Aldo E. GCSE Maths tutor, A Level Maths tutor, A Level Further Mathem...

Aldo E.

£20 - £22 /hr

Currently unavailable: for regular students

Studying: Engineering (Masters) - Cambridge University

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4 reviews| 8 completed tutorials

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About me

Hi, I'm Aldo and I'm a fourth year engineering postgraduate student at the University of Cambridge.

Engineering is a scientific discipline in which maths and physics are used to first understand the world around us, and then sculpt it to the way we want. I've been passionate about the sciences, in particular maths and physics, for as long as I can remember, and I would like nothing more than to pass on that passion, enthusiasm and technical ability to others.

I have previous tutoring and teaching experience both online on MyTutor and in person. While I was in my second year of sixth form I helped out in AS maths classes as well as GCSE maths and physics classes. I also led a year 9 maths class.

If you need any help with maths, further maths or physics at A-level or GCSE, please do not hesitate to get in touch! I look forward to hearing from you!

Hi, I'm Aldo and I'm a fourth year engineering postgraduate student at the University of Cambridge.

Engineering is a scientific discipline in which maths and physics are used to first understand the world around us, and then sculpt it to the way we want. I've been passionate about the sciences, in particular maths and physics, for as long as I can remember, and I would like nothing more than to pass on that passion, enthusiasm and technical ability to others.

I have previous tutoring and teaching experience both online on MyTutor and in person. While I was in my second year of sixth form I helped out in AS maths classes as well as GCSE maths and physics classes. I also led a year 9 maths class.

If you need any help with maths, further maths or physics at A-level or GCSE, please do not hesitate to get in touch! I look forward to hearing from you!

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Ratings & Reviews

5from 4 customer reviews
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Martina (Parent)

January 15 2017

Good lesson, good explanation about questions and topics I don't understand but I know do.

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Mauro (Parent)

January 14 2017

Really good at explaning concepts and formulas!

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Martina (Parent)

January 28 2017

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Pierre (Student)

December 28 2016

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
ChemistryA-level (A2)A*
Biology (AS)A-level (A2)A
EngineeringDegree (Bachelors)FIRST

General Availability

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Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£22 /hr
Further MathematicsA Level£22 /hr
MathsA Level£22 /hr
MathsA Level£22 /hr
PhysicsA Level£22 /hr
PhysicsA Level£22 /hr
MathsGCSE£20 /hr
MathsGCSE£20 /hr
PhysicsGCSE£20 /hr
PhysicsGCSE£20 /hr

Questions Aldo has answered

How do you calculate the Earth's escape velocity?

The escape velocity is the speed that an object must have in order to have enough energy to escape the Earth's gravitational field. To calculate the escape velocity, we can apply the principle of Conservation of Energy, which states that energy cannot be created or destroyed. We equate the total energy that the body has when it is on the surface of the Earth with the total energy it has when it is infintely far away and no longer experiences the Earth's gravity:

 

Energy on the surface of Earth = Energy at infinity

 

The total energy at each point will be the sum of the kinetic energy of the body (0.5*m*v2) and its gravitational potential energy (-G*M*m/R).

 

On the surface of the Earth, it's kinetic energy will be 0.5*m*ve2, where ve is its escape velocity. It's gravitational potential energy will be -G*M*m/R0, where R0 is the radius of the Earth. We assume that the object has just enough energy to reach infinitely far away from the Earth, which means that it has no kinetic energy once it has reached infinity, and has therefore come to rest. It's gravitational energy will also be zero at infinity, because that is how gravitational potential energy is defined (you can see from the expression for gravitational potential energy that as the distance R becomes very large, i.e. infinitely large, it becomes infinitely small).

 

We place all of these quantities in the expression above to obtain:

 

0.5*m*ve2 – G*M*m/R0 = 0 + 0

 

We can cancel the mass of the body, m, from everything on the left hand side:

 

0.5*ve2 – G*M/R0 = 0

 

And move the G*M/R0 onto the right hand side:

 

0.5*ve2 = G*M/R0

 

We multiply everything by 2 and take the square root to make ve the subject of the formula:

 

ve = (2*G*M/R0)0.5

 

Substituting in some numbers (G = 6.67x10-11 m3 kg-1 s-2, M = 5.97x1024 kg, R0 = 6.271x106 m):

 

ve = 11.2 km/s

 

It is important to note here that we are neglecting the effects of air resistance as is often done in these calculations. Air resistance would transfer energy from the escaping object into thermal energy in the atmosphere, reducing the object's total energy as time passed. Therefore, more energy would be required at the start, and hence the speed at the start would need to be higher.

The escape velocity is the speed that an object must have in order to have enough energy to escape the Earth's gravitational field. To calculate the escape velocity, we can apply the principle of Conservation of Energy, which states that energy cannot be created or destroyed. We equate the total energy that the body has when it is on the surface of the Earth with the total energy it has when it is infintely far away and no longer experiences the Earth's gravity:

 

Energy on the surface of Earth = Energy at infinity

 

The total energy at each point will be the sum of the kinetic energy of the body (0.5*m*v2) and its gravitational potential energy (-G*M*m/R).

