Answer:
I think option d is the answer
Answer:
1.1⁰C
Explanation:
Width W = 5mm = 0.005
Thickness t = 1 mm = 0.001
K = thermal conductivity = 150W/m.K
P = q = heat transfer rate = 4W
We are to find the steady state temperature between the back and the front surface
We have to make these assumptions:
1. There is steady state conduction
2. The heat flow is of one dimension
3. The thermal conductivity is constant
4. The heat dissipation is uniform
We have:
∆t = t*P/k*W²
= (0.001m x 4W)/150x(0.005)²
= 0.004/0.00375
= 1.06667
This is approximately,
1.1⁰C
Thank you!
Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
(E. Call the hospital to take them away
