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Rus_ich [418]
3 years ago
7

What is one way to take advantage of creating even more kinetic energy on this particular technologically advanced field

Physics
1 answer:
ch4aika [34]3 years ago
7 0

Incomplete question. However, I provided a brief about Kinetic energy generation.

<u>Explanation:</u>

Interestingly, Kinetic energy in simple terms refers to the energy possessed by a body in motion.

It is often calculated using the formula E =  \frac{1}{2}MV^{2}

A good example of creating even more kinetic energy is a hand crank toy car that moves after you wind it a little, when the car moves it is generating another measure of K.E.

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The term you are looking for is "electric field".
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A 730-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 120°, the kinetic energy of th
romanna [79]

Answer:

Energy of scattered photon is 232.27 keV.

Kinetic energy of recoil electron is 497.73 keV.

The recoil angle of electron is 13.40°

Explanation:

The energy of scattered photon is given by the relation :

E_{2}=\frac{E_{1} }{1+(\frac{E_{1} }{m_{e}c^{2}  })(1-\cos\theta) }     .....(1)

Here E₁ is the energy of incident photon, E₂ is the energy of scattered photon, m_{e} is mass of electron and θ is scattered angle.

Substitute 730 keV for E₁, 511 keV for m_{e} and 120° for θ in equation (1).  

E_{2}=\frac{730 }{1+(\frac{730 }{511  })(1-\cos120) }

E₂ = 232.27 keV

Kinetic energy of recoil electron is given by the relation :

K.E. = E₁ - E₂ = (730 - 232.27 ) keV = 497.73 keV

The recoil angle of electron is given by :

\cot\phi=(1+\frac{E_{1} }{m_{e}c^{2}  })\tan\frac{\theta}{2}

Substitute the suitable values in above equation.

\cot\phi=(1+\frac{730 }{511  })\tan\frac{120}{2}

\cot\phi=4.20

\phi = 13.40°

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3 years ago
Work and power!
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Answer:

The work done in a unit of time is C. Power.

Explanation:

Hope this helps

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Explain what is meant by the term mechanical advantage.
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Read 2 more answers
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
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