What happens to the heat energy when you increase the length of an object

1 answer:

8 0

Answer: When the temperature of an object increases, the average kinetic energy of its particles increases. When the average kinetic energy of its particles increases, the object's thermal energy increases. Therefore, the thermal energy of an object increases as its temperature increases.

You might be interested in

Research the potential for a large earthquake off the coast of Oregon. What are two of the biggest concerns with a large earthqu

Sphinxa [80]

There is a 33% chance of a large earthquake off the coast of Oregon.

The two main concerns of a large earthquake are Gas fires and Tsunami .

Explanation:

Everytime a large earthquake hits in any other part of the world, We find a small earthquake hitting off the coast of Oregon.

So scientists feel that as United States is prone to Earthquakes, the people have to be prepared for such eventualities.

The main concern for this large earthquake is, that it might lead to gas leakages at homes that can start a gas fire, which might spread. So people have to be careful and check on any gas leakages.

Such large earthquakes can also lead to tsunami waves. So people have to take precautions and evacuate, if they are near the coastline.

6 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$