Recent content by Sarlacc

I'm not entirely sure what you're asking but when you for example, touch something, it will activate the nerves in your hand which send an electrical signal through nerve fibres into your brain where the information is processed. If you're asked to show that this is true or something you could...

Well that's pretty much your first "consideration" answered. In reality a bungee cord can only stretch as far as its tensile strength allows. Since it asks you to estimate you might have to assume the Chancellor weighs the average human weight (I think it's 70kg) which means you can get the...

Well gravity will accelerate all objects at the same speed regardless of mass. I'm not entirely sure I'm going the right way with this but you have the initial velocity (0), the final velocity (also zero because at max extension he'll pause for a second), the distance and an acceleration. So...

This actually varies between countries and even states. For example in Victoria, Australia, x is used for distances. However Queensland uses s. I can't remember where I saw it but there's a webpage out there that lists the most common ones.

Huh. That's pretty cool. I think it's just a coincidence though. Maybe.

Turns out I was right! I knew it'd come down to me not noticing something simple, so thanks for pointing out how that a/t graph didn't make sense. Thanks for the help guys! I'd edit the title and stick this update in my last point but looks like I can't. But I just wanted to let anyone...

I have class tomorrow. Hopefully I'll find out the final, definite, for sure, no-really answer then.

Wha? The question states that the car accelerates from rest (0) over 400m for 19 seconds. And then it decelerates at a constant rate for 5.1 seconds. OHHHHHHHHHHH I think I see what you're saying. I'll try asking him that tomorrow.

That's two triangles?

A friend of mine did it with a graph. at 19s the acceleration is 2.22. then it goes down to zero which takes 5.1 seconds. half-base*height for the first side: (1/2)(19)*2.22= 21 and then (1/2)(5.1)*2.22=5.66 Which is of course 26-ish in total. That works. And my teacher said it's the right...

"A car accelerates from rest over 400 metres in 19 seconds. The driver then brakes and the car stops in 5.1 seconds with constant deceleration." Part one of the question is "calculate the acceleration for the first 400m" which went just fine: x=ut+(1/2)at^2 400=180.5a a=2.22 m/s Now the...