Recent content by kubaanglin

I accepted my offer to MSRP and am super excited. I'm probably going to be working on the newly initiated SPARC fusion reactor at the MIT Plasma Science and Fusion Center.

I just received an acceptance letter to MIT's MSRP. I applied for nuclear engineering/plasma physics and fusion. I was also accepted to SULI PPPL. I don't know which one to choose!

I accepted an offer from SULI PPPL. However I also applied to MIT's MSRP, but they won't release decisions until next week. If I get into MIT, I would have to withdraw my acceptance to SULI.

@Aidan Davis I would recommend joining the fusor forum. We are a group of amateur and professional scientists interested in fusion technology and IEC fusion reactor construction/research. Richard Hull has written many FAQs regarding reactor construction and theory. If there is information you...

The vial is a neutron bubble dosimeter. It contains superheated fluid droplets suspended in a gel. When a fast neutron strikes the liquid droplets, they vaporize to form visible bubbles. The number of bubbles generated is proportional to the neutron dose experienced by the detector. D-D fusion...

Homework Statement [/B] What is the shortest time required for a harmonic oscillator to move from ##x = A## to ##x = \frac{A}{2}##? Express your answer in terms of the period ##T##. Homework Equations [/B] ##x(t)=Acos(\omega t)=Acos(2\pi\frac{t}{T})## The Attempt at a Solution ##A=Acos(0)##...

Oh, that makes sense. ##Q=Av=(wy)\int_0^y (\sqrt{2gy})=(wy\sqrt{2g})\int_0^y \sqrt{y}=(wy\sqrt{2g})(\frac{\sqrt{y^3}}{\frac{3}{2}})=\frac{2}{3}w\sqrt{2gy^3}## Thank you!

Homework Statement [/B] When the level of water in a reservoir is too high, the water spills out over a spillway, as illustrated in the figure below Neglecting viscosity, show that the water flow ##Q## over the spillway is given by $$Q=\frac{2}{3}w\sqrt{2gy^3}$$ Homework Equations [/B]...

Homework Statement Stretchable ropes are used to safely arrest the fall of rock climbers. Suppose one end of a rope with unstretched length ##l## is anchored to a cliff and a climber of mass m is attached to the other end. Ehrm the climber is a height ##l## above the anchor point, he slips and...

Thank you so much for your help!

Oh, ##ω=\frac{2π}{T}\neq\frac{2πr}{T}## Then this should be the answer: $$T=\frac{2π}{\sqrt{\frac{GM}{R^3}}}$$ $$T=\frac{2π}{\sqrt{\frac{(6.67⋅10^{-11})(5.96⋅10^{24})}{(6.37⋅10^6)^3}}}=5066 s=84.44 min$$

Okay, so then ##a=ω^2x=G\frac{Mx}{R^3}## ##ω=\sqrt{\frac{GM}{R^3}}=\frac{2πr}{T}## ##T=\frac{2πr}{\sqrt{\frac{GM}{R^3}}}## Is this correct?

So ##\cos(θ)=\frac{x}{r}## and then ##F = G\frac{mMx}{R^3}##?

The angle would be ##0°## so then ##\cos(0°)=1##. That leaves me with ##F = G\frac{rmM_{Earth}}{R^3}##. How would that account for the component of the force if the tunnel was not through the center of the Earth?

##F = G\frac{rmM_{Earth}}{R^3}\cosθ## where ##θ## is the angle between ##r## and the direction of motion. ##\cosθ = \frac{x?}{r}##.