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iogann1982 [59]
3 years ago
15

The coach has not seen improvement as expected in his team and plans to change practice session. To help the players he changes

the Frequency, Intensity, Time and Type of their exercise and uses the fitness principle of frequency
overload
progression
reversibility​
Physics
1 answer:
Allushta [10]3 years ago
7 0

Answer:

progession

Explanation:

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
2 years ago
Read 2 more answers
A water wave has a speed of 22.0 meters/second if the wave frequency is 0.0680 hertz what is the wavelength
serg [7]
V = f(wavelength)
22.0 = 0.0680 (wavelength)
wavelength = 323.52 m
8 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant
ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt    ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt     ..............2

so  area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r  = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt    ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91  10^{-2}cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 10^{-2} cm

so now from equation 3 we find emf

emf = - (0.500 )  2π(9.08 10^{-2} )   1.91  10^{-2}

emf = - 0.005445

and magnitude of emf = 0.005445 V

so

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0
2 years ago
Un pendul este suspendat de un ax cu o tijă subțire foarte ușoară.
FromTheMoon [43]

Answer:

A)

B)

C)

Explanation:

Given that a pendulum is suspended by a shaft with a very light thin rod.

Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²

We can determine the answer to these questions using angular kinematics.

Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.

Here are the formulas for angular kinematics:

  • θ = ωt
  • ∆w =
  • L [Angular momentum] = mvr [mass × velocity × radius]

A) What is the minimum speed required for the pendulum to traverse the complete circle?

We can use the formula v = √gL derived from

B) The same question if the pendulum is suspended with a wire?

C) What is the ratio of the two calculated speeds?

4 0
2 years ago
Which of the following statements accurately describes oceanic electric fields?
vichka [17]

Answer:

i sure it is D

Explanation:

3 0
2 years ago
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