Author | Maths and Physics (Problems) |

*http://prntscr.com/6kcuiw*
It looks as if we have an area of 6. |

#29
4 |

*- I suppose the answer to post 29 is 16. I tried to solve it in my head first but it became too messy.*
- Nah that's wrong ;) And well as I said, it's easier than it looks :)
Opps, I obviously forgot the 1,3,4 product. I would like to change my answer to 4.
You make it sound like something that should be easy to solve in your head. I'm unfortunately far from being able to do that. I was even wrong when using pen and paper. :D |

Yea, 4 is the right answer :)
I think the only small trick that people might get caught into is being careful with when taking - sign, and when the + sign :)
How did you solve the area problem? |

just trying the coke problem
assume q is heat change
mcq=mcq+mL+mcq
m*4.2*(60-4)=1*4.2*(0--18)+1*334+1*4.2*(4-0)
m*235.2=426.4
m=1.813L
I am assuming coke has the same specific heat capacity as water and same density as water
p/s just trying |

Post 44: *How did you solve the area problem? *
Just extract a, b and c from the given equation. Substitute parts to make it less cumbersome. Insert them into Heron's formula. Watch them cancel each other out.
Or: Brute force it assuming whole numbers.
Post 45:
Well done Lord spiral-doom. I have some objections to answering with 4 digits when your input data only got 2, but your answer is correct. |

for Lord STB:
Extract as in? I'll try again though.
Well yea, Gonlador gave me a solution involving *Brute force it assuming whole numbers.*. |

Okay my brother found another solution, I don't know if it's shorter than yours but works for me :)
Taking 20 to RHS and then splitting it into -2 + 4 -3 + 9 - 4 + 16, we get
(2a-2) + 4 + (3b-3) + 9 + (4c-4) + 16 = (2x2)Sqrt(2a-2) + (2x3)sqrt(3b-3) + 2x4sqrt(4c-4)
Thus we can take these terms to LHS to get 3 square numbers -
[sqrt(2a-2) - 2]^2 + [sqrt(3a-3) - 3]^2 + [sqrt(4c-4) - 4]^2 = 0
Each square has to be zero, so we get a = 3, b =4 and c =5
:)
Nice question I must say, whoever made it.
So no question pending as of now? I'll get some more! :) |

Sorry for triple post. Here are some more :)
Have only posted Maths till now, sorry for that, will soon post some Physics :)
http://prntscr.com/6knchr
http://prntscr.com/6kncp3 - Have almost done this, am stuck at the step which involves - Spoiler Alert - Fermat's Point, which I was unable to calculate. |

Hmm, 49 part 2..
Getting Emin as 19.5
Not sure how to represent that in asked form.. |

Part 1 answer is CosA |

for The One Ring:
Part two I haven't got the answer but I got the condition that it lies between 15 and 17 ;)
Part one that's the write answer, but can you explain why? I was unable to solve it but this is the right answer :) |

You use that Tan(a-b) formula.
Just pure simplification. |

ayan the answer to that probability questian is 22% |

a bit of a mistake |

Problems unsolved till now - #49 Part 1.
Here's another -
A chain hangs on a thread and touches the surface of a table by its lower end. Show that after the thread has been burned, the force exerted on the table by the falling part of the chain at any moment is twice the weight of the part already resting on the table.
So 2 unsolved questions now :) |

More physics, 3 unsolved now :)
http://prntscr.com/6l26zw
What force (F) should be applied for the block to move -
1)Upwards with an acceleration of 'a'
2)Downwards with an acceleration of 'a'
The weight of the block (Triangle) is G.
Please give solution, thank you! |

Post 57
*What force (F) should be applied for the block to move -*
1)Upwards with an acceleration of 'a'
2)Downwards with an acceleration of 'a'
Assumptions: G is not a weight but a mass of G kg.
The system is in a gravity g m/s^2.
1) F = gG/2 + aG/2
2) F = gG/2 - aG/2 |

Post 56 * A chain hangs...*
Assumptions:
Each link of the chain got a mass m and is in a gravitational field g. When falling, each link is decelerated linearly until next link reaches the table. Each link got the length l. The links are numbered 1 to N and we chose to study link number n.
The force applied by n links resting at the table is:
f1 = gmn
The energy stored in a link falling from height nl is:
e = gmnl
The energy must be released when decelerating the distance l:
f2 = e/l = gmnl/l = gmn
My conclusion is that the statement in the question is wrong. The force coming from deceleration of the chain is equal to the force coming from part resting on the table.
(I really dislike "weight". It does not have any place i physics because it can be interpreted as either force or mass.) |

Post 48 * Taking 20 to RHS and then splitting it into -2 + 4 -3 + 9 - 4 + 16, we get *
Without saying how and why you did this spit, you are not actually providing an answer.
Using Heron's formula gives you a huge expression and I filled a page before I even got it halfway decent. I quit there because it didn't seem like a fun solution. I also suppose that the result wouldn't help much because the problem likely got an infinite number of valid solutions. A single equation with multiple variables usually do.
Using brute force under the assumption that all numbers must be whole during the entire calculation makes it very easy. Look at:
2a-2
It must be a square for the square root to be whole. So we just have to look for an 'a' that give a result of 1, 4, 9, 16, 25.... It is rather obvious that a=3.
Use the same logic for b and c. Problem solved. |