questions that I got wrong Flashcards

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1
Q
A

A for ductile metal

D for rubber polymer

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2
Q

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

A

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

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3
Q

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

A

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

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4
Q
A

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

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5
Q
A

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

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6
Q

what is direction of electric field the same as

A

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

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7
Q
A

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

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8
Q

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

A

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

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9
Q

when does %uncertainty add up

A

adds up all the time

when something is squared you double it

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10
Q
A

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

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11
Q
A

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

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12
Q

stability of protons and neutron

decay equations where applicable

A

neutron is unstable so turns into proton

protons stable on their own but not in nucleus

n -> p + e- + antineutrino

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13
Q

what is the grad of flux linkage graph

A

gradient is negative emf

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14
Q
A

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

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15
Q
A

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

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16
Q
A

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

17
Q

how to Δp / T from f = ma

A

f = ma = Δp

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

18
Q
A

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

19
Q
A

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

20
Q

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

A

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

21
Q

how to rewrite electrical potential energy equation

A

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

22
Q

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

why is a high temp required

A

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

23
Q

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

A

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

24
Q
A

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

25
Q
A

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

26
Q
A

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

27
Q
A

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

28
Q

binding energy released = 18MeV

total number of particles = 2.32 * 1020

what is one practical difficulty with obtaining this

A

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

29
Q
A

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

30
Q

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

A

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

31
Q
A