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larisa86 [58]
2 years ago
12

Each step in any energy conversion process will _____. gain energy create energy dissipate energy destroy energy

Physics
2 answers:
Damm [24]2 years ago
7 0
The correct answer is to dissipate energy. Each step in any energy conversion process will dissipate energy. The energy conversion is the process in which a form of energy is being converted into a new or another form of energy.
fomenos2 years ago
7 0

The correct answer to the question is : Energy dissipate.

EXPLANATION:

Before coming into any conclusion, first we have to understand law of conservation of energy.

As per law of conservation of energy, energy can neither be created nor be destroyed. It can only change from one form to another form, and the total energy of the universe is always constant.

As per the question, there is energy conversion. One form of any energy may be converted into various types of energy. One particular type of energy is not obtained always.

Hence, the correct answer to the question is energy is dissipated during energy conversion.

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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long ran
makvit [3.9K]

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

7 0
3 years ago
A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi
Ahat [919]
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0
3 years ago
What type of clouds usually accompany cold fronts?
ASHA 777 [7]

Answer: Cumulus

Explanation: Most large cloud fronts are made up of cumulus clouds, large storm clouds are cumulonimbus clouds.

6 0
3 years ago
Read 2 more answers
What are the physics terms behind watching TV?
GalinKa [24]
Everyone knows that one of their favorite past times is sitting in front of the television and watching movies, shows, or playing video games. However with this almost motionless, lazy activity comes a great deal of static physics and mechanics.

When you are sitting down enjoying whatever show it is you may be watching, you actually have several forces acting on you concurrently. For example, by sitting on the couch with no extra weight on you, your weight is equivalent to the normal force, or the force of the couch on you. In addition to the force of the couch of you, if you are leaning on an arm or laying down, a similar force acts on you, except at an angle or incline. The general rule for laying on the couch watching television is that whatever force you exert on an object, that object exerts the same force in the opposite direction, or 180 degrees around.

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3 years ago
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