, we use the fact that the ball's direction of motion changes by 5.0° in 1.0 s.

ω = v/r where ω = Δθ / Δt = (5*π/180) / 1.0 s (Must convert degrees to radians for angular motion)
ω =  0.0873 s⁻¹

Finally:
q = mv / rBsinθ = (2.10 g * 1 kg / 1000 g)(0.0873 s⁻¹) / (0.90 T)
q = 2.04×10^-4 C

Proton Orbit

A proton is sailing through the outer region of the Sun at a speed of 0.275 c. It traverses a locally uniform magnetic field of 0.345 T at an angle of 29.8^°. What is the radius of its helical orbit? (Hint: v|| and v⊥ can be considered separately.)

Only v⊥ is influenced by B, so we can completely ignore v||. As such, it becomes a simple plug in values question:


The helical movement is defined the same as circular motion for v⊥:

mv² / r = qvBsinθ
r = mv / qBsinθ

q =  1.6E-19 C
m = 1.67E-27 kg
c = 3.0E+8 m·s⁻¹

We have enough information to solve for r now:


r = mv / qBsinθ = (1.67E-27 kg)(0.275*3.0E+8 ms⁻¹) / (1.6E-19 C)(0.345 T)sin(29.8)
r = 1.24 m 

Current in Parallel Wires

An infinitely long wire lies along the z-axis and carries a current of I=2.95 A in the positive z-direction. A second infinitely long wire is parallel to the z-axis and lies along the plane x=+12.0cm. Find the current in the second wire if the net magnetic field at x=+7.50cm is zero.

Long wire indicates this formula:

B = µ₀I / 2πr
µ₀ = 4π×10⁻⁷ N·A⁻²

Find B of the first wire, using the distance at which B of the second wire is zero:

B = µ₀I / 2πr
B = (4π×10⁻⁷ N·A⁻²)(2.95 A) / (2π)(7.50 cm * 1 m / 100 cm) = 7.87E-6 T

Now we can use this B to find I in the second wire (Using the premise that B1 = B2) by using the distance d - r where d is the separation of the two wires and r is the distance at which B2 = 0

I = B*2πr / µ₀
I = (7.87E-6 T)(2π*(12.0 cm - 7.50 cm) * 1 m / 100 cm)) / (4π×10⁻⁷ N·A⁻²)
I = 1.77 A

Parallel Wires

The diagram above depicts two long horizontal straight parallel wires that are a distance d=22.30cm apart and each carries a current of 3.90A in the same direction, out of the page. What is the magnitude of the magnetic field at a point that is a perpendicular distance r=27.04cm from both wires?

A simplified vector question. Since both wires carry the same amount of current in the same direction, by vector addition, the total B will be 2Bcosθ. The vertical component of each vector cancels, and the horizontal component adds (Since one B points diagonally northwest while the other points diagonally southwest).

The angle θ is found by trigonometry:

arcsin((22.3/2) / 27.04) = 24.35

B = µ₀I / 2πr
B = (4π×10⁻⁷ N·A⁻²)(3.90 A) / (2π)(27.04 cm * 1 m / 100 cm) = 2.88E-6 T

By vector addition:

Btotal = 2Bcosθ = 2(2.88E-6 T)cos(24.35)
Btotal = 5.26×10^-6 T 

What is the direction of the resultant magnetic field at point P? Express its direction, θ, in degrees as follows: a vector pointing left corresponds to 0 degrees, a vector pointing up corresponds to 90 degrees, a vector pointing right corresponds to 180 degrees, and so on.


In the first part, it was determined that the vertical components of the vectors canceled; that means the resultant magnetic field travels parallel to the normal; hence the answer is 0 degrees.


Hydrogen Atom


A rather simplistic model of the hydrogen atom has a single electron revolving around a nuclear proton with an orbital radius of 5.20E-9 m at a speed of 4.00E+6 m/s. Determine the magnetic field at the proton due to the electron.

