Recent content by hb20007

Yeah, makes sense... Thanks :biggrin:

Okay, now how about a reflection over y = -x?

f(-x) is a reflection over the y axis -f(x) is a reflection over the x axis Now, how do we represent a reflection over y=x?

Your answer did the trick, thanks

Yup, these are introductory-level formulas for materials obeying Hooke's Law. Can you please explain why it's Force/Area and not 1/2 * Force/Area?

I am familiar with the formula for energy density: \frac{1}{2} * \frac{Force*Extension}{Area*length} and also the formula for elastic potential energy: \frac{1}{2} * Force*Extension. I noticed that there is a 1/2 in both formulas because we are concerned with the average force in each case...

That's quite interesting, definitely going to read up on it

Thanks Drakkith, your answer is quite insightful.

Perhaps virtually 100% of physicists out there believe that we live in an 'accelerating universe', but as a person who's been researching the topic for a while, it seems to me the only evidence we have is based around redshift. I know redshift is a good enough argument to convince even...

Thanks once again. The question actually wasn't difficult at all but I struggled with it because I expected it to be so. Once you showed me how to eliminate the concepts of angular acceleration and frequency I could think clearly and I just got the answer

That's a really useful simplification, thanks. However, what would be the linear motion counterpart for frequency?

Homework Statement A turntable is a uniform disc of mass m and radius R. The turntable is initially spinning clockwise when looked down on from above at a constant frequency f_0. The motor is turned off at t=0 and the turntable slows to a stop in time t with constant angular deceleration...

Oh. Does this mean that Eagle's answer was wrong and that the reason answer 2 is false is not because the formula of work is not mgh as Eagle argued but because change in potential energy is equal to - (work done by gravity) and not equal to work done by gravity itself?

Another Question ! I was thinking of applying this logic to a pendulum situation. Let's say the pendulum is in motion and is at a certain moment in time at an angle of 45° to the vertical. I know that the work done by the tension is 0 as it's tangential to the direction of motion. What...

Got it. Thanks