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DanielleElmas [232]
3 years ago
14

Which of the following causes your hands to feel warm when you rub them together?

Physics
2 answers:
Sholpan [36]3 years ago
8 0

Answer:

Friction

Explanation:

Hi there! I'm glad I was able to help you!

Friction would be your best option.

I chose friction as the answer after using the process of elimination.

Normal force, in a simple definition, is basically when nothing happens and an object is sitting on top of another, so the net force is zero. In our case, we're rubbing our hands together, so option A is out of the question.

Spring force, which is much different than normal force, requires the use of an actual spring. It's a force that compresses AND decompresses a spring.

Gravity is the force that keeps us (and other objects), from floating into the air. Gravity is like a giant human magnet!

Finally, the force of friction acts in the complete and total opposite direction of the way an object, any object (in this case, our hands), want to slide. When we rub our hands together, it kind of activates the molecules and stuff inside of them to turn on, or activate, just a little bit quicker than before your hands even touched together. The faster the molecules move, the higher the temperature becomes!

I hope this helped you! Comment below if you have any other questions! :)

tekilochka [14]3 years ago
7 0

Answer:

Friction

Explanation:

<em>Friction causes heat because it forces particles to move at a quicker speed and gain more energy.</em>

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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

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$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

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Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

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