The electron orbits the proton; this is analogous to the magnetic field at the center of a loop:

B = µ₀I / 2r

To find I, recall that:

I = q/T ; current is defined as charge over time. q is proton charge: 1.6E-19 C

To find time T, which is really the period due to circular motion:

T = 2π / ω and ω = v / r so
T = 2π / (v / r) = 2π / (4.00E+6 ms⁻¹ / 5.20E-9 m)
T = 8.17E-15 s

Now find I:

I = q/T = (1.6E-19 C)(8.17E-15 s) = 1.31E-33 C

Finally, find B:

B = µ₀I / 2r
B = (4π×10⁻⁷ N·A⁻²)(1.31E-33 C) / 2(5.20E-9 m)
B = 2.37×10^-3 T 

Small Solenoid

A small diameter, 11- cm long, solenoid has 283 turns and is connected in series with a resistor of 137 Ω. Calculate the magnitude of the magnetic field in the middle of the solenoid when a voltage of 55 V  is applied to the circuit.

Simply use the solenoid equation:

B = µ₀*N*I / L and I = V / R
B = (4π×10⁻⁷ N·A⁻²)(283)(55 V / 137 Ω) / (11 cm * 1 m / 100 cm)
B = 1.30E-3 T

Power of the Sun

Part 1:

Given:

Flux: 137 W
Area: 0.1 m^2 (Area of solar panel which captures the sun's rays)
d = 1.5 x 10^11 m (Distance from the surface of the sun to the solar panel)

Find the total output power of the sun.

The general formula for power in terms of flux is:

P = SA * Φm / A

In words, the power equals the surface area times the magnetic flux density.

We have enough information to find those unknowns:

Φm / A = 137 W / 0.1 m^2 = 1370 W / m^2


SA: Since the sun is a sphere, we use the surface area formula of a sphere

SA = 4πd^2 = 4π(1.5 x 10^11 m)^2 = 1.88 x 10^23 m^2


Finally:

P = SA * Φm / A = 1370 W / m^2 * 1.88 x 10^23 m^2 = 3.9 x 10^26 W


Part 2:


Given:


λ = 500 nm

Find the number of protons per second.

Energy of a photon is defined as:

E = hf = h(c/λ)
 
Where E is the Power found in part 1 and h is Planck's Constant (6.626 x 10^-34)

Recall that frequency and wavelength are related by the speed of light: c = λf

That's the energy of 1 photon; however we are interested in the number of photons per second. Therefore we add a constant n in our equation and solve appropriately:

E = nh(c/λ)
n = (E*λ)/(h*c) = (3.9 x 10^26 W * 500 x 10^-9 m) / (6.626 x 10^-34 * 3.0 x 10^8 m/s)
n =  9.81 x 10^44

Writing about Writing

Me and my flowchart fetish;

磁束 / Magnetic Flux

Two problems regarding Magnetic Flux.

1) Find the induced current of the loop of wire below:

Given:

r = 10.0 cm
B = From 1 T to 5T
N = 15
Δt = 10s
R = 10 Ω

This question is easily answered using Faraday's Law and Ohm's Law.

Faraday's Law:


ε = -NΔΦ_m / Δt
Φm = BAcosθ

ε = -(15*(5-1)πr^2cos30) / 10
ε = -15π(0.10)^2 * √3/2 * 4/10
ε = 0.0290757133 V

Ohm's Law:

ε = IR
I = ε/R

I = 0.0290757133 / 10
I = 0.0029075713 A

2) Find the induced force on the orange bar.

 l = Length
B = Magnetic Field

No numbers given, so we are finding a general solution. Once again, use Faraday's Law and Ohm's Law:

a) ε = IR

b) ε = ΔΦ_m / Δt = BΔA / Δt = BlΔx / Δt = Blv


Equate a and b:

IR = Blv
I = Blv / R


Finally:


F = IlB = Blv / R * I * B
F = B^2l^2v / R

Math 155 Practice Midterm

Did this for practice; real one was of course harder.

Underworld 4 - Pictures

Just some pictures around campus; they'll probably be setting up for a while.

