questions that I got wrong Flashcards
A for ductile metal
D for rubber polymer
what happens when source of rays placed at focal length from lens
they have negative curvature cancelled out by positive curvature meaning you get parallel rays with the image at infinity distance
what happens when source of rays place at 2f from lens
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
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
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
what is direction of electric field the same as
the direction of the electric field is the direction of movement of a positive charge
we are looking for electric field between plates so we’re using E=v/d to work out EFS and then F = Eq
how to use flemmings left hand rule to explain why charge moves in circle
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
when does %uncertainty add up
adds up all the time
when something is squared you double it
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
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
stability of protons and neutron
decay equations where applicable
neutron is unstable so turns into proton
protons stable on their own but not in nucleus
n -> p + e- + antineutrino
what is the grad of flux linkage graph
gradient is negative emf
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
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 / T1/2
input data to get 2
Q2: A = λ N
A1 / A2
we worked out N = 2
A = log2 / 15 / log2 / 10 * 3 = 4/3
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
how to Δp / T from f = ma
f = ma = Δp
m(v-u) / Δt = mv - mu / Δt = mΔv = Δt = Δp / t
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
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
explain why T must be higher for -E / Kt = 15 then = 30
f = e-Ea / Kt
rewrite as 1 / eEa / 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
how to rewrite electrical potential energy equation
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
why does r need to be small between 2 protons to overcome repulsion
why is a high temp required
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
what does a solid line over some parts of lines of a TEM image mean
a line over some atoms means a dislocation, there’s no atoms there
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
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
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
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
binding energy released = 18MeV
total number of particles = 2.32 * 1020
what is one practical difficulty with obtaining this
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
So that negligible current passes through it meaning it does not affect the value it’s trying to measure
Suggest and explain one reason for the difference between the desntiy obtained in (iii) and the measured density of copper.
in a real solid there are small spaces between atoms so in a real solid has a smaller mass per unit volume
