Sam C. A Level Physics tutor, GCSE Physics tutor, A Level Maths tutor...

Sam C.

Currently unavailable: for new students

Degree: MEng Engineering Science (Masters) - Oxford, New College University

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About me

ABOUT ME:

I am studying Engineering Science at New College, Oxford. I am also a key member of the prestigious University Boat Club, coxing the Oxford Blue boat in the 2016 Boat Race. I am passionate about academic success and about Maths, but I also love helping other people to succeed. 
I am privileged to have been taught by many exceptionally talented teachers and have learnt a lot about teaching styles and approaches from them. I always go above and beyond to achieve results, and would love to share my passion and enthusiasm with you.

EXPERIENCE:

I have both a younger brother and sister, and have helped them extensively throughout their school careers, especially with Maths which they both now excel at. 
In my final years at school I was appointed coach and mentor for year 11 students to help with their Chemistry GCSE’s. Also, in my role as an experienced coxswain for the Oxford University Boat Club I have used my skills to coach, teach and inspire other team mates, to help them achieve their best.

THE SESSIONS:

During the sessions you will take charge, we will keep the lessons active and make sure you do plenty of thinking through things for yourself, that’s the best way to learn. And of course, we will have plenty of fun, maths and physics can be really fascinating and I hope you find new things each session that really interest you. Make sure to let me know things you’re struggling with before our sessions.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Physics GCSE £18 /hr
.PAT. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
SpanishA-LevelA*
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5from 7 customer reviews

Namita (Student) October 13 2016

Namita (Student) October 2 2016

Ashley (Parent) September 28 2016

Ashley (Student) September 28 2016

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Questions Sam has answered

A capacitor discharge circuit of time constant 45ms includes a capacitor and resistor. The capacitor has a capacitance of 18µF What is the resistance of the resistor?

This question is in one of the CIE Physics Pre-U sample papers, and if you know the definition of the time constant for a capacitor circuit (=RC), its very easy, however even if you didnt it canbe derived fairly easily. Consider a circuit containing just a capacitor C and resistor R, with the ...

This question is in one of the CIE Physics Pre-U sample papers, and if you know the definition of the time constant for a capacitor circuit (=RC), its very easy, however even if you didnt it canbe derived fairly easily.

Consider a circuit containing just a capacitor C and resistor R, with the capacitor initially storing some charge Q0 with a voltage V0 across it. Call the curent flowing i, the voltage across the capacitor as time progresses Vc and the changing charge Q. By Kirchoff's law,

Vc + iR=0

Then we know that dQ/dt = i and C=Q/Vc, so we can say:

Q/C + RdQ/dt=0 and so RCdQ/dt + Q=0 

This has solution Q=Q0exp(-t/RC) and comparing this with the standard form of a time decaying property, X=X0exp(-t/k), where k is the time constant, gives k=RC, and then the problem is trivial.

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6 months ago

196 views

How do I evaluate something like 3070^2-3069^2?

This looks pretty daunting, but the key to this sort of a problem, which has come up a couple of times in previous PAT tests, is to recognise that this expression is actually the difference of two squares. It can be generalised to the problem of evaluating: x2-y2 which has as its solution: ...

This looks pretty daunting, but the key to this sort of a problem, which has come up a couple of times in previous PAT tests, is to recognise that this expression is actually the difference of two squares. It can be generalised to the problem of evaluating:

x2-y2

which has as its solution:

x2-y2 = (x+y)(x-y)

so in this case:

30702-30692 = (3070+3069)(3070-3069)=(6139)(1)=6139

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6 months ago

173 views
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