Answer: D(t)= 50(4/5)^t
Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).
If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.
Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration, 
We have to find the final velocity at the end of the 100.0 m.
We know that

Using the formula






Hence, her final velocity at the end of the 100.0 m=6.5 m/s
<span>92.96 million mi..........</span>
The elastic potential energy of a spring is given by

where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.
The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:

In our problem, initially the spring is uncompressed, so

. Therefore, the work done by the spring when it is compressed until

is

And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
The answer is C. An occluded front.