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Emma H.

Currently unavailable: for new students

Degree: Earth Sciences (Masters) - Oxford, Exeter College University

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About me

Hi, I'm Emma. I'm a Masters' student at The University of Oxford, studying Earth Sciences, and I love all things Maths and Science! I've been tutoring for over five years, one to one for both GCSE and A Level and I love helping people achieve their full potential, as well as helping people realise subjects like the sciences and maths can actually be quite fun once you get the hang of them! I am very patient, and will always try to see things from the most abstract point of view, so as to help anyone and everyone to understand key concepts and learn key information to achieve the highest grades possible across the board. I believe exam technique is key to achieving top grades, and so my methods focus heavily on past paper questions, and really understanding what you have been asked so that all answers are clear and concise. Having gone through the process of applying to Oxbridge, I am also available to give any advice or tips on Personal Statements and to dispel any myths you may have heard about the dreaded interviews, as well as going through practice interview questions for a physical sciences course.

About my sessions

I like to get in touch with the pupil before each session so that we can work together to produce some goals for the tutorial. This may involve the student giving me a topic that they simply want to be taught for an hour, or emailing me some work that they have done for me to look at and then we can discuss it together in the tutorial and look at way to improve. I am very open to suggestions from students because not everyone learns in the same way, and varying the way material is taught can also be very beneficial for understanding and memory of that material.

Subjects offered

SubjectQualificationPrices
Biology A Level £24 /hr
Maths A Level £24 /hr
Physics A Level £24 /hr
Biology GCSE £22 /hr
Chemistry GCSE £22 /hr
Maths GCSE £22 /hr
Physics GCSE £22 /hr
Science GCSE £22 /hr

Qualifications

SubjectQualificationLevelGrade
MathematicsA-levelA2A*
PhysicsA-levelA2A*
BiologyA-levelA2A*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

23/10/2012

General Availability

Currently unavailable: for new students

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Ratings and reviews

5from 5 customer reviews

Adam (Student) July 25 2017

Very good tutor, makes maths fun again!

Alison (Parent) November 23 2015

Great lesson. Just what I need. Thanks

Megan (Student) June 19 2016

Alison (Parent) June 19 2016

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Questions Emma has answered

Describe the structure of DNA

DNA is the molecule within cells that contains genetic information, in the form of a sequence of bases that form a code that makes up a protein. DNA is a polymer, meaning that it is a macromolecule (large molecule) made up of many repeating subunits known as monomers. In the case of DNA, the m...

DNA is the molecule within cells that contains genetic information, in the form of a sequence of bases that form a code that makes up a protein. DNA is a polymer, meaning that it is a macromolecule (large molecule) made up of many repeating subunits known as monomers. In the case of DNA, the momomers are known as nucleotides, and many nucleotides are linked together form a polynucleotide chain. Each nucleotide is a molecule made up of three individual units: a Phosphate group, the pentose (five carbon) sugar Deoxyribose and a Nitogen-containing base. There are four types of nucleotide base that are named differently due to subtle differences in their chemistry. These are Adenine, Guanine, Cytosine and Thymine. Many mononucleotides join together to form a polynucleotide chain by means of Phosphodiester bonds, that form between the phosphate group of one nucleotide and the deoxyribose group of another. The base then sticks out from this phosphate-sugar backbone. However, the polynucletide chain alone is not enough to form DNA, as the DNA molecules's structure is that of a double helix made up of two (double) polynucleotide chains wound round eachother in a helical shape (helix). The two chains are joind by hydrogen bonds that form between complimentary base pairs. The word complimentary here refers to the fact that bases that will only pair with one other base: Adenine will only pair with Thymine (by means of two hydrogen bonds) and Cytosine will only pair with Guanine (by means of three hydrogen bonds). 

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2 years ago

1608 views

If a car is travelling over a curved hill, what is the maximum speed it can travel before losing contact with the road surface?

This is a question about circular motion that also includes some knowledge of Newton's laws. The car is travelling over a curved surface, which you can imagine as part of the circumference of a circle. In order for the car to be travelling in a circular path, the resultant of all the forces ac...

This is a question about circular motion that also includes some knowledge of Newton's laws. The car is travelling over a curved surface, which you can imagine as part of the circumference of a circle. In order for the car to be travelling in a circular path, the resultant of all the forces acting on it must provide the centripetal force (which acts towards the centre of the circle). The forces acting on the car are its weight, equal to its mass times the gravitational field strength (mg) acting vertically downwards, and the normal reaction force from the road surface, acting perpendicularly to the road surface. We shall call this R. The resultant of the two forces is equal to mv2 /r, where r is the radius of curvature of the circular path, and this force is the centripetal force. When the car is at the top of the hill, the reaction force is vertically upwards and the weight is vertically downwards, thus the resultant of the two that acts towards the centre of the circular path = mg - R. This resultant force provides the centripetal force and so mg -R = mv2 /r. If the car is to lose contact with the road, then there will no longer be a normal reaction force from the road acting on the car, therefore R=0. When this happens when the car is at the top of the hill, the centripetal force, mv2/r = mg. There is mass (m) present on both sides of the equation here which cancel out, leaving us with v2/r = g. We can rearrange this to make v, the speed at which the car loses contact with the road surface, the subject of the equation. We end up with v= (gr)1/2. The reason this is the maximum speed of the car when it loses contact is because this is when the entirety of the car's weight contributes to the centripetal force (since it acts in the same line as the centripetal force when the car is at the top of the hill) rather than some component of it.)

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2 years ago

2335 views
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