作业帮shows the temperature of a cup of coffee during a ten minute period.(Each of your answers in (a) – (c) should have the units "degrees per minute.")(a) What was the average rate of cooling from minute 0 to minute 10? (b) What was the ave
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/06 04:59:42
shows the temperature of a cup of coffee during a ten minute period.(Each of your answers in (a) – (c) should have the units "degrees per minute.")(a) What was the average rate of cooling from minute 0 to minute 10? (b) What was the ave
shows the temperature of a cup of coffee during a ten minute period.(Each of your answers in (a) – (c)
should have the units "degrees per minute.")
(a) What was the average rate of cooling
from minute 0 to minute 10?
(b) What was the average rate of cooling
from minute 7 to minute from
minute 8 to minute
(c) What was the rate of cooling at
minute at minute
(d) When was the cold milk added to
the coffee?
shows the temperature of a cup of coffee during a ten minute period.(Each of your answers in (a) – (c) should have the units "degrees per minute.")(a) What was the average rate of cooling from minute 0 to minute 10? (b) What was the ave
a) It takes 10 minutes for the temperature to fall from 150 to 10. So the average cooling rate is (150-70)/10 = 8 degrees per minute;
b) Same method. From minute 7 to 8, it's 80 to 75, so (80-75)/(8-7) = 5;
From minute 8 to 9, it's 75 to 70, so (75-70)/(9-8) = 5;
c) At minute 8, to be exact you have to draw a tangent at the point on the curve, I'm sure you can do this by yourself, then just extent it long enough to measure its tangent, that's the rate.
Another approximate method is just to take the average of the two numbers you got in question b) which is 5.
At minute 2, same methods apply. The approximate way gives you 10.
d) Apparently, it's at minute 6, cas the way how the temperature lowers exhibits a turning point here.