Notes - Fisika - Finals

I should take a walk for a bit

Alternating Current

ƒ=1/T

y=A*sin(ω*t)

E=(E_m)*sin(ω*t)=N⋅B⋅A⋅ω⋅sin(ω*t)

ƒ is frequency

T is period of the current

A is amplitude

ω is angular frequency

(V_effective)=(V_max)/√(,2)

(I_effective)=(I_max)/√(,2)

Oscilloscopes measure maximum voltage (V_max), voltmeters measure effective voltage (V_E). Effective voltage is basically the amount of DC voltage needed to match the AC voltage.

How do you read an oscilloscope?

(V_max)=(s_y)⋅A

T=(s_x)⋅λ

The vertical axis uses a scale based on volts per centimeter.

The horizontal axis uses a scale based on miliseconds per centimeter.

R-L-C

(X_L)=ω*L=2*π*ƒ*L

(X_C)=1/(ω*C)=1/(2*π*ƒ*C)

V=√(,(V_R)2+((V_L)-(V_C))2)

I⋅Z=√(,(I*R)2+(I*(X_L)-I*(X_C))2)

Z=√(,R2+((X_L)-(X_C))2)

I=V/Z=(V_L)/(X_L)=(V_R)/R=(V_C)/(X_C)

tan(θ)=((X_L)-(X_C))/R

(X_L)>(X_C) → Circuit is inductive

(X_C)>(X_L) → Circuit is capacitive

(X_L)=(X_C) → Resonance with ƒ

ƒ=1/(2*π√(,L*C))

I(t)=(I_max)*sin(ω*t)

V(t)=(V_max)*sin(ω*t+θ)

P=I2*R

Z is impedance, basically the total opposing resistance.

R is resistance from a resistor

(X_L) is inductive resistance

(X_C) is capacitive resistance

(V_L) is voltage across the inductor

(V_C) is voltage across the capacitor

P is dissipation, only taking into account R

In a phasor diagram, (V_R) and I have an angle of 0 (right). (V_L) has an angle of 90 (up) and (V_C) has an angle of 270 (down). This means that (V_L) and (V_C) have opposing directions.

Highlighted Problems

Problem 1

∴(V_max)=3⋅80=240

T=8⋅2.5=20

∴ƒ=1/(20⋅10)=1000/20=50

(B)

Problem 2

Voltmeters measure effective voltage.

(V_max)=4⋅10=40

∴(V_effective)=40/√(,2)=20√(,2)

T=2⋅8=16

∴ƒ=1/(16⋅10)=1000/16=125/2=62.5

Drawing is easy, so I won't showcase it. The vertical scale is basically how many squares a peak or a valley takes up, and the horizontal scale is how many squares 1 wave takes up.

Problem 3

(V_R)=80

(V_max)=100

80=(I_max)⋅160

(I_max)=0.5

(V_RL)=100=√(,80+(V_L)2)

(V_L)=√(,100-80)=60=(X_L)⋅I

(X_L)=60/0.5=120

Z=100=√(,80+(Δ(V))2)

100-80=(Δ(V))2

Δ(V)=√(,100-80)=60

Δ(V)=(V_C)-(V_L)=60

You can swap the order of (V_C) and (V_L) to find a nonzero solution.

(V_C)=60+60=120

120=0.5⋅(X_C)

(X_C)=240

Problem 4

Well, let's just get right to it.

V(t)=100√(,2)*sin(100*t)

(X_L)=2⋅100=200

(X_C)=1/(10⋅100⋅10)=1000 (it's supposed to be micro faradays)

tan(θ)=(200-1000)/600=-4/3

∴θ=-53 (a)

Rangkaian bersifat kapasitif karena (X_C)>(X_L) (b)

Z=√(,600+(200-1000)2)=1000

100√(,2)=1000⋅(I_max)

(I_max)=0.1√(,2)

I(t)=0.1√(,2)*sin(100*t+53) (c)

(V_R)=600⋅0.1=60

(V_L)=200⋅0.1=20

(V_C)=1000⋅0.1=100

(d)

P=1/10⋅600=6 (e)

Black Body Radiation

Use a calculator.

