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# expotential growrth and decay watch

1. The question is a radioactive substance decays at a rate proportional to its mass m. Initially its mass was 2 x 10^-3 and after 50 days it was reduced to 1.5 x 10^-3 g. How long woul it take for the mass to reach half its original size.

So my first equation would be m = 2 x 10^-3 multiplied by e^-kt

I got K as 1/50 ln 4/3 which is right, kinda stuck on what to do next?
2. (Original post by SunDun111)
The question is a radioactive substance decays at a rate proportional to its mass m. Initially its mass was 2 x 10^-3 and after 50 days it was reduced to 1.5 x 10^-3 g. How long woul it take for the mass to reach half its original size.

So my first equation would be m = 2 x 10^-3 multiplied by e^-kt

I got K as 1/50 ln 4/3 which is right, kinda stuck on what to do next?
What is is half it's original size? hint: (2 * 10^(-3))/2.

Then equate that to your equation for m and solve for t.
3. (Original post by Zacken)
What is is half it's original size? hint: (2 * 10^(-3))/2.

Then equate that to your equation for m and solve for t.
Yeah got it, can you help me out with mechanics?

The question is Two cables are attached to a cable car with a mass of 450kg. They are at 30 degrees to the horizontal, find the tension in each cable. Right I drew out a sketch and I did T ( Sin 30 + sin 30 ) = 4410
and i got T as 4410 which is right! The next part is the cable now accelerates at 0.5ms, find the tension in each cable now.

I thought I'd find the resultant force which is F = MA so it was 225n. I then thought 225 = TCos 30 - T Cos 30 But my answer just cancels out any ideas?
4. (Original post by SunDun111)
Yeah got it, can you help me out with mechanics?

The question is Two cables are attached to a cable car with a mass of 450kg. They are at 30 degrees to the horizontal, find the tension in each cable. Right I drew out a sketch and I did T ( Sin 30 + sin 30 ) = 4410
and i got T as 4410 which is right! The next part is the cable now accelerates at 0.5ms, find the tension in each cable now.

I thought I'd find the resultant force which is F = MA so it was 225n. I then thought 225 = TCos 30 - T Cos 30 But my answer just cancels out any ideas?
Why are you subtracting them from each other?

From my sketch, I resolve horizontally to get T cos 30 - T cos 30 = 0. So no horizontal force.

Resolving vertically I get T sin 30 + T sin 30. Which means that 2Tsin 30 vertically upwards.
5. (Original post by Zacken)
Why are you subtracting them from each other?

From my sketch, I resolve horizontally to get T cos 30 - T cos 30 = 0. So no horizontal force.

Resolving vertically I get T sin 30 + T sin 30. Which means that 2Tsin 30 = 225?
I subtracted because to get the resultant force i did Forces right - forces left? the answers are 4540N and 4280N
6. (Original post by SunDun111)
I subtracted because to get the resultant force i did Forces right - forces left? the answers are 4540N and 4280N
Yes, you resolved horizontally and proved there was no force horizontally. Good.

Now, resolve vertically, what is force acting vertically upwards? (hint: tension) What about vertically downwards? (hint: weight)

Now equate the net force upwards/downwards to .
7. (Original post by Zacken)
Yes, you resolved horizontally and proved there was no force horizontally. Good.

Now, resolve vertically, what is force acting vertically upwards? (hint: tension) What about vertically downwards? (hint: weight)

Now equate the net force upwards/downwards to .
Ok but im a bit confused, in F = MA is the F the resultant force horizontally, or both combined?
8. (Original post by SunDun111)
Ok but im a bit confused, in F = MA is the F the resultant force horizontally, or both combined?
F is the resultant force. Since there is no horizontal force, the resultant force is vertical.
9. (Original post by Zacken)
F is the resultant force. Since there is no horizontal force, the resultant force is vertical.
ok thanks so i'd get 225 + T (Cos 30 + cos 30) = 4410? But i'd get the same answer for T here?
10. (Original post by SunDun111)
ok thanks so i'd get 225 + T (Cos 30 + cos 30) = 4410? But i'd get the same answer for T here?
Why are you still using cos if you're resolving vertically...?
11. (Original post by Zacken)
Why are you still using cos if you're resolving vertically...?
Oh yeah.. Doing it, with Sin's i got 4410 - 225 / sin 30 + sin 30
And my answer was 4185, but that dosent work does it?
12. (Original post by SunDun111)
Oh yeah.. Doing it, with Sin's i got 4410 - 225 / sin 30 + sin 30
And my answer was 4185, but that dosent work does it?
I would agree with that answer, so either I'm missing something blatantly obvious/being dumb or the given answer is wrong.
13. (Original post by Zacken)
I would agree with that answer, so either I'm missing something blatantly obvious/being dumb or the given answer is wrong.
Maybe i told the question wrong, the 0.5 ms of acceleration acts to the right? Does that make any difference?
14. (Original post by SunDun111)
Maybe i told the question wrong, the 0.5 ms of acceleration acts to the right? Does that make any difference?
15. (Original post by Zacken)
i will when i get time
16. (Original post by SunDun111)
i will when i get time
Okay.
17. (Original post by Zacken)
Okay.
Thought you can help me out, generally with M1, im fine with the questions on forces/ newtons laws unless the particle is on a slope, e.g. a question is a child slides down a steep straight slide that is incline 60 degrees to the horizontal, the child has a mass of 30kg and the coefficiant of friction between the slide and child is 0.6 If i want to calculate the normal reaction force how do i go about doing this? it puts me of that the slide is inclined 60 degrees, ?
18. (Original post by SunDun111)
Thought you can help me out, generally with M1, im fine with the questions on forces/ newtons laws unless the particle is on a slope, e.g. a question is a child slides down a steep straight slide that is incline 60 degrees to the horizontal, the child has a mass of 30kg and the coefficiant of friction between the slide and child is 0.6 If i want to calculate the normal reaction force how do i go about doing this? it puts me of that the slide is inclined 60 degrees, ?
Draw a sketch and then resolve perpendicular to the slope. There will be a component of weight acting perpendicular into the slope and the reaction will act perpendicular out of it. Equate those two.

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