Currently unavailable: for regular students
Degree: Materials Science (Masters) - Oxford, Corpus Christi College University
Hi! I'm an Undergraduate studying Materials Science at the University of Oxford, about to begin my second year of study. I have a passion for the Physical Sciences - in particular Maths, Physics and Chemistry, and am keen to encourage such a passion in others while helping them with anything they find particularly tough.
I have previous experience with both tutoring and classroom teaching at Fiwilia school in Zambia, as well as working as a classroom assistant at my local school on Friday afternoons.
I really enjoy tutoring and hope that any students would enjoy my tutoring too!
What I can teach:
My degree in Materials Science focuses predominantly on the three subjects I took at Higher Level for the IB: Physics, Maths and Chemistry. I am happy to tutor these up to IB Higher Level Standard.
I am also more than happy to tutor anyone interested in Materials Science and who would potentially like to apply for the subject - including help with Interviews and PAT tests for Oxford, or with Personal Statements/Interviews for other Universities (I also applied succesfully to Imperial College, Manchester and Exeter).
Finally I spent three years at a French International School, and studied French up to IB Standard Level, so would be keen to help any students with their French.
How I Teach:
I find tutoring works best when the student leads the sessions with the questions that they have. If possible I encourage students to send me their questions in advance so that I can provide the most comprehensive answers possible when we work through problems together, however I am also very happy to take any spotaneous questions as we go along. Succesful sessions for me tend to include plenty of active discussion on the topic, and working through plenty of past paper questions together.
I'm available for as much tutoring as possible, excluding 21-25 July and 8-13 September, and can work flexible hours throughout the summer. I would also be keen to continue tutoring through the following academic year(s), whether it is for regular or one-off help sessions. Please get in touch and book a free meeting if you're interested!
|French||13 Plus||£18 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|-Personal Statements-||Mentoring||£20 /hr|
|.PAT.||Uni Admissions Test||£25 /hr|
|Preliminary Examinations in Material Science||Higher||First with Distinction|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Tessa (Student) November 23 2016
Tessa (Student) August 18 2016
So the best way that I have found of impressing an examiner, is through the use of interesting phrases and tenses. Knowing some unique vocab is also good but is more of a sign of having memorised words and can also be hard to work into your passage.
For example, you can use 'si clauses' to work two tenses into one phrase:
The most common use of the Si clause is to use:
Si + Verb in Imperfect Tense + Verb in Conditional Tense
eg: Si j'étais riche, j'achèterais une voiture!
Translation: If I were rich, I would buy a car!
You could also use some idiomatic phrases that really show an understanding of french:
"tomber dans les pommes" translates literally to "To fall in the apples", but is a french idiom that actually means "to faint"
"faire l'andouille" translates literally to "To make the sausage" but actually means "to do something ridiculous"
"Avoir un poil dans la main" translates literally to "to have a hair in the hand" but means "To be lazy"
These sorts of phrases show you really know your french inside-out and are likely to impress the examiner and encourage them to give you a higher mark!see more
From the rules of logarithms, we know that:
log(A) - log(B) = log(A/B)
log(1-x) - log(x) = log[(1-x)/x]
Therefore from the question, we know:
log[(1-x)/x] = 1
If we then take both sides of the equation as a power of 10:
(1-x)/x = 10^1
and then multiply both sides through by x:
Solving for x:
We can check our answer by inserting it into the original equation:
log(1-x) - log(x) = log[1-(1/11)] - log[1/11]
and using the rule log(A) - log(B) = log(A/B):
log[1-(1/11)] - log[1/11] = log(10/11)-log(1/11)
Thus we know x=1/11see more
To start with, we should work out all of the quantities we need:
Mass of Trolley A: MA= 5kg
Mass of Trolley B: B weighs 10kg more than A
--> Therefore MB= 10 + 5 = 15kg
Mass of Both Trolleys together: MAB = 15 + 5 = 20kg
Initial Velocity of A: VA= 15 m/s
Initial Velocity of B: VB= 0 m/s
By the conservation of momentum, we know that the total momentum of a closed system is unchanged for any process. Thus the initial and final momenta of the two trollies combined are the same, ie:
PA + PB = PAB where P is momentum
We know that momentum P = mv
MAVA + MBVB = MABVAB
From above, we know all of these quantities except for VAB which is the quantity we need to find for the answer. Thus we can substitute in all of the values and find VAB:
(5x15) + (15x0) = (20xVAB)
75 = 20VAB
VAB = 75/20 = 3.75 m/ssee more