Recent content by Banker

I myself am a student who recently finished high school and took physics to the highest level. I am the student who would ask those questions no one thinks about every now and then and would usually score high in exams and tests. You then have students who ask questions, but very basic...

Ahh thank you for the clarification! Yeah the Scottish qualifications board are subpar compared their English counterparts.

F = mrw^2 F = 0.2(0.35)(6)^2 Central force = 2.5 N Let Tension = T and the angle theta = x From the diagram, Tcosx = mg Tsinx = central force hence Tsinx = 2.5 Dividing the first eq by the second: Tsinx/Tcosx = 2.5/mg Simplifying gives tanx = 2.5/mg Weight, mg, = 0.2 x 9.8 (0.2kg is the...

I did that and I got 52 degrees as my answer. The mark scheme says 36 but I'm not too sure if they used the correct value for mass.

Homework Statement Question 2.b. from this paper - http://www.sqa.org.uk/files_ccc/PhysicsEQPAH.pdf Homework Equations let theta = x Tcosx = mg Tsinx = 2.5 (2.5 is the centripetal force) The Attempt at a Solution I rearranged the above equations and got 52 degrees as my answer. The...

Next time you attempt these questions, remember when a body is at rest, you can equate vertical and horizontal components of force to work out tension or whatever quantity you are trying to find.

I worked out the uncertainty in my gradient and y-intercept using the parallelogram method and my final answer is 164+ 3.07T, with the uncertainty in gradient being plus or minus 1.30 and the uncertainty in the y-intercept being plus or minus 44.8. So the closest I could get to the actual...

So how would I conclude my experiment? Because obviously my results don't fall in line with the standard. Is it scientifically accurate to call this experiment 'inaccurate'?

Ahh. I misinterpreted a graph I saw online and thought I had to convert into K. I plot another graph and this time I got the result v = 164 + 3.07T

I found the equation online(v = 331.4 +0.6T). That is similar to y=mx + c, the equation of a straight line. So I am going under the assumption that that is what my results should mirror.

I only measured the speed at five intervals between 25 and 40 degrees celsius. The line I found is Excel's best fit line, but yes, I recorded something similar to 286 for 40 degrees.

I calculated the speed of sound using v=d/t where d is the fixed distance between each of the two mics and t is the time taken for the sound wave to travel between them(fast timer was used). I plotted this and their respective temperatures(in K).

Alright, I plotted speed of sound against temperature in Kelvin. I tried to replicate the formula v = 331 +0.6T Mine went horribly wrong. I got v = -675 + 3.07T

Homework Statement I did an experiment involving the speed of sound at different temperatures. I placed two microphones at a fixed distance apart and I measured the time taken for a sound wave to travel between the two mics. I repeated this for different temperatures. I want to make a graph for...

@mfb Thanks for the reply I did a little experimenting and I combined the random uncertainty for each of my times with the digital reading uncertainty(using ∆w^2 = ∆x^2 + ∆y^2 + ∆z^2, x = random uncertainty, y = scale/digital reading uncertainty, z= calibration uncertainty ) and also did the...