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### Author Topic: re: energy producing experiments  (Read 121558 times)

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #285 on: December 16, 2021, 03:26:35 PM »
Relationship between momentum and kinetic energy

ke = 1/2 * m * V^2            kinetic energy = 1/2 * mass * velocity squared
ke = 1/2 * m * v * v           kinetic energy = 1/2 * mass * velocity * velocity
p = m * v                            momentum = mass * velocity
ke = (m * v) * (1/2 * v)      kinetic energy = momentum * 1/2 * velocity
An object's kinetic energy due to its motion is equal to the product of its momentum
and 1/2 its velocity.    ke = p * (1/2 * v)
also
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity.    p = ke / (1/2 * v)

#### Delburt Phend

• Full Member
• Posts: 208
##### Re: re: energy producing experiments
« Reply #286 on: December 16, 2021, 04:44:05 PM »
The collision does not take 4 sec; I am pointing out that the quantity of Ft to make the initial motion is equal to the quantity of force * time needed to produce the final motion.

You mention "moment of inertia" is this moving in a circle?

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #287 on: December 16, 2021, 09:12:57 PM »
@ Delburt Phend

? 1.  What is the momentum of 3 Newtons of force    times   25 seconds of time ?
? 2. What is the difference between an apple ?

#### Delburt Phend

• Full Member
• Posts: 208
##### Re: re: energy producing experiments
« Reply #288 on: December 16, 2021, 09:32:37 PM »
The apple will have 75 kg m/sec of momentum.

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #289 on: December 17, 2021, 12:27:56 AM »
Relationship between momentum and kinetic energy

ke = 1/2 * m * V^2            kinetic energy = 1/2 * mass * velocity squared
ke = 1/2 * m * v * v           kinetic energy = 1/2 * mass * velocity * velocity
p = m * v                            momentum = mass * velocity
ke = (m * v) * (1/2 * v)      kinetic energy = momentum * 1/2 * velocity

An object's kinetic energy due to its motion is equal to the product of its momentum
and 1/2 its velocity.    ke = p * (1/2 * v)
also
An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity.   p = ke / (1/2 * v)
... ... ... ... ... ... ...
... ... ... ... ... ... ...
scenario 0

force = 3 N
time  = 25 seconds
mass = zero

force  = mass times acceleration
momentum = mass times velocity

momentum = zero
... ... ... ... ... ... ...
scenario 1
force = 3 N
time = 25 seconds
mass = 15 kg

force  = mass times acceleration
momentum = mass times velocity

momentum = 15 kg * 5m/s ..  "75 momentum"

kinetic energy = 1/2 * m * v^2

velocity = 5 m/s
because
1 Newtons of force will cause 1 kg to accelerate to 1 meter / second in 1
second  (by definition)
and
1 Newton of force will cause 15 kg to accelerate to 1/15 (0.0666666666666666
6666666666666667)  meter / second in 1 second
and
3 Newtons of force will cause 15 kg to accelerate to ( 0.0666... times 3) which is
0.2 meter/second in 1 second
and
3 Newtons of force will cause 15 kg to accelerate to, 0.2  times 25, in 25 seconds
and
this is equal to 5 meters per second velocity (0.2 * 25 = 5)

A 15 kg mass moving at 5 m/s has 187.5 joules of  kinetic energy
Ek = 1/2 * Mass * Velocity ^2
kinetic energy = 1/2 * 15 kg * 5m/s^2 ... 1/2 * 15 kg * 25 =  187.5 joules

5 m/s ^2 = 25
15 kg * 25 = 375
375 / 2 = 187.5 joules

An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity.    p = ke / (1/2 * v)
187.5 joules / 1/2 * 5 m/s   as in  187.5 joules / 2.5 m/s  as in  "75 momentum"

momentum = mass * velocity
15 kg of mass * 5 m/s =                                                             "75 momentum"

