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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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Determine whether each of the following combinations of forces can be in equilibrium if their directions
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The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward
ch4aika [34]

The object's <u>initial velocity</u> is equal to -10.59\frac{m}{s}

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

So, isolating the initial velocity, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}

Finally, substituting and calculating, we have:

v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}

Hence, we have that the <u>initial velocity</u> of the object is -10.59\frac{m}{s}

Have a nice day!

7 0
4 years ago
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