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pav-90 [236]
3 years ago
10

We are just a bunch of molecules that hang out together but don't touch each other. Is this believable? If this is the case, wha

t goes on in the body that allows us to "feel" when we touch something? How is this related to an object applying a force to another object?
Physics
1 answer:
inn [45]3 years ago
3 0
While the idea that we are a bunch of molecules that do not ever "touch" may seem counterintuitive, it is indeed the case. If we think about the structure of an atom, it is made of a nucleus composed of protons and neutrons, and surrounded by electrons. The electrons are moving at very high speeds in random directions all around the nucleus creating electromagnetic fields, defined by various wave functions. These electron clouds are essentially what do the "touching" between molecules.

When two molecules come in contact with each other, rather than solid particles physically touching, it is the two electron clouds of each molecule that are interacting and essentially repelling each other. If we try to bring two negatively charged species together, they will naturally repel one another. Therefore, when two electron clouds try to interact, the electrons will electrostatically repel each other. When this occurs, the force that each electron cloud exerts electrostatically on the other cloud is transmitted to their nuclei.This is how it is related to general objects applying forces to one another. The force of repulsion felt by the electron clouds is ultimately transmitted to the nerve cells in our skin which is perceived as pressure and a sense of touch, despite the fact that no particles are physically touching.
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3 years ago
Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the
Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

m=180 kg the mass of the raft

F=75 N the force applied by Sawyer

V=2 m/s the raft's final speed

V_{o}=0 m/s the raft's initial speed (assuming it starts from rest)

We have to find the time t

Well, according to Newton's second law of motion we have:

F=m.a (1)

Where a is the acceleration, which can be expressed as:

a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}} (2)

Substituting (2) in (1):

F=m\frac{V-V_{o}}{t-t_{o}} (3)

Where t_{o}=0

Isolating t from (3):

t=\frac{m(V-V_{o})}{F} (4)

t=\frac{180 kg(2 m/s-0 m/s)}{75 N}

Finally:

t=4.8 s

6 0
3 years ago
While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

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