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Aloiza [94]
2 years ago
12

While it’s impossible to design a perpetual motion machine, that is, a machine that keeps moving forever, come up with ways to k

eep some type of periodic motion going for a very long time. Explain the limits on perpetual motion. Discuss what slows down a machine and how you might minimize those effects for at least one specific form of periodic motion, such as the motion of a spring or pendulum, or circular motion. Come up with an inventive solution!
Physics
2 answers:
Svetlanka [38]2 years ago
8 0

Answer:

We can not create a perpetual motion machine because this kind of motion violates two laws of thermodynamic.

Explanation:

We can not create a perpetual motion machine because this kind of motion violates two laws of thermodynamic.

1. In any isolated system we can not create and destroy energy, it is a law of conservation of energy, all the energy is transformed, so the thermal efficiency, that is the produced work divided by the input heating can not be greater than one.

A perpetual motion machine needs to create energy to remain its movement constant along the time.

2. The second law of thermodynamic is related to the entropy, it is always positive. A natural process runs only in one sense, and is not reversible. Therefore the output work power of heat of a engine for instance is always smaller than the input heating power ans the rest of the heat energy supplied is wasted as heat to the ambient surroundings. The thermal efficiency has a maximum, given by the Carnot efficiency, which is always less than one.

That is way we can not create a perpetual motion machine.  We can create a machine that minimize those effects and maximize the efficient. We can reduce the air resistance for example.

I hope it helps you!

MissTica2 years ago
7 0
A perpetual motion machine is (as the name implies) a machine that moves perpetually; it never stops. Ever. So if you created one today and set it going, it would keep on going until the Big Freeze<span>. Calling that “a long time” is an understatement of epic proportions</span>
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<span> (a) does an electric field exert a force on a stationary charged object? 
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</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
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</span><span>(h) does a magnetic field do so?
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6 0
2 years ago
Which formula is used to find fluctuation of the shape of body
Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0
2 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the jun
AlexFokin [52]

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

i_1+i_3=i_2                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

i_3=i_2-i_1=0.75A-0.40A=0.35A

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

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2 years ago
A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.
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2 years ago
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

ωf = ω₀+ α*t  Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
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