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Tutor profile: Ewan K.

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Ewan K.
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Questions

Subject:Physics (Electricity and Magnetism)

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Question:

The filament in a 6V filament bulb is commonly made from Tungsten. If this filament has a length of 1.5cm and a cross sectional area of 8x$$10^{-11}m^{2}$$ and has a current of 1.3A applied to it, what is the bulbs resistivity?

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Ewan K.

Since we are asked to find resistivity, the main equation we will use is R=$$\frac{\rho L}{A}$$ where R is resistance, L is the length of tungsten, A is the cross sectional area and $$\rho$$ is the resistivity. The first thing to do is to make sure we use SI units so we can convert our length, L, into metres giving L=0.015m. Next, we need to find a value for resistance, R, by using V=IR. We're told values of 6V for V and 1.3A for I in the question, giving a value of R=4.6$$\Omega$$. We then rearrange the equation for resistivity to make $$\rho$$ the subject so that $$\rho = \frac{RA}{L}$$ and by then substituting in our values for R, L and A, we obtain $$\rho$$ = 2.46x$$10^{-8} \Omega$$m

Subject:Nuclear Physics

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Question:

Briefly describe how photon emission tomography (PET) is used to make images of the body based on the distribution of positron emitting tracers in the body.

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Ewan K.

For a PET scan, the subject is injected with a substance that emits positrons. These positrons are only able to travel a short distance before being annihilated by an electron. When this annihilation occurs, in order to conserve energy and momentum, photons are emitted in pairs with each photon in this pair travelling in an opposite direction. A ring of scanners around the body is used to detect these photons and model their origin, forming an image of the body from the emission of these photons.

Subject:Physics

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Question:

A golf club strikes a ball. The ball leaves the golf club at an angle of 28° to the horizontal and at a speed of $$40 m s^{-1}$$. Calculate the horizontal distance the golf ball should travel before reaching the ground.

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Ewan K.

There are several aspects to this question. We want to find the horizontal distance travelled and to do this we can use $$speed = \frac{distance}{time}$$. However, to do this we need to know the time the ball is in motion. This can be found by examining the vertical component and using SUVAT equations. In the vertical plane, if we take the case of S=0, we know $$V = -U = -40\sin(28)$$ and $$A=-g$$. We then use the equation $$V=U+AT$$ and rearrange in the form $$T=\frac{V-U}{A}$$. This gives a value of T, the time in which the ball is in motion, as 3.83 seconds. We then return to the horizontal component of the problem. We need to find the speed travelled in this plane, found by doing $$speed=40\cos(28)$$ which gives a value of $$35.32 m s^{-1}$$. With this, we then substitute our values of speed and time to find $$distance=35.32 m s^{-1}* 3.83 s$$ so that $$distance=135.27m$$

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