P=e⋅σ⋅A⋅T4

I=P/(4*π*r2)

P is radiation, joules per second, aka watts.

e is coefficient of emissivity

σ is the Stefan Boltzmann constant, 5.67⋅10

A is the area

T is temperature in Kelvin

I is intensity, with r being distance

Highlighted Problems

Problem 1

A=(10⋅10)2=1⋅10⋅2

P=5.67⋅10⋅8/10⋅1⋅10⋅2⋅(727+273)4=2⋅453.6

Problem 2

The given variables are R, r, I

I=P/(4*π*r2)

P=4*I*π*r2=σ*A*T4

4*I*π*r2=σ⋅(4*π*R2)⋅T4

I*r2=σ*R2*T4

T4=(I*r2)/(σ*R2)
=(1400⋅(1.5⋅10)2)/(5.67⋅10⋅(7⋅10)2)
≈1.13⋅10

T≈5800

Photoelectric Effect

E=h⋅ƒ=(h⋅c)/λ

(KE_max)=e⋅(V_s)

(KE_max)=E-(W_s)

(W_s)=h⋅(ƒ_s)=(h⋅c)/(λ_s)

E is energy from photon, joule or eV

h is 6.62⋅10

ƒ is light frequency

(KE_max) is maximum kinetic energy

e is 1.6⋅10

(V_s) is stopping potential

(W_s) is work function / energi ambang

(ƒ_s) is work frequency / frekuensi minimum / frekuensi ambang

c is the speed of light

1 eV is 1.6⋅10 joule

Highlighted Problems

Problem 1

(ƒ_s)=2.30

2.30⋅1.6⋅10=(6.62⋅10⋅3⋅10)/(λ_s)

(λ_s)≈5.39⋅10 (a)

E=(6.62⋅10⋅3⋅10)/(150⋅10)=1.324⋅10

(KE_max)=1.324⋅10-2.30⋅1.6⋅10=9.56⋅10 (joules) (b)

9.56⋅10=1.6⋅10⋅(V_s)

(V_s)=5.975 (c)

Problem 2

(W_s)=1.6=2.56⋅10 (a)

2.56⋅10=(6.6⋅10⋅3⋅10)/(λ_s)

(λ_s)=7.73⋅10=773 (b)

(0.4⋅1.6⋅10)+(2.56⋅10)=6.6⋅10⋅ƒ

ƒ=4.84⋅10 (c)

Compton Effect

Photons scatter off electrons.

Δ(λ)=(λ^′)-λ=h/(m⋅c)⋅(1-cos(θ))

E=(h⋅c)/λ

(E_k)=E-(E^′)

At this point the symbols should be obvious, but here they are.

Δ(λ) is change in wavelength

h is 6.6⋅10

c is the speed of light

m is mass of an electron, 9.1⋅10

λ and (λ^′) are the wavelengths before and after scattering

θ is the angle of scattering

E is an energy of a photon

(E_k) is the amount of kinetic energy given to the electron

Highlighted Problems

Problem 1

λ=3.5⋅10⋅10=3.5⋅10

cos(37)≈4/5

Δ(λ)=(6.6⋅10)/(9.1⋅10⋅3⋅10)⋅(1-4/5)=4.83⋅10

(λ^′)=4.83⋅10+3.5⋅10=3.54⋅10=0.0354 (a)

(E^′)=(3⋅10⋅6.6⋅10)/(3.54⋅10)=5.59⋅10 (b)

E=(3⋅10⋅6.6⋅10)/(3.5⋅10)=5.65⋅10

(E_k)=5.65⋅10-5.59⋅10=0.06⋅10 (c)

Problem 2

λ=1.000⋅10

Δ(λ)=(6.6⋅10)/(9.1⋅10⋅3⋅10)⋅1=2.41⋅10=0.00241 (a)

(λ^′)=1⋅10+2.41⋅10=1.024⋅10

(E_k)=3⋅10⋅6.6⋅10⋅(1/(1.000⋅10)-1/(1.024⋅10))
=4.640⋅10

4.64⋅10=(m⋅v2)/2
=1/2⋅9.1⋅10⋅v2

v=10099096=1.0099096⋅10