David  M. A Level Physics tutor, GCSE Physics tutor, A Level Maths tu...

David M.

Currently unavailable: until 25/11/2016

Degree: Physics (Masters) - Warwick University

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About me

Background

I am a third year physics student studying at the University of Warwick. I am passionate about helping people learn and my enthusiasm for science and mathematics will make our sessions focused and fun, so you can attain the high grades you deserve.

I am friendly, approachable and patient and my number one aim is for the student to feel as comfortable and happy as possible so as to maintain the most practical working environment. 

Session Outline

The direction of the session comes from the student, so tutoring is tailored to the most important areas for you. If at any point you do not know right away what you need help with I can suggest topics to cover. Before the session I will have prepared some possible examples and questions to look at so we can get straight into the content without wasting time!

We will go through theory, explaining concepts in fresh new ways at a suitable pace before looking at examples. When you are 100% happy we will look at past paper questions.

I can guide students on what past papers and excercises they can do in their own time in order to be well prepared for the exam and we can look at exam technique when appropriate.

I will also provide students with other resources they can utilise; websites, vidoes and problems which they can look at inbetween sessions. 

Subjects offered

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

Qualifications

QualificationLevelGrade
PhysicsA-LevelA
ChemistryA-LevelA*
MathematicsA-LevelA*
English LiteratureA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

25/11/2016

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Questions David has answered

How do I show two lines are skew?

Skew lines in 3 dimensions are those which are not parallel and do not intersect. Let's take two example lines: l1 = (2, 2, -1) + λ(3, 1, -3) and l2 = (1, 0, 1) + µ(6, -4, 9) First we need to show that they are not parallel. To do this we take the direction vectors (the second part with λ or...

Skew lines in 3 dimensions are those which are not parallel and do not intersect.

Let's take two example lines: l1 = (2, 2, -1) + λ(3, 1, -3) and l2 = (1, 0, 1) + µ(6, -4, 9)

First we need to show that they are not parallel. To do this we take the direction vectors (the second part with λ or µ constats) and check that one is not a multiple of the other. You can probably see by inspection that this is the case here. To be sure you can divide the first component of one by the first of the other and then check that this is not the same for both of the other components. 

3/6 = 1/2, 1/-4 = -1/4 and -3/9 = -1/3. Here dividing the components by eachother shows that one direction vector is not a multiple of the other since the values are not all the same. So l1 and l2 are not parallel.

Next we need to show that they don't intersect. To do this we can set up three simultaneous equations. Equating the x component of one line to the other and the same for y and z . For example with l1 and l2:

2 + 3λ = 1 + 6µ, equation 1,
2 + λ = 0 - 4µ, equation 2,
-1 - 3λ = 1 + 9µ, equation 3.

To show these don't intersect we need to show that these three equations aren't consistent (so the lines can't cross). One simple way to do this is to determine the value of λ and µ using two of the three equations, then substitute these values into the third equation you haven't used. If the equation is incorrect then they don't intersect. 

For example here I used equations 1 and which gave µ = -1 and λ = -7/3 (I did this by subtracting 1 from 3). If we substitute these values into equation we get 2*(-7/3) = -4*(-1) which gives -1/3 = 4, which is ofcourse incorrect and so these equations aren't consistent and the lines don't intersect. Therefore in this example l1 and l2 are skew.

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8 months ago

244 views

If the highest frequency a song is 10 kHz and it is encoded at 16 bits per sample what is the minimum number of bytes needed to encode the 3 minute song?

The key to this question is to remember the Nyquist rate of a signal. This is the lowest sample rate which can be used for a signal without losing valid frequencies or gaining incorrect frequencies. This is equal totwice the highest frequency. Therefore the sample rate needs to be 20 kHz. Sin...

The key to this question is to remember the Nyquist rate of a signal. This is the lowest sample rate which can be used for a signal without losing valid frequencies or gaining incorrect frequencies. This is equal to twice the highest frequency.

Therefore the sample rate needs to be 20 kHz. Since there are 16 bits per sample the number of bits per second is 16 multiplied by 20 000 which is 320 000 bits per second.

To calculate the number of bits in 3 minutes we need to multiply 320 000 by the number of seconds in 3 minutes. Which gives:
320 000 x 3 x 60 = 57 600 000 bits 

Remember to divide by 8 to get it in bytes, since there are 8 bits in a byte. This finally gives:
57 600 000/8 = 7 200 000 bytes = 7.2 Megabytes

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8 months ago

209 views

What is the De Broglie wavelength of an electron given it has a kinetic energy of 1 eV? You are given the mass of an electron is 9.11x10^-31 kg and Planck's constant is 6.63x10^-34

The De Broglie wavelength equation is as follows: λ=h/p We know the value of Planck's constant h and so to calculate the wavelength all we need is the momentum, which is equal to mv. The kinetic energy is given as 1 eV. Remember 1 eV is equal to 1.6 x 10-19  Joules. Using the equation for kin...

The De Broglie wavelength equation is as follows:

λ=h/p

We know the value of Planck's constant h and so to calculate the wavelength all we need is the momentum, which is equal to mv.

The kinetic energy is given as 1 eV. Remember 1 eV is equal to 1.6 x 10-19 Joules. Using the equation for kinetic energy and the given mass of the electron we can determine the velocity of the electron as follows:

K.E = 0.5*m*v2

Which can be rearranged to be in terms of velocity v:

v = (2*K.E*m)0.5

By substituting in 1.6 x 10-19 for K.E and 9.11 x 10-31 for m we get v = 5.93 x 105 ms-1 (remember to keep the full non-rounded value in your calculator!)

Then using the initial equation for the wavelength and remembering p = mv, we can substitute in our values for h, m and v as follows:

λ = 6.33 x 10-34 / (9.11 x 10-31 x 5.93 x 105)  

λ = 1.17 x 10-9 m

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8 months ago

204 views
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