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Draw a diagram

Let the temperature (as a function of time) be T = a sin (nt + b) where t is the time in hours. Set t = 0 as 4 AM and find the values of a, n and b. Substitute the temperature values into the LHS of the equation and solve for t in each case.

In a particle accelerator, a fast moving particle (with v = 0.99c) decays into two photons, of total energy of 6.2 × 10−27 J. What is the rest mass of the fast moving particle?

there's a way to do it without using a substitution: use the fact that a = eln(a) i.e. 2 = eln(2) So 2ln(x) = eln(x)*ln(2) then integrate the RHS to get the answer

i got the first one ok but i still need help with the second one

i got the first one ok but i still need help with the second one

thanks

yes i do need help

see attachment:

http://www.scribd.com/doc/49281225/The-Student-s-Guide-to-HSC-Physics http://tianjara.net/hsc/notes/HSC_Phys_Notes.pdf

Thanks!

Prove by induction: 5n + 6n - 1 is divisible by 10 for all integers n

thanks

Find the limit of (1 - cosx)/x2 as x approaches zero

6M/k (k + 3) = 6 [M/k][k + 3] But M/k might not be a whole number

Prove by induction that n(n + 1)(n + 2) is divisible by 6 When n = 1, n(n + 1)(n + 2) = 6, which obviously works Assume true for n = k, i.e. k(k + 1)(k + 2) = 6M Consider n = k + 1 (k + 1)(k + 2)(k + 3) = 6M/k (k + 3) <--- how do i prove this is divisible by 6?

1 + 3 + 5 + ... + (2n + 1) = n/2 (2a + (n-1)d) = n/2(2 + (n-1)2) = n(1 + n - 1) = n^2, which is divisible by n

Then it would just be (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is obviously divisible by 4

You can start by finding the sum of consecutive odd integers: 1 + 3 + 5 + ... + (2n - 1) Sn = n/2 (2a + (n-1)d) = n/2 (2x1 + (n - 1)x2) = n/2 (2 + 2n - 2) = n * n = n2 But this is not always divisible by 4

There must be something missing in question 1 otherwise it wont work