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# Tutor profile: Marta R.

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Marta R.
Maths and Physics Tutor
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## Questions

### Subject:Astrophysics

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Question:

A $$30M_\odot$$ star reaches the end of its life and explodes. A $$10M_\odot$$ black hole is left, and the rest of the star is ejected by the explosion. If the total explosion energy is $$10^{46}J$$ and 10% is converted to the kinetic nergy of the ejecta. What is the initial expansion speed of the supernova remnant after the explosion?

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Marta R.

To solve this problem, we need to set the variables. The ejected mass ($$m_{E}$$) is $$20M_\odot$$ once the star was originally $$30M_\odot$$ and the black hole is $$10M_\odot$$. The efficiency of the explosion ($$\epsilon$$) is 10% and the supernova energy ($$E_{SN}$$) is $$10^{46}J$$. With the Kinetic Energy formula and calling the initial velocity $$V_{EJ}$$, the formula becomes: $$\frac{1}{2}m_{E}V_{EJ}^{2}=\epsilon \times E_{SN}$$ and rearranging the equation, $$V_{EJ}=({\frac{2\times E_{SN} }{m_{E}} )}^{\frac{1}{2}}$$. Converting from solar masses to KG, (1$$M_\odot$$=2\times 10^{30}KG) we get $$V_{EJ}=7.1\times 10^{3} kms^{−1}$$.

### Subject:Physics

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Question:

A car is travelling at $$15 ms^{-1}$$. The car accelerates uniformly to $$33 ms^{-1}$$. The car's acceleration is $$1.2 ms^{-2}$$. What is the distance covered while the car is accelerating?

Inactive
Marta R.

This problem is solved using S.U.V.A.T. We have $$u=15 ms^{-1}$$, $$v=33 ms^{-1}$$ and$$a=1.2 ms^{-2}$$. The formula we can use is: $$v^{2}=u^{2}+2as$$. Rearraging this, we get, $$s=\frac{v^{2}-u^{2}}{2a}$$. So, $$s=\frac{33^{2}-15^{2}}{2\times1.2}$$=360m. The car covered a 360m distance while accelerating.

### Subject:Partial Differential Equations

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Question:

Find u(x,y) from the following $$u_{y} = x^{2}+y^{2}$$

Inactive
Marta R.

To find this, we have to first integrate both x and y with respect to y and as it is $$u_{y}$$ there will be a function of x (I will call it f(x)) that will not appear in the above eqaution but has to be accounted for when constructing u(x,y). So, integrating $$x^{2}$$ with respect to y, we get $$x^{2} y$$ and integrating $$y^{2}$$ with respect to y, we get $$\frac{y^{2}}{3}$$. So $$u(x,y)=x^{2} y+\frac{y^{2}}{3}+f(x)$$ where f(x) is the function of x that is not accounted for in $$u_{y}$$

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