questions that I got wrong

Question 1

Answer 1

A for ductile metal

D for rubber polymer

Question 2

what happens when source of rays placed at focal length from lens

Answer 2

they have negative curvature cancelled out by positive curvature meaning you get parallel rays with the image at infinity distance

Question 3

what happens when source of rays place at 2f from lens

Answer 3

as source moves from f to 2f, image is moving back from infinity so image which is huge at this point is getting smaller until at 2f image size = object size

Question 4

Answer 4

between f and 2f image is really big which is why source placed at distance from lens in cinema projectors

so between f and 2f

Question 5

Answer 5

you need avagadro constant as you have free electrons from each atom, molar mass, density

you can find moles from mass / molar mass

then moles * avagadro constant = total number of electrons

Question 6

what is direction of electric field the same as

Answer 6

the direction of the electric field is the direction of movement of a positive charge

Question 7

Answer 7

we are looking for electric field between plates so we’re using E=v/d to work out EFS and then F = Eq

Question 8

how to use flemmings left hand rule to explain why charge moves in circle

Answer 8

thumb is direction of force

index is direction of field

middle is direction of charge

direction of force felt is at 90 degrees to direction of field so 2 forces make at move at 45 degrees, so it moves in a circle

Question 9

when does %uncertainty add up

Answer 9

adds up all the time

when something is squared you double it

Question 10

Answer 10

to find x think of conservation of energy

they gave us m and g so we can find h(x)

1/2kx² = mgx

inout and solve for 4

Question 11

Answer 11

pv = 1/3nmc² (c is mean square speed)

rearrange for v = (nm / 3p) c²

brackets is constant

if you doubled the mean square speed, v doubles

Question 12

stability of protons and neutron

decay equations where applicable

Answer 12

neutron is unstable so turns into proton

protons stable on their own but not in nucleus

n -> p + e- + antineutrino

Question 13

what is the grad of flux linkage graph

Answer 13

gradient is negative emf

Question 14

Answer 14

gradient of flux linkage is negative emf

0 to t₀ there’s a constant flux induced so a constant emf is induced so nothing is changing so grad doesn’t change

t₀ to time no change in grad so no emf induced

answer is d

Question 15

Answer 15

2 sources of equal mass so N0 is constant

to find ratio N = N₀e^-λ ^{* t} / N₀e^-λ ^{* t}​

N0 is constant so e^-λ ^{* t}​ / e^-λ ^{* t}​

can get λ from Log2 / T_1/2

input data to get 2

Q2: A = λ N

A1 / A2

we worked out N = 2

A = log2 / 15 / log2 / 10 * 3 = 4/3

Question 16

Answer 16

P = f/a

f = Δmomentum = m * Δ velocity

particles absorbed so Δv = speed we’ve been given

mΔv = Number of particles * mass of each * v = mΔv = Δp

we need Δp over one second so Δp / 1 = Δp = f

pressure = f / a

Question 17

how to Δp / T from f = ma

Answer 17

f = ma = Δp

m(v-u) / Δt = mv - mu / Δt = mΔv = Δt = Δp / t

Question 18

Answer 18

remember that springs in series (where they’re on top of each other) have a k that gets smaller than if they were separate

this means they have a longer oscillation

A = 1/2k B = 1/3K C = 2k d = 3k

B will therefore have the longest oscillation and so the longest time period, f = 1 / t so then it will have the longest frequency

for the second question, they switched around T = 2pi root m / k

but you can see that k has been replaced by 2k so its c which also has 2k

Question 19

Answer 19

since activation energy is constant for any process so pick any point

f = 0.2 at 800k

0.2 = e^{-E / 800k}

solve for Ea

Question 20

explain why T must be higher for -E / Kt = 15 then = 30

Answer 20

f = e^{-Ea / Kt}

rewrite as 1 / e^{Ea / Kt} = 1: e^{-Ea / Kt}

so if you put in -E / Kt = 15 you have 1 : e¹⁵

so one particle for every e¹⁵ activates

but when -E / Kt = 30 you have 1 : e^{<span>30</span>}

so one particle for every e³⁰ activates

so T must be higher in e¹⁵ for this to happen because e¹⁵ << e³⁰

Question 21

how to rewrite electrical potential energy equation

Answer 21

EPE = kQq / r

well Q and q are the same so write as charge² = e²

k = 1 / 4pi epsilon

and r = r

so EPE = e² / 4pi epsilon * r

Question 22

why does r need to be small between 2 protons to overcome repulsion

why is a high temp required

Answer 22

the strong force means protons can grab each other overcoming repulsion but for this r has to be very small as gluons have a very short range

to get r small you need a high kinetic energy Ke = KT in this case so you need a high temp for a high kinetic energy

Question 23

what does a solid line over some parts of lines of a TEM image mean

Answer 23

a line over some atoms means a dislocation, there’s no atoms there

Question 24

Answer 24

phase difference between maxima is one wavelength, 1 wavelength is the equivalent to 2pi and 360 degrees

here however the phase difference is pi, so every other arrow points in one direction, one between is opposite

Question 25

Answer 25

power in watts is equivalent to J s^-1 so P = 3.5 \* 10^-3 J so in one second 3.5 \* 10^-3 J is given out E = hc / wavelength = energy of photon divide energy given out in one second by the energy of a photon to find out number of photons

Question 26

Answer 26

m = 0.12 t = 0.04 v = 10(in the up direction) but ke conserved so v down is 10 as well impulse = f \* change in time = change in momentum = m\*change in velocity so change in velocity = 10 - - 10 = 20 ms^-1 f = mv / t 0.12 \* 20 / 0.04 = 60N

Question 27

Answer 27

1mm = 10^-3m 1mm³ = 10^-9m³ v = 4.2 \* 10^-9m³ we have a ratio of 1:1 as there's equal amounts of 2H:3H we need the total number of particles, we need average molar mass 2 + 3 / 2 = 2.5g d \* v = mass / molar mass = moles \* avagadro constant = total number of particles E = nkt = total number of particles \* boltz constant \* temp = 1.3Mjoules

Question 28

binding energy released = 18MeV total number of particles = 2.32 \* 10²⁰ what is one practical difficulty with obtaining this

Answer 28

fusion takes place between 2 particles so half the number of particles to get how many reactions happen in one reaction 18MeV released so in this many particles: (0.5 \* 18MeV \* 2.32 \* 10²⁰ ) \* 1.6 \* 10^-19 = 334MJ a lot more energy produced than used need very highy energy lasers multiple lasers that all need to hit at the same time

Question 29

Answer 29

So that negligible current passes through it meaning it does not affect the value it’s trying to measure

Question 30

Suggest and explain one reason for the difference between the desntiy obtained in (iii) and the measured density of copper.

Answer 30

in a real solid there are small spaces between atoms so in a real solid has a smaller mass per unit volume

Question 31

Answer 31