Thomas C. 11 Plus Maths tutor, A Level Maths tutor, 13 plus  Maths tu...

Thomas C.

Currently unavailable: for new students

Degree: Natural Sciences (Physical) (Bachelors) - Cambridge University

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About me

Hi! I`m Tom, a 2nd year Natural Sciences undergraduate at Cambridge University. I am a very positive and empathetic person with a love for the Sciences.

I am an experienced tutor in Maths, Physics and Chemistry from GCSE to A level and beyond. I studied Maths, Further Maths, Physics and Chemistry for A level  at Merchant Taylors’ School, Northwood and achieved 4 A*s. This means I am very familiar with the GCSE and A level syllabuses for these subjects and I understand what needs to be done to push grades up to an A or A* in the exams.

I can appreciate that the Science subjects can be very challenging. I hope to combine my passion for the Sciences with my patient and gentle approach to make these daunting subjects much more accessible for you. During the sessions I hope to create an environment where learning is fun and interesting, rather than dull and frustrating. I am a friendly, positive tutor and I would love to be able to help you with your studies.

I have lots of experience volunteering for charities that encourage the inclusion of teenagers with disabilities into society and one of the key skills this required was patience, of which I have a lot.

Availability: I`m currently on my summer holiday from university which lasts until September 15th. During this time I am available to tutor throughout the day on most days.

During term time I am available to tutor in the evenings and on the weekends.

I will also be available to tutor during the Christmas and Easter holidays (5 weeks each).

Please contact me for more information as I`m very flexible and am happy to work around what is best for you.

Experience: I have tutored various students for Maths and Physics at GCSE, and Maths, Physics and Chemistry at A level. This is my main area of expertise. However I have also tutored at levels below and above this. I have given Maths and Science tuition to a student for the Common Entrance 13+ exam and violin lessons to a beginner student age 9.

I am also able to extend your studies beyond the school syllabus having given tutoring for the Physics entrance test to Oxford University, trained students for the British Chemistry Olympiad and tutored students for STEP, the additional Maths exam required for Cambridge University admissions.

If you are interested in any of these options, please get in touch. Thank you

Subjects offered

SubjectLevelMy prices
Chemistry A Level £20 /hr
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Chemistry GCSE £18 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Science GCSE £18 /hr
Maths 13 Plus £18 /hr
Science 13 Plus £18 /hr
Maths 11 Plus £18 /hr
-Oxbridge Preparation- Mentoring £20 /hr
-Personal Statements- Mentoring £20 /hr
.PAT. Uni Admissions Test £25 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Please get in touch for more detailed availability

Ratings and reviews

5from 2 customer reviews

Emma (Student) September 21 2016

Very patient, and in depth explanations - Thank you!

Nikki (Parent) August 24 2016

Very good tutor. My son finds him clear, and intelligent. Always keen to explore question further and in more depth. Friendly and committed. Certainly want to continue and will be a great support and help.

Questions Thomas has answered

A 0.20 kg mass is whirled round in a vertical circle on the end of a light string of length 0.90 m. At the top point of the circle the speed of the mass is 8.2 m/s. What is the tension in the string at this point?

A diagram would be very beneficial for this problem. We can draw a free body force diagram of the mass. At the top of the circle the two forces acting on it are its weight and tension from the string. Both are acting vertically downwards. This problem is an example of circular motion, so the ...

A diagram would be very beneficial for this problem. We can draw a free body force diagram of the mass. At the top of the circle the two forces acting on it are its weight and tension from the string. Both are acting vertically downwards.

This problem is an example of circular motion, so the equation to use will be:

F = (m*v2) / r

where F is the centripetal force (acting towards the centre of the circle), m is mass, v is velocity and r is radius

Therefore we can calculate what the centripetal force will be:

F = (0.2*8.22) / 0.9

F = 14.94222...N

As we said earlier, there are two forces acting on the mass towards the centre of the circle: its weight and the tension. We can calculate the weight from the body's mass using W = m*g

W = 0.2*9.81

Then F = weight + tension

tension = F - weight

tension = 14.942 - 0.2*9.81

tension = 13.0N

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4 months ago

394 views

Differentiate the function: y = sin(x^2)*ln(5x)

We are tasked with differentiating y = sin(x2)*ln(5x) This function is actually a product of the functions: sin(x2) and ln(5x) Therefore the product rule will be required. First let's calculate the derivatives of our individual functions before combining them. The derivative of sin(x2) is 2...

We are tasked with differentiating y = sin(x2)*ln(5x)

This function is actually a product of the functions:

sin(x2) and ln(5x)

Therefore the product rule will be required.

First let's calculate the derivatives of our individual functions before combining them.

The derivative of sin(x2) is 2x*cos(x2) using the chain rule.

The derivative of ln(5x) is 1/x.

Now to combine these using the product rule. Our answer will be:

2x*cos(x2)*ln(5x) + sin(x2)*1/x

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4 months ago

147 views

How many moles of Magnesium must react with excess Oxygen to produce 80g of Magnesium oxide?

This question looks at the reaction: Magnesium + Oxygen goes to Magnesium Oxide Mg + O -->  MgO The first step is to calculate the moles of MgO produced using the mass in the question and the relative formula mass of MgO. The RFM of MgO can be found using a periodic table. The atomic mass of...

This question looks at the reaction:

Magnesium + Oxygen goes to Magnesium Oxide

Mg + O -->  MgO

The first step is to calculate the moles of MgO produced using the mass in the question and the relative formula mass of MgO.

The RFM of MgO can be found using a periodic table. The atomic mass of Mg is 24 and the atomic mass of Oxygen is 16. Therefore the RFM of MgO is 24+16=40.

Now we can find the moles of MgO using the formula: moles = mass / RFM

moles MgO = 80 / 40

moles MgO = 2

Looking back to the equation for the reaction:

Mg + O --> MgO

We can see that one unit of Mg goes to produce one unit of MgO.

Therefore to produce 2 moles of MgO, we must use 2 moles of Mg.

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4 months ago

158 views
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