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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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A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t
notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, F_f, backward

The frictional force can be written as

F_f = \mu mg

where

\mu=0.03 is the coefficient of kinetic friction

m=150 kg is the mass of the motorbike

g=9.8 m/s^2 is the acceleration of gravity

Therefore the net force is given by

\sum F = F - F_f = F - \mu mg

And substituting, we find

\sum F=250 - (0.03)(150)(9.8)=205.9 N

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

\sum F = ma

where

m is the mass

a is the acceleration

In this problem, we have

\sum F = 205.9 N is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

a=1.37 m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s

Learn more about Newton's laws of motion and accelerated motion:

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6 0
3 years ago
What is the effective resistance of a car’s starter motor when 150 A flows through it as the car battery applies 12.0 V to the m
QveST [7]

Answer:

From ohms law,

V=IR

R=V/I =12.0/150 =0.08 ohm.

8 0
3 years ago
How do you label <br> Time= energy • power
Lana71 [14]

First of all, that equation is not correct, which may be the reason
that you're having trouble assigning units to the quantities.

Power is defined as [energy / time],  so  [Energy] = [ power x time ],
and
         [Time] = [ energy / power ].

Unit-wise, these equations are correct just as they appear here,
with no proportionality constants or conversion factors, when ...

[ Power ] = watts
[ Energy ] = joules
[ Time ] = seconds .
 

5 0
3 years ago
Who exerts more pressure? a) A girl of 50 kg, wearing heels with an area of 1 cm2. b) An elephant of 4000 kg with foot area of 2
Mrrafil [7]

Answer:

The girl exerts more pressure.

Explanation:

Pressure can be defined as the force exerted normally or perpendicularly per unit area.

i.e P = F/A

<u>Girls</u>

Area of the heel = 1cm² = 10^(-4) m²

Force = mg = 50 × 10 = 500N

Pressure =

\frac{500}{10 ^{ - 4} }

= 5 \times  {10}^{6}

<u>Elephant</u>

<u>Area</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>0</u><u>cm</u><u>²</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>x</u><u> </u><u>1</u><u>0</u><u>^</u><u>(</u><u>-</u><u>2</u><u>)</u><u>b</u><u> </u><u>m</u><u>²</u>

<u>Force</u><u> </u><u>=</u><u> </u><u>mg</u><u> </u><u>=</u><u> </u><u>4</u><u>0</u><u>0</u><u>0</u><u>0</u><u>N</u>

<u>Pressure</u><u> </u><u>=</u><u> </u>

<u>\frac{40000}{2.5 \times  {10}^{ - 2} }</u>

<u>= 1.6 \times  {10}^{6}</u>

5 0
3 years ago
The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb
Delvig [45]

Answer:

v =  \sqrt{20h}

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (<em>divide by m throughout</em>)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (<em>substitute g=10</em>)

v²= 20h (<em>×2 on both sides</em>)

v= √20h (<em>square root both sides</em>)

4 0
3 years ago
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