Momentum = force * time
3 Newtons  * 25 seconds =                                                          "75 momentum"
... ... ... ... ... ... ...
... ... ... ... ... ... ...
scenario 2

force = 3 N
time  = 25 seconds
mass = 4 kg

force  = mass times acceleration
momentum = mass times velocity

momentum = zero
... ... ... ... ... ... ...
force = 3 N
time = 25 seconds
mass = 4 kg

momentum = 4 kg * 0.75 m/s = 3

kinetic energy = 1/2 * m * v^2

velocity =  18.75  m/s
because
1 Newtons of force will cause 1 kg to accelerate to 1 meter / second in 1
second  (by definition)
and
1 Newton of force will cause 4 kg to accelerate to 1/4 (0.25)  meter / second in 1 second
and
3 Newtons of force will cause 4 kg to accelerate to 0.25 times 3 which is
0.75 meter/second in 1 second
and
3 Newtons of force will cause 4 kg to accelerate to 0.75  times 25, in 25 seconds
and
this is equal to 18.75 meters per second velocity (0.75 * 25 = 18.75)
... ... ... ... ...
A 4 kg mass moving at 18.75 m/s has  ?  joules of  kinetic energy
Ek = 1/2 * Mass * Velocity ^2
kinetic energy = 1/2 * 4 kg * 18.75 m/s^2 ... 1/2 * 4 kg * 351.5625 = 703.125 joules

18.75 m/s ^2 = 351.5625
4 kg * 351.5625 = 1406.25
1406.25 / 2 = 703.125 joules

An object's momentum is the quotient of the division of its kinetic energy
by 1/2 of its velocity.    p = ke / (1/2 * v)
703.125 joules / 1/2 * 18.75 m/s  or stated as 703.125 joules / 9.375 m/s or stated
as in                                                "75 momentum"

momentum = mass * velocity
4 kg of mass * 18.75 m/s =           "75 momentum"

Momentum = force * time
3 Newtons  * 25 seconds =           "75 momentum"

Momentum = force * time

cool !

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #290 on: December 17, 2021, 04:14:12 PM »
Floor quote: "The kinetic energy the object has (due to its motion), is the same amount of energy as was needed to cause its acceleration it to that specific velocity."

How is this statement consistent with ballistic pendulums?

Good explanation here

http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

"The ballistic pendulum is a classic example of a dissipative collision in which conservation
of momentum can be used for analysis, but conservation of energy during the collision
cannot be invoked because the energy goes into inaccessible forms such as internal
energy. After the collision, conservation of energy can be used in the swing of the
combined masses upward, since the gravitational potential energy is conservative."

#### seychelles

• Hero Member
• Posts: 931
##### Re: re: energy producing experiments
« Reply #291 on: December 17, 2021, 08:00:06 PM »
ALIAS TINMAN AND OR OZSOLAR  POWER HAS JUST COME UP
WITH A SUPER GREAT IDEA AND OR INVENTION. CHECK IT OUT.

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #292 on: December 18, 2021, 04:26:08 AM »

Guys check it out

It looks promising

A cross between the Flynn brothers ? and the MagMirror engine ?
Kenneth C. Kozeka, Ph.D. KEDRON CORPORATION  (EDEN PROJECT) ?

But definitely new,  and all his own invention ultimately.

P.S.
Apologize / don't mean to be stepping on your topic

#### lota

• Jr. Member
• Posts: 66
##### Re: re: energy producing experiments
« Reply #293 on: December 18, 2021, 10:54:52 AM »
Hi, very good. But how high is the COP?

#### seychelles

• Hero Member
• Posts: 931
##### Re: re: energy producing experiments
« Reply #294 on: December 18, 2021, 11:30:00 AM »
LOTA HOW MANY MILLIONS OF DOLLARS HAVE YOU IN YOUR BANK ACCOUNT.
THEN WE CAN TALK.

#### Floor

• Guest
##### Re: re: energy producing experiments
« Reply #295 on: December 18, 2021, 02:52:42 PM »
Back on topic ...

How much energy is delivered to an object when 1 Newton of force
is applied to it causing that object to accelerate (given that 100% of the
energy applied is transferred as acceleration) ?

This depends upon the length of time the Newton of force is applied for.

See the "Magnets Motion and Measurement part 1-9 PDF file, attached below,
for detailed explanation.