 

On the surface of the Earth, it's kinetic energy will be 0.5*m*ve2, where ve is its escape velocity. It's gravitational potential energy will be -G*M*m/R0, where R0 is the radius of the Earth. We assume that the object has just enough energy to reach infinitely far away from the Earth, which means that it has no kinetic energy once it has reached infinity, and has therefore come to rest. It's gravitational energy will also be zero at infinity, because that is how gravitational potential energy is defined (you can see from the expression for gravitational potential energy that as the distance R becomes very large, i.e. infinitely large, it becomes infinitely small).

 

We place all of these quantities in the expression above to obtain:

 

0.5*m*ve2 – G*M*m/R0 = 0 + 0

 

We can cancel the mass of the body, m, from everything on the left hand side:

 

0.5*ve2 – G*M/R0 = 0

 

And move the G*M/R0 onto the right hand side:

 

0.5*ve2 = G*M/R0

 

We multiply everything by 2 and take the square root to make ve the subject of the formula:

 

ve = (2*G*M/R0)0.5

 

Substituting in some numbers (G = 6.67x10-11 m3 kg-1 s-2, M = 5.97x1024 kg, R0 = 6.271x106 m):

 

ve = 11.2 km/s

 

It is important to note here that we are neglecting the effects of air resistance as is often done in these calculations. Air resistance would transfer energy from the escaping object into thermal energy in the atmosphere, reducing the object's total energy as time passed. Therefore, more energy would be required at the start, and hence the speed at the start would need to be higher.

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3 years ago

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How do you find the cube root of z = 1 + i?

Firstly we express z in polar form:

 

z = R*ei*θ

 

where |z| = (Re2 + Im2)0.5 = (12 + 12)0.5 = 20.5

 

θ = arg z = tan-1(Im/Re) = tan-1(1/1) = π/4

 

Therefore z = (20.5)*ei*π/4

 

We can add on any multiple of 2π to the argument of z without affecting the value of the complex number:

 

z = (20.5)*ei*(π/4 + 2*π*n)

 

where n is an integer

 

We then take cube roots of both sides (not forgetting to cube root the modulus R as well as the exponent):

 

z1/3 = (21/6)*ei(π/12 + 2*π*n/3) = (21/6)*ei(π + 8*π*n)/12

 

Because we are calculating the cube root, we expect three solutions. To find these three roots, we substitute in three consecutive integers into n. We will choose n = 0, 1, 2.

 

Solution 1 (with n=0): z1/3 = (21/6)*ei(π/12)

Solution 2 (with n=1): z1/3 = (21/6)*ei(3π/4)

Solution 3 (with n=2): z1/3 = (21/6)*ei(17π/12)

 

We can convert these back into Cartesian form using:

 

z = R*(cosθ + i sinθ)

 

We find that:

 

Solution 1: z1/3 =(21/6)*(cos(π/12) + i sin(π/12)) = 1.08 + 0.291i

Solution 2: z1/3 = (21/6)*(cos(3π/4) + i sin(3π/4)) = -0.794 +0.794i

Solution 3: z1/3 = (21/6)*(cos(17π/12) + i sin(17π/12)) = -0.291-1.084i

Firstly we express z in polar form:

 

z = R*ei*θ

 

where |z| = (Re2 + Im2)0.5 = (12 + 12)0.5 = 20.5

 

θ = arg z = tan-1(Im/Re) = tan-1(1/1) = π/4

 

Therefore z = (20.5)*ei*π/4

 

We can add on any multiple of 2π to the argument of z without affecting the value of the complex number:

 

z = (20.5)*ei*(π/4 + 2*π*n)

 

where n is an integer

 

We then take cube roots of both sides (not forgetting to cube root the modulus R as well as the exponent):

 

z1/3 = (21/6)*ei(π/12 + 2*π*n/3) = (21/6)*ei(π + 8*π*n)/12

 

Because we are calculating the cube root, we expect three solutions. To find these three roots, we substitute in three consecutive integers into n. We will choose n = 0, 1, 2.

 

Solution 1 (with n=0): z1/3 = (21/6)*ei(π/12)

Solution 2 (with n=1): z1/3 = (21/6)*ei(3π/4)

Solution 3 (with n=2): z1/3 = (21/6)*ei(17π/12)

 

We can convert these back into Cartesian form using:

 

z = R*(cosθ + i sinθ)

 

We find that:

 

Solution 1: z1/3 =(21/6)*(cos(π/12) + i sin(π/12)) = 1.08 + 0.291i

Solution 2: z1/3 = (21/6)*(cos(3π/4) + i sin(3π/4)) = -0.794 +0.794i

Solution 3: z1/3 = (21/6)*(cos(17π/12) + i sin(17π/12)) = -0.291-1.084i

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3 years ago

1359 views

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