Zac T. A Level Chemistry tutor, A Level Further Mathematics  tutor, A...

Zac T.

Currently unavailable: until 03/10/2016

Degree: Natural Sciences (Bachelors) - Cambridge University

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About me

About me

I am a student at the University of Cambridge where I study Natural Sciences, a course which allows me to maintain my interest in multiple areas of science and maths. I take modules in Chemistry, Physics, Earth Science and Maths and am able and willing to help in any of those areas.

I have found from being taught in supervisions at university that the best thing you can take from the experience is an appreciation for the subject so I aim to try to convey the joy I find in the sciences and hope that some of my enthusiasm rubs off on you. 

Supervisions

I have had some tutoring experience before so I know that an understanding of the subject is far more valuable than simply knowing the textbook answers so this will be my focus. However, I will defer to you for the content you wish to cover and will be happy to help with exam questions if that is what you are looking for.

By the end of a supervision, I will ensure that you are confident enough in your knowledge of the topic that you can explain it to me via whichever method you find most useful (be it diagrams, words or maths).

What next

I you have any questions or think I can help you, feel free to contact me or book a session, both through the website. It will help if I know your exam board and what I can help with.

I look forward to helping you.

Subjects offered

SubjectLevelMy prices
Chemistry A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Biology GCSE £18 /hr
Chemistry GCSE £18 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Science GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
ChemistryA-LevelA*
PhysicsA-LevelA*
BiologyA-LevelA
AEA MathsUni Admissions TestA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

03/10/2016

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Questions Zac has answered

Find the shortest distance between the line L: x=1+t, y=1+2t, z=1-t and the point A: (2,3,4)

First, write the line in vector form r=(1 1 1) + t(1 2 -1). Consider a point P on the line such that the line connecting P and A is perpendicular to L. The vector P->A is (2 3 4) - (1 1 1) - t(1 2 -1) = (1 2 3) - t(1 2 -1). To make P->A perpendicular to L, the dot product of P->A and the direc...

First, write the line in vector form r=(1 1 1) + t(1 2 -1). Consider a point P on the line such that the line connecting P and A is perpendicular to L. The vector P->A is (2 3 4) - (1 1 1) - t(1 2 -1) = (1 2 3) - t(1 2 -1).

To make P->A perpendicular to L, the dot product of P->A and the direction vector of L must be zero. This means (1 2 -1) . (1-t 2-2t 3+t)=0 so 1-t+4-4t-3-t=0 or 2-6t=0 so t=1/3. This means that P is (4/3 7/3 -2/3) and P->A is (2/3 4/3 10/3). Therefore the shortest distance between A and L is sqrt(4/9+16/9+100/9)=2sqrt(10/3).

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5 months ago

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How can the average speedx of a gas molecule be derived?

To start with, some assumptions must be made to simplify the problem. Firstly the N molecules are assumed to have no size, be identical and not interact with each other. The second assumption is that the molecules move randomly and collide elastically with the container which is a cube of side...

To start with, some assumptions must be made to simplify the problem. Firstly the N molecules are assumed to have no size, be identical and not interact with each other. The second assumption is that the molecules move randomly and collide elastically with the container which is a cube of side length L.

When a particle collides with the container in the x direction, it rebound in the opposite direction at the same speed meaning the change in momentum is 2mvx. The momentum change of n particles is therefore 2nmvx.

The force exerted on the wall F=Impulse/Time so next we need to find the number of particles colliding with the wall per second. To do this, consider that all particles less than v metres from the wall will collide before one second. This means that any particle in the volume vxL2 will collide. However, only half the particles will be travelling towards the wall so a factor of 1/2 is needed. The number of particles in this volume is vxL2/2*N/L3=vxN/2L. This means F=2mvx+vxN/2L=mNvx2/L. As all directions are equivalent, v2=vx2+vy2+vz2=3vx2 so F=mNv2/3L.

The pressure is therefore P=F/L2=mNv2/3L3=mNv2/3V. This means that PV=Nmv2/3. Also, PV=NkBT so NkBT=Nmv2/3 so kBT=mv2/3. The average speed can now be found in terms of temperature and mass only: v=sqrt(3kBT/m).

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5 months ago

166 views

Integrate (x+4)/(x^2+2x+2)

At first glance, this may look like an inverse trig integral but as the top contains an x term, we must use a different method. First, rewrite the top in two parts, one that is a multiple of the derivative of the bottom and one that is just a number. In this case the derivative of the bottom ...

At first glance, this may look like an inverse trig integral but as the top contains an x term, we must use a different method.

First, rewrite the top in two parts, one that is a multiple of the derivative of the bottom and one that is just a number. In this case the derivative of the bottom is 2x+2 so rewrite as (2x+2)/2+3.

Now split the fraction in two: (2x+2)/2(x2+2x+2)+3/(x2+2x+2). The first part is now a log integral integrating to 1/2 ln(x2+2x+2).

To integrate the second we must complete the square on the bottom. x2+2x+2=(x+1)2+1. Now using the substitution u=x+1, dx=du, we can see that the second part is an inverse tan integral integrating to arctan(x+1).

The overall integral is therefore 1/2 ln(x2+2x+2) + arctan(x+1).

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5 months ago

144 views
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