Been at this for hours..

the temperature ,P, of a conductor at time, t, is given by

P=P^0[1-e-^-t/T]

Where P^0 is the initial temperature and T is a constant.

Express T in terms of P,P^0 and t and determine its value when p^0 = 190, P= 20 and t = 40

Any help would be much appreciated...Thanks

2. $\dfrac{P}{P_0} = 1-e^{-t/T}$

Subtract 1 from both sides: $\dfrac{P}{P_0} -1 = -e^{-t/T}$

Multiply through by -1: $1 -\dfrac{P}{P_0} = e^{-t/T}$

Can you finish off? (use logs)

3. Thanks very much for the help and the hint....

Hopefully I have it sussed!

4. If you want to check your answers I've put mine in a spoiler

Spoiler:
$T = -\dfrac{t}{\ln \left(1 - \dfrac{P}{P_0}\right)}$

$T = -\dfrac{40}{\ln \left(1 - \dfrac{20}{190}\right)} \approx 359.6$

Your answer will have whatever unit $t$ is in

5. Thanks for the spoiler, Matches what I had thankfully!

I am stuck with final question - After this hopefully thats me done with algebra!!

Show using factor theorem,that 2x-1 is a factor of...

2x^4-x^3-6x^2+5x-1

and hence express 2x^4-x^3-6x^2+5x-1 as a product of a linear & cubic factor

6. let $f(x)=2x^4-x^3-6x^2+5x-1$
If $2x-1$ is a factor, $f(\frac{1}{2}) = 0$. That is to say, when $x = 0.5$, the quartic should equal 0.
Then, once you've proven that, take out a factor of $(2x-1)$.
$
2x^4-x^3-6x^2+5x-1=(2x-1)(x^3... ....+1)$

Can you find the $x^2$ and $x$ coefficients?

7. Thanks, so would I be right in saying..

f(0.5)=2(0.5)^4-(0.5)^2-6(0.5)^2+5(0.5)-1??

8. Originally Posted by scott6162
Thanks, so would I be right in saying..

f(0.5)=2(0.5)^4-(0.5)^3-6(0.5)^2+5(0.5)-1??
Just a small typing mistake, but I'm confident you have the right idea. If that is equal to 0 (and you'll have to test it) then you've proven that $(2x-1)$ is a factor. You then need to factorize the expression, and I've started you off in my above post.