I understand that taking logs of a negative number isn't possible because no number to any power produces a negative number. But why not a negative base?
Say, log-10(100) = 2. Rewritten, -102 = 100, which is accurate.
You could suggest that you may as well just ignore the negative because -x2 =...
1. Homework Statement
The light from a bulb shines equally in all directions. If 20W of light is given off, what will the intensity be 12m from the lamp to 2 significant figures? (Consider the shape of the region illuminated if the light hits this surface after travelling 12m in all...
What was your answer? What did you plug in for b2 - 4ac? Your approach is correct, but I'd draw a graph to be sure. (Hint: you know the parabola is negative, and you know the y-intercept is -10.)
Right. At 30s drag from air resistance is less than the force of weight, so Baumgartner is accelerating.
At 45s drag = force of weight, terminal velocity.
At 70s drag is greater than force of weight, hence deceleration.
Not sure if that's exactly the right terminology.
Area under graph =...
Right, sometimes the obvious answer is the right one. 95 * 7 = 665N.
So questions that ask you something like this where you end up calculating mg for upwards force are neglecting air resistance.
1. Homework Statement
a)ii) Calculate the upward force F acting on Baumgartner at this point. (3 marks)
total mass of Baumgartner = 95 kg
g = 9.8 ms-2
b) Describe the shape of the graph between 30s and 70s. Explain the velocity changes in terms of changes in the air through which Baumgartner...
1. Homework Statement
The curve C with equation y = f(x) passes through the point (5, 65).
Given that f'(x) = 6x2 -10x - 12,
a) use integration to find f(x)
b) Hence show that f(x) = x(2x+3)(x-4)
3. The Attempt at a Solution
I have no problem with this question, except it seems the given...
Helpful, thanks. This answers 1) and 2), still not sure about 3) because the final velocity would still be 15m/s. That wasn't questioned earlier.
Also, I take it you're happy with my answer and working?
I'm familiar enough with suvat, I just wasn't expecting to use it in this question, partly because of its mark count and partly because this is the first instance I've needed to use it outside of projectile motion.
Thanks for the explanation. It's beginning to make sense, though I'm still...
I apologise in turn. I was cavalier about it, because I thought that this was getting seriously overblown for a 1 mark. But I've just learnt that this question has been taken from a practice guide, not an exam paper. It would probably pass for more in an exam.
Anyway, I thought about this some...
The distance s travelled after 1 second is s = 0.5 * a
The distance s travelled after 2 seconds is s = 2u + 0.5 * a * 22
2u + 0.5 * a * 22 - 0.5 * a = 2u + 1.5a = 15 (second distance - first distance)
Is that what you meant?
This is really too much for a 1 mark.
Throwing a cheap blow over a technicality? It was bad enough that you were useless.
The only variable would be acceleration, but I don't know the initial velocity between t = 1 and t = 2?
15 = u + 0.5 * a (for 2nd second, t = 1 so neglected)
s = 0.5 * a (for 1st second)
I need to understand...
I'm baffled. They will meet at the same time, which is 16s. Therefore plugging this into the acceleration should give me the same distance of 192m. Otherwise the units do not work.
Maybe, but now I have to understand why all my other working was wrong. :P Why was the 7.5 wrong?
I edited my post just before yours, is that progress?
It's 0, because I'm looking at the first second as you suggested, and the car starts at rest.
If you were to look at the second second, you'd have s = v1 + 0.5v1 = 1.5v1, assuming that v1 = a.
EDIT: Ah, s = 15. So v1 = 10m/s?
Just a note, I didn't think the question would be this long-winded...
This seems to go against the fundamentals.
If the motorcycle travels a certain distance after time t, t should give you that distance when plugged into ms-2. The seconds squared cancels out. Otherwise the units don't make sense.
It's t2 multiplied by the acceleration of the motorcycle. Since the acceleration is m/s2, shouldn't that function give you a value of some significance?
1. Homework Statement
The carriage and its passengers start at rest at A. At B, the bottom of the ride, the maximum speed of the carriage is 20ms-1. The vertical distance between A and B is 110m. The length of the track between A and B is 510m. The mass of the carriage and the passengers is...