Recent content by wizzkids

Now I know it is a bit hard for high school students to obtain apparatus for a first-hand investigation. Your research can be a first-hand investigation or a second-hand investigation of other people's data; they are both valid for depth studies. You could investigate modern devices that exploit...

To gain a solid understanding of the evidence for the photoelectric effect, I recommend students watch this video. It's an oldie but a goodie. Don't worry about the wooden Physics professor who does the demonstration; he knows his stuff, that's all that matters.

Yes there is. Go to https://pubchem.ncbi.nlm.nih.gov/ The Published Chemistry Database of the National Library of Medicine. From there you can branch to all the other chemical databases.

Don't be too disappointed if you discovered your control specimen was contaminated or behaved in some what that you did not anticipate. It happens all the time in science. You should try to track down the identity of the contaminant or the source of the uncontrolled variable and discuss how it...

That is correct. The answer is an irrational number, but in an exam you have to choose the closest option which is (C).

Actually in the solution it would have been quicker to state the law of triangles, sum of the interior angles must be 180o and therefore the gradient of the line CB must be -1.

The free-body diagram has three vectors as shown below.

This is the Double Slit Experiment. Go look up the Physics formula sheet and you will find a formula that links the positions of constructive interference in the interference pattern with the wavelength of light and the spacing d between the slits. Then you just have to interpret the information...

Yes, the hypothesis needs to be in there, otherwise what are you investigating? Your hypothesis creates the inquiry question, and your hypothesis must be referred to in your closing remarks to determine if you achieved what you set out to do. Spoiler Alert: drag force is not proportional to...

I chose my point of reference as the top of the cliff, so distances are measured from the top of the cliff.

@kkk579 The answer given in the paper is wrong, btw. There is nothing fundamentally wrong with your calculation, but there is no need to include the mass of the block. All objects fall at the same rate regardless of their mass. You can use the standard equations for uniformly accelerated motion.

Yes, you could do the calculation that way. If the object has half it's final velocity then it has a quarter of its final kinetic energy, and therefore it has lost a quarter of its gravitational potential energy, so it has fallen 1/4 x 50 metres, and it is 37.5 metres above the ground.

@kkk579 In question 6 (c) you performed a calculation based on energy. The correct way to calculate the height is to use v2 =u2 +2as Make v = 15.65 m/s, let u =0 and rearranging terms we can say S = 15.652 / 2 x 9.8 and the answer is 12.49 m

Question 15 (a) is also badly worded. I think what the examiner meant to say was, "Draw the force-vector diagram at the point Y." If you answer the question literally, the free-body diagram for the mass A is trivial, as @kkk579 has shown, although quantifying the forces is not quite so trivial...

There is no correct answer. All the options are false. The best way to analyse this force-time graph is to recognise that the area under the force-time graph is equal to the Impulse, which is also equal to the Change of Momentum. Break the graph into three regular shapes, a triangle, a...