Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)

Explanation:

count given by old sample = .97 disintegrations per minute per gram

count given by fresh sample = 6.68 disintegrations per minute per gram

Half life of radioactive carbon = 5568 years

rate of disintegration

dN / dt = λ N

In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .

Let initial no of radioactive be N₀ and after time t , number reduces to N

N₀ / N = 6.68 / .97

Now

λ is disintegration constant

λ = .693 / half life

= .693 / 5568

= .00012446 year⁻¹

Putting the values in the equation above

1.929577 = .00012446 t

t = 15503.6 years .  

why is cellular respiration important for homeostasis?

it changes food into energy that the organism can use.

which of the following is most likely to happen if a cell is not able to divide?

the cell will not be replaced when it is damaged.

explanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.