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It happens in a vacuum. Once the photoelectrons are emitted from the metal surface (the photocathode), they accelerate towards the anode (using the kinetic energy provided by the photons. ie, KE = hf - work function).

In relation to Hertz, he only achieved a spark in the secondary loop when the discharge gaps were in a parallel plane with each other. From this, he observed that EMR travelled in straight lines and were in fact transverse waves, since only these can be polarised. So basically, I remember the...

I just used the answers in the back of the book (surfing physics - shadwick *cringe*). I used his answer for r to solve the rest of the question. I understand all that stuff about equating the magnetic force with that of centripetal, so I think my only problem is solving part a :mad1: - the...

I'm still a little unsure about two things... For a), how do you solve for the force when you're not given the charge? And for d) I keep getting 0.086m (using 0.018m as my r from part b)

A single positively charged particle, mass 4.6x10^-27kg, enters a 4.0T magnetic field into the page at 2.5x10^6m/s. The field covers an area of 0.1mx0.1m. It enters at 90 degrees, 0.05m up from the bottom left corner. a) Calculate the force on the particle while it is in the field. b) Calculate...

10 - 20 turns should be sufficient (I got mine to work using just 12 turns) perhaps your motor didn't work due to other factors, such as the brushes not touching the commutator properly (too much friction, or not enough contact). or maybe the armature cannot spin around the axle freely, or the...

just to add, in an ideal motor (where there is nothing attached to slow it down) the speed of the armature coil increases to a point where the back emf = supply emf. thus, there will be no voltage across the coil -> no current flow -> no net force acting -> armature coil rotates at a constant...

I usually only complete two titrations - one test run (let the solution in the burette flow out quickly and gain a rough idea of the equivalence point), and one proper run (I'm pretty lazy when it comes to titrations, so I put all my time and effort into gaining just the one result and using...

from what module?

would the answer be 3.75x10^-2 N? or 5.625 N? ...or neither? lol

Re: First round assessments hahahahaahhahahahahahaha

First round assessments How did you go in the first round of HSC assessments? What were they? What were your marks? I'll start, English Advanced: 'Belonging' speech (related texts), 9/15 Mathematics: Locus + Geometrical Applications of Calculus exam, 32/40 Chemistry: Current chemistry...

A: upwards the magnetic field created by the charged particles inside the transmission line would interact with an external magnetic field (the earth's poles). if the current runs along the equator, then the force would act perpendicular to it. i.e. right-hand push rule

you are correct. if a satellite with an altitude of approx 35,800 km isn't above the equator, it won't remain at a fixed point. if we were trace it's location from earth, we would see a 'figure 8' pattern, which has a period equal to that of the earth's - hence, geosynchronous

How about the 'Spotlight' physics textbook. Has anyone come across it? If so, how is it?

Jacaranda? MacMilllian? Spotlight? Any others?