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3 years ago

The blank approach emphasizes the scientific study of observable actions and/or responses and their environmental determinants

Physics

1 answer:

lakkis [162]3 years ago

8 0

Answer:

"behavioral" is the right answer.

Explanation:

The behavioral approach would be founded on the principle of generating information thru all the extrapolation as well as the believing which our ecosystem might be what tends to create us to differ greatly as well as experience severe sickness.
Throughout several other phrases, it is indeed an analysis of the relationship among both our thoughts as well as our attitudes or actions.

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A pendulum is released from rest at point A. If the horizontal line represents the reference point, the pendulum has energy at p

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Only kinetic

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3 years ago

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The electron dot diagram shows the arrangement of dots without identifying the element.

olasank [31]

Answer: tellurium (Te)

atomic number = 52 ,

Number of energy levels = 5;

First energy level = 2

Second energy level = 8

Third energy level = 18

Fourth energy level = 18

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4. Locate the data and observations collected in your lab guide. What are the key results? How would you best summarize the data

melisa1 [442]

Answer:

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2 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

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2 years ago

What is magnetic shielding?

AysviL [449]

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2 years ago