Recent content by Daniel Luo

So I should just go ahead with my answer which corresponds to approx. the charge of 8 electrons?

Am I not supposed to find the actual electric charge of an electron? Isn't that he point of the experiment?

It just doesn't seem right? I get the no. of electron in the drop to be 8 (rounded).

How do I find out? Do I use the mass, molar mass and Avogadro's number?

Homework Statement When an oil drop falls freely, the velocity first increases, but afterwards it quickly reaches a constant speed, since the air resistance becomes equal to the weight of the oil drop. The air resistance is given by: Fair=6\pir\etav, where \eta is the viscosity of air...

Couldn't you also say that the radius is the opposite side in the right angled triangle. The adjacent side is the distance from CN to Earth, and the 0.083/2=0.0415 degrees is the angle. Since tan(0.0415) = radius/distance we have that the distance is radius/tan(0.0415). This gives the right answer.

How can the radius be the distance from CN to Earth? Isn't the radius simply the radius of crab nebula? And how is the arc length how much CN has expanded the last 960 years?

I simply do not understand what you just wrote. The diameter is 2 * 1.18*10^6 m/s * 960 years * 31557600 years/s = 7.15 * 10^16 m. What do I do from here?

I don't think it will be much help to you. It's reversed. Please can't you just tell me what I'm doing wrong?

I did that already... And I get the same answer... What am I doing wrong?

Homework Statement Hi. The Crab Nebula which can be seen with even a small telescope, is the result of a supernova. All parts of the nebula are moving away from the centre at 1.18*10^6 m/s. The angle that the Crab Nebula forms with the Earth is 0.083 degrees. This supernova was...

#CAF123 OF COURSE! It is the square of the y-component of the initial velocity! So it's sin squared theta. Thanks for pointing it out and letting me think my self :-).

Ok so I tried this: The max. height is: D = [(6gD)*sin(θ)] / (2g) which simplifies to: D = 3Dsin(θ) Hence: sin(θ) = 1/3. I tried using this for the horizontal distance: R = [(6gD) * (2 cos(θ) sin(θ))] / g I found cos(θ) by: √(12-(1/3)2) = (√8)/3 Next, R = [ 6gD * 2...

#Haruspex Thanks for the correction. It is corrected now.

Homework Statement You are playing catch with a friend in the hallway of your dormitory. The distance from the floor to the ceiling is D, and you throw the ball with an initial speed v0=√(6gD). What is the maximum horizontal distance (in terms of D) that the ball can travel without bouncing...