# Tutor profile: John H.

## Questions

### Subject: Applied Mathematics

A man of mass 80 kg jumps out of a plane. The force due to gravity mg pulls him down but an opposing Air resistance is proportional to his velocity, R = Kv where K = 20 Ns/m What will be the man`s terminal velocity?

He reaches terminal velocity when the force due to gravity is equal to the Air resistance - so the net force is zero and he is no longer accelerating. That means $$R = mg $$ or $$Kv = mg. $$ So taking g as $$10 m/s^2$$ $$v = mg/K = 80 x 10/20 = 40 m/s$$

### Subject: Basic Math

A geometric series has first term a = 360 and common ratio r = 7/8 Giving your answers to 3 significant figures where appropriate, find the 9th term of the series.

It is a geometric series which means each term is a multiple of the previous term. The amount multiplied is called the common ratio. And the first term is 360 so the series looks like this 360, 360x7/8, 360x7/8x7/8, .... or we can write 360, $$360 x (7/8^1)$$, $$360 x (7/8^2)$$, $$360 x (7/8^3)$$ So we can see that the 9th term will be $$360 x (7/8^8)$$ Which a calculator tells us is 123.7 or 124 to three significant figures.

### Subject: Physics

1. A helium–neon laser (λ = 632.8 nm) is used to calibrate a diffraction grating. If the first-order maximum occurs at 20.5°, what is the spacing between adjacent grooves in the grating?

Using the formula we derived for the maxima from a grating (or a pair of slits) nλ = d Sinθ where θ is the angle to the perpendicular at the grating, n is the order of the (n = 1,2, etc) d is the spacing between the lines on the grating and λ is the wavelength of light. re-arranging, d = nλ/Sinθ = 1x 632.8 x 10^-9/(Sin 20.5°) = 1.81 x 10^-6 metres. That is, about a thousandth of a millimetre!

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