I'm new here and can anyone help me out with this please :)

[Rimsha Noor1]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w03_qp_2.pdf
question 8 b. can anyone show me how to solve this along with an explanation pleasee?

 

[sweetjinnah]

5^y= k and find ,in terms of k, the value of 5^y+1
Alright this is the question.. wat i think is when u just imagine 5^2 = 25 this shows y=2 and k= 25
next equation 5^2+1 = 5^3 = 125 y=2 and k=25 when u multiply 25 by 5 u get 125 (i.e 5k)
hence the answer should be 5k. I hope u understand.

 

[Rimsha Noor1]

ohkay thanks alot

 

[sweetjinnah]

No prob!

 

[sj0007]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Physics (0625)/0625_s12_qp_32.pdf
q 11 b help me wid dis asap

 

[Rimsha Noor1]

(a) this is because the sample has almost completely decayed and the reason why the count rate increases at times, it might be due to background radiation.

 

[sj0007]

(a) this is because the sample has almost completely decayed and the reason why the count rate increases at times, it might be due to background radiation.

i need ans for part b........anyway thanx for considering and sparing ur precious time to help me......fikr na karo aaj kal mujhe acha bane ka dora pada hai!!!

 

[Rimsha Noor1]

ohh achaaa ..
half life of background plus source is about 2.8 days ( 52 counts)
count rate of only background is almost 14. so 52-14 = 38. use the graph to find the half life of this 38 ( source) , it would be around 1. so 2.8-1 = 1.8
that's the answer.

 

[sj0007]

bohat bohat shukriya aapka

ohh achaaa ..
half life of background plus source is about 2.8 days ( 52 counts)
count rate of only background is almost 14. so 52-14 = 38. use the graph to find the half life of this 38 ( source) , it would be around 1. so 2.8-1 = 1.8
that's the answer.