Yes dude.did u calculate the distance for subsequent when string broke ??
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Yes dude.did u calculate the distance for subsequent when string broke ??
what was the tension ?Yes dude.
hey can u show me how u solved for tension and acceleration in the first place?I just solved it and I'm getting 1.9524m for my length of string.
T=1.0666...
a=2/3 (for 0.7m), then 2.8 (after the string breaks)
v (when string breaks) = *root* (2 x 2/3 x 0.7) = 0.96609
then it increases to 2
So
2^2 = 0.96609^2 + 2(2.8)s
s = 0.5476
So the length of string will be 2.5-0.5476 = 1.9523
yes your tension and acceleration is right !! i was scared that my tension and acc is wrong ....hey can u show me how u solved for tension and acceleration in the first place?
no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??yes your tension and acceleration is right !! i was scared that my tension and acc is wrong ....
same as yours !!!! dont worry our answers are 100% correct !!!no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??
add me on fb !! [email protected]no i dont think so, the other guy in this thread posted some different value of that. What was yours btw??
Okayy..add me on fb ... [email protected]
same as yours !!!! dont worry our answers are 100% correct !!!
yes .. thatss rightHaha I'm not wrong
Mass of A (on the ramp) was 0.5kg, mass of B was 0.1 kg
The ramp was 2.5m long and 0.7m high, so that means that sin(Theta)=7/25
Component of A down the ramp will be mgsin(theta), and hence you get 1.4N
Use newtons second law, F=ma
1.4 - T = 0.5a
T - 1 = 0.1a
Add to get rid of T,
0.4 = 0.6a, hence a = 2/3 and T = 1.06666....
42 or 43 ... 35 m was from the floor in q2 ... right ??What will be the GT any idea?
I gave variant 42. I think it was moderately easy.
OMG. This question is a disaster for me.Haha I'm not wrong
Mass of A (on the ramp) was 0.5kg, mass of B was 0.1 kg
The ramp was 2.5m long and 0.7m high, so that means that sin(Theta)=7/25
Component of A down the ramp will be mgsin(theta), and hence you get 1.4N
Use newtons second law, F=ma
1.4 - T = 0.5a
T - 1 = 0.1a
Add to get rid of T,
0.4 = 0.6a, hence a = 2/3 and T = 1.06666....
what was ur tension and acceleration bro ??OMG. This question is a disaster for me.
I don't know.. I'm losing that 8 marks for sure then. :'(
main marks are for main logic applied to the question !!! yes it is possibleIs there error carried forward in question 2 because in i) I wrote all the steps correct but i suppose i put the values incorrect on the calculator so ii) also will be wrong .i'm hoping that i'll lose 1 mark in this question only.Is it possible?
What will be the GT any idea?
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