#### Delburt Phend

• Full Member
• Posts: 208
##### Re: re: energy producing experiments
« Reply #296 on: December 18, 2021, 05:15:23 PM »
No it does not.

Force does not produce consistent quantities of energy.

Force produces momentum,  F = ma   a = v/t Therefore Ft = mv  it is not Ft = 1/2 mv²

If 7 N is applied to 3 kg for 4 seconds you will get;  7 N * 4 sec = 3 kg * v  = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.

If 7 N is applied to 30 kg for 4 seconds you will get;  7 N * 4 sec = 30 kg * v  = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.

Force does not produce energy it produces linear Newtonian momentum.

#### sm0ky2

• Hero Member
• Posts: 3784
##### Re: re: energy producing experiments
« Reply #297 on: December 18, 2021, 06:02:24 PM »
No it does not.

Force does not produce consistent quantities of energy.

Force produces momentum,  F = ma   a = v/t Therefore Ft = mv  it is not Ft = 1/2 mv²

If 7 N is applied to 3 kg for 4 seconds you will get;  7 N * 4 sec = 3 kg * v  = 9.33 m/sec This gives you 28 units of linear momentum and 130.66 J of energy.

If 7 N is applied to 30 kg for 4 seconds you will get;  7 N * 4 sec = 30 kg * v  = .933 m/sec This gives you 28 units of linear momentum and 13.066 J of energy.

Force does not produce energy it produces linear Newtonian momentum.

It is actually an integral equation. If you want to truly understand this problem,
I suggest learning calculus and trigonometry.

FT = ma = m* (delta)v
That is the change in velocity, integrated over time

Since mass is on both sides, we take that out and we see
that the FT results in a value of acceleration.

Anything off the horizontal constraint (with respect to the gravitational field)
looks something like [FT +{<+\-}9.8m/s/s]
Which is a vectored quantity

So the vector is also integrated over time
What we see mathematically, is that both the
vector and the acceleration are changing over time.

4 seconds takes up about a full sheet of notebook paper
if you write it all out in standard academic format.

Then you will understand that every second, you are inputting
different values for acceleration and vector (change in vector changes acceleration)

The summation of which can be estimated at 1/2q^2

This is simple shorthand physics, we do this in most “public equations”
Because the average person can’t perform pages of calculus for a simple problem
in the real world.

To get from one to the other is where trig comes in
A triangulation of the acceleration curve
You are correct in saying that is not the actual equation
It is an average of the change in velocity over time

The results are the same, because one works out the velocity
the other starts with the velocity and works it backwards

#### sm0ky2

• Hero Member
• Posts: 3784
##### Re: re: energy producing experiments
« Reply #298 on: December 18, 2021, 06:17:07 PM »
Now understand that “one second” is an arbitrary value.

We can instead use a value of 1-trillionth of a second.

We see that for every sec/ 1,000,000,000,000 the acceleration and vector change
so 1 trillion sheets of paper later, we realize that the best of our equation are only an approximation.

We could divide the trillionth of a second another trillion times.

and all of this only applies to the “stationary” inertial reference of our position on the earth’s surface
Which moves at 1000mph relative to the sun, which moves at 25Mmph relative to the inertial
reference of our galactic center, which moves at 1 trillion mph with respect to another arbitrary value
representing our sector of expanding space-time. After which time itself becomes  an integral quantity.

#### sm0ky2

• Hero Member
• Posts: 3784
##### Re: re: energy producing experiments
« Reply #299 on: December 18, 2021, 06:25:50 PM »
Even the 9.8m/s/s of gravity is not entirely accurate
We can integrate gravity using tSecs (trillionths of a second)
and derive many decimals which (were it not for terminal velocity)
would become significant over long drops.

Using this level of accuracy in tests to verify the equation,
gives rise to variances in the true value of gravitational acceleration
from location to location.
Mapping the earth’s gravity for the purposes of physics,
We see that even “sea level” is not the appropriate altitude in every location.
We have no consistency for our standard tests, beyond a certain accuracy.

Not to mention sea level in 1706 is meters under water today.