Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Explanation:
Average velocity = displacement / time
v = (25 m − 2 m) / 4.5 s
v = 5.11 m/s
1) Size of the image: 2 cm
In order to calculate the size of the image, we can use the following proportion:

where
p = 80 m is the distance of the tree from the pinhole
q = 20 cm = 0.2 m is the distance of the image from the pinhole
= 8 m is the heigth of the object
is the height of the image
By re-arranging the proportion, we find

2) Magnification: 0.0025
The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

so, in this problem we have

You have to do the math of each and see which one adds up to 66.5