もう一度式シートない

But unlike last time, I needed it really badly. Maybe it would have been less awful if I did have one. Doubt it though. Suddenly, Physics has lost its spot as my most hated subject, with Statistics taking the crown now.

Underworld 4 at SFU

As an interesting aside, Underworld 4 is being filmed at SFU. It started today although all of that was just setting up:

統計情報を勉強する 二日目

You know the drill.

統計情報を勉強する 初日目

This week will be all self study of Statistics. Will post example questions and solutions as a way of studying.

式シートない

No formula sheet, Brain only, Final Destination:

キルヒホッフの法則 / Kirchoff's Laws

The diagram below shows a circuit where; R₁=6.00 Ω, R₂=2.00 Ω, R₃=3.00 Ω, V₁=2.00 V, V₂=9.00 V, and V₃=15.0 V. Find I₁, I₂, and I₃.

Problems like these aren't really hard, you just have to do the math correctly (Which I tend to mess up on).

Start with the Junction Rule, and pick a junction to define your current equation. Here we pick the top junction.

The Junction Rule states that all current entering a junction must equal all current exiting the junction. In order words:


I₁ + I₂ + I₃ = 0 (Equation 1)

It doesn't really make sense that all 3 currents go in the same direction, but the Junction Rule works regardless. Since they give you directions in the problem, we have to work with them, even if they seem counter-intuitive.

Now we use the Loop Rule (The sum of all potential changes is zero) to derive two more equations.

We start at the red dot and move counterclockwise, taking note of all potential changes. From the red dot to the orange dot, there is a potential increase, since we are moving through a resistor and we are opposing the current I_1.  From the orange dot to the yellow dot, there is a potential drop, since we are moving from the positive terminal of a battery to the negative terminal. From the yellow dot to the green dot, there is a potential increase, since we are moving from a negative terminal of a battery to the positive terminal. Finally, from the green dot to the blue dot, there is a potential drop, since we are moving through a resistor and we are moving with the current I_2.

In equation form then:

I₁R₁ – V₁ + V₂ – I₂R₂ = 0 (Equation 2)

Similarly, we do the same procedure again, with the other half of the circuit:

Using the same logic as before, we derive the equation:

I₂R₂ – V₂ + V₃ – I₃R₃ = 0 (Equation 3)

Notice that all 3 equations equal 0. This means we can equate them and solve for a particular variable, then use that variable to solve the rest.

Also notice that the variable I₂ is in each equation. Let us rearrange our equations so that they each have I₂ on one side:

I₁ + I₂ + I₃ = 0 (Equation 1)

-I₂ = I₁ + I₃

I₁R₁ – V₁ + V₂ – I₂R₂ = 0 (Equation 2)

I₁ = (V₁ – V₂ + I₂R₂) / R₁

I₂R₂ – V₂ + V₃ – I₃R₃ = 0 (Equation 3)

I₃ = (I₂R₂ – V₂ + V₃) / R₃

Finally, plug equation 2 and 3 into equation 1 and solve for I₂:

-I₂ = (V₁ – V₂ + I₂R₂) / R₁ + (I₂R₂ – V₂ + V₃) / R₁
-I₂ = 2/6 - 9/6 + (2/6)I₂ + (2/3)I₂ - 9/3 + 15/3
I₂ = -5/12 A (-0.417 A) 

We are almost done the problem. With I₂, we can find I₁ or I_3. Let's find I₃ first:

 I₃ = (I₂R₂ – V₂ + V₃) / R₁
 I₃ = [(-5/12)(2)]/6 - 9/6 + 15/6
 I₃ = 31/18 A (1.72 A)


All that's left is to find . We can do the same procedure as above, or we can use equation 1 and solve for the remaining variable. We will do the latter:


I₁ + I₂ + I₃ = 0
I₁ = I₂ - I₃ = 0
I₁ = -(-0.417) - 1.72
I₁ = -1.303 A


One final thing to note is that there is no such thing as a negative current; the negative signs merely denote that our initial guess in direction was incorrect. So a properly drawn diagram would therefore be: