# Tutor profile: Jack G.

## Questions

### Subject: Applied Mathematics

State the Navier-Stokes equations for incompressible viscous fluid flow, defining the variables and constants used.

$$\rho \frac{D\vec u}{Dt}= \rho \vec F - \vec \nabla p + \mu \nabla^2 \vec u$$ $$\vec \nabla \cdot \vec u=0$$ Here, $$\rho$$ is the density of the fluid, $$\vec u$$ is the velocity field of the flow, $$t$$ is time, $$\vec F$$ is the resultant force field on the fluid, $$p$$ is the scalar pressure field of the fluid, and $$\mu$$ is the fluid's viscosity.

### Subject: Environmental Science

What is Ocean Acidification? Describe the cause, effects, and likely projected damage in the 21st century.

The Ocean absorbs 90% of excess atmospheric heat and around 30% of the carbon dioxide emitted into the atmosphere. When carbon dioxide is dissolved in the ocean, it reacts with water to create carbonic acid, as follows: $$CO_2 + H_2O ⇌ H_2CO_3$$ This lowers the pH of our oceans, making them more acidic, with potentially catastrophic effects on marine ecosystems. Through the absorption of both excess heat and carbon dioxide, increasing the temperature and decreasing the pH, coral reefs around the tropics are bleaching at an alarming rate. The Great Barrier Reef, for example, experienced back-to-back extreme bleaching events for the first time in history in 2016 and 2017. This loss of essential high-biodiversity ecosystems is felt in the health of our oceans worldwide. Although ocean acidification is still green, if we follow RCP8.5, we could see consecutive extreme bleaching events every year by 2100.

### Subject: Physics

A projectile is launched from ground-level, at an angle $$\theta$$ to the horizontal. For what value of $$\theta$$ is the maximum horizontal range of the projectile reached? Note, $$0<\theta<\frac{\pi}{2}$$ (in radians).

Using $$suvat$$ we split these quantities into vertical and horizontal components. Vertical: $$s=0, u=usin\theta, v=-usin\theta, a=-g, t=t$$ (1) Horizontal: $$s=s, u=ucos\theta, v=ucos\theta, a=0, t=t$$ (2) Using: $$v=u+at$$ (1): $$2usin\theta=-gt$$ Using: $$s=ut+\frac{1}{2}at^2$$ (2): $$s=ucos\theta t$$ We can now sub the value of t from (1) into (2): $$s=ucos\theta \times \frac{-2usin\theta}{g}$$ Hence $$s=\frac{u^2sin(2\theta)}{g}$$ (3) Now we have found the horizontal displacement, we must find the maximum by differentiating (3) w.r.t $$\theta$$, setting equal to zero, and solving for $$\theta$$: $$\frac{ds}{d\theta}=\frac{2u^2cos(2\theta)}{g}=0$$ Hence: $$cos(2\theta)=0$$ Thus $$2\theta=\frac{\pi}{2}+2\pi k$$, for $$k = 1,2,3.....$$ Which gives $$\theta=\frac{\pi}{4}$$, given the constraints on $$\theta$$ provided. Hence the angle of projection that achieves the maximum horizontal distance is $$\frac{\pi}{4}$$ radians, or 45 degrees to the horizontal.

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