Alexander N. GCSE Physics tutor, IB Physics tutor

Alexander N.

Currently unavailable: for regular students

Degree: Physics (Bachelors) - Durham University

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About me

About me:

I am a physics student, studying at Durham University and I have always loved teaching people. I initially started it as my favourite way to learn and study. I would try to teach anything I had learnt to a friend, my parents or my little sister in order to make sure I really understood it. This then developed into taking part in tutoring sessions at school for those in lower years in order to help them in subject areas they were struggling with and just couldn't get a grasp of. 

Unlike normal school or tutor classes, I have the benefit having been in a very similar position you will be in, a couple of years ago, and so any areas that you may be struggling with I can present them in a different light that helped me understand them which will hopefully work for you too.

The sessions:

During our sessions what we cover is completely up to you. In maths and science understanding comes as a precedent to knowledge and so we will work to build your understanding until you too can explain it easily back to me, your friends, family or even your pets! A lot can be achieved in 55 minutes and hopefully with more interactive teaching methods than just working through boring questions, hopefully you'll finish loving science and/or maths even more than I do!

What next?

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website). Remember to tell me your exam board and what you're struggling with.

I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Physics GCSE £18 /hr
Physics IB £20 /hr

Qualifications

QualificationLevelGrade
PhysicsBaccalaureate7
MathsBaccalaureate6
Design TechnologyBaccalaureate6
FrenchBaccalaureate6
EnglishBaccalaureate6
EconomicsBaccalaureate6
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

Questions Alexander has answered

A lamp has a rating of 18V 9W. How much energy is transferred to the bulb in 5 minutes? Calculate the current through it when connected to a 18V supply.

The first question comes from the realisation that the power rating, "9W" represents 9 joules being supplied every second and so can be written as 9J/s  and so in 5 minutes (or 300 seconds) the energy supplied is: 9J/s * 300s = 2700J  The second part comes from the equation P=IV and we already ...

The first question comes from the realisation that the power rating, "9W" represents 9 joules being supplied every second and so can be written as 9J/s and so in 5 minutes (or 300 seconds) the energy supplied is:
9J/s * 300s = 2700J 

The second part comes from the equation P=IV and we already have numbers for power and voltage:
9 = 18 * I 
I = 9/18 = 0.5Amps

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1 year ago

362 views

On a see-saw Mary, weight 600N balances John, weight 200N when she sits 1.5m away from the pivot. How far from the pivot is John?

Using the knowledge that the moments are balanced, we can create an equation to calculate the distance John is from the pivot: 600N*1.50m = 200N*(distance) (600/200)*1.5 = distance distance = 4.5m

Using the knowledge that the moments are balanced, we can create an equation to calculate the distance John is from the pivot:

600N*1.50m = 200N*(distance)
(600/200)*1.5 = distance
distance = 4.5m

1 year ago

337 views

Resistors of 5 ohms and 10 ohms are connected in series with a battery supplying 3 volts. What is the total resistance ? And calculate the current in the circuit.

For the first part of the question we note that the resistors are in series, this means we can simply add together the resistance, giving a value of: 15 ohms For the second part we use the equation V=IR,  substituting in the known voltage and calculated resistance: 3 = 15 * I I = 3/15 = 0.2 Amps

For the first part of the question we note that the resistors are in series, this means we can simply add together the resistance, giving a value of:
15 ohms

For the second part we use the equation V=IR, 
substituting in the known voltage and calculated resistance:
3 = 15 * I
I = 3/15 = 0.2 Amps

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1 year ago

555 views
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