Radiation is different than the other types of heat transfer because

Explanation:

Heat transfer by radiation tends to be minimal in many everyday instances. This is in part, due to the 4th power exponentiation of the heat transfer equation that governs radiative heat transfer.

For instance, a bar of metal of 75 ºF will interact with your body, from a distance, when your body is at 98 ºF. But the temperature difference of 14 ºF will be so low, that it is difficult to feel or detect. However, if that bar was heated to 120º F and set close to you, you may begin to feel the heat from radiation. Then, if the bar was heated to the point that it began to glow red, several hundred degrees, not only would you feel it, but it may make you uncomfortably hot very quickly. If it was then heated to being white hot, you would almost certainly be burned by it, not by touching it, but simply by the radiative heat transfer. Again, its the 4th power exponent that really makes radiative heat transfer a powerful force when the temperature increases to a high level.

Another example from a text that I once used set an example of a person in a room with their body temperature at 98 ºF. In one case, the walls of the room surrounding the body are said to be 50º F, but in the other case, the walls are said to be 100º F. In both cases, the air in the room is said to be 75ºF. The question is asked: why is the person colder in the room with 50 ºF walls, when the air is the same temperature in both cases.

The answer is that the 100 ºF walls are acting to slightly warm the body, whereas the 50º F walls are actually taking heat away from the body by way of radiative heat transfer. You can do the equations to see exactly how much the different temperatures of the walls will effect this situation.

One final example - if you look into a thermos, you will see a highly reflective liner inside of the thermos. This is a radiant barrier, and it is put into the thermos to keep the liquid inside from losing heat by way of radiation. The reflective film will transfer back most of the heat that would otherwise be lost by radiating to the outer surface of the thermos, after which it would conduct through the side of the thermos. So a thermos provides a good example of a solution for heat transfer in the form of radiation.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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it is a solid

explanation: