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Xelga [282]
3 years ago
6

Although he did not present a mechanism, what were the key points of Alfred Wegener’s proposal for the concept of continental dr

ift?
Physics
1 answer:
valentinak56 [21]3 years ago
4 0

Answer: Alfred Wegener provided some of the important points that supported the theory of continental drift. They are as follows-

  1. The continents were once all attached together, and this can be proved by studying the coastlines of some of the continents that perfectly matches with one another.
  2. The appearance of similar rock types and similar fossils (including both animals and plants) has also contributed much information that continents were once all together.
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What type of charge will an object have if the object contains less protons than electrons?
Burka [1]

Answer:

Hello, I believe it would have a negative charge considering protons have a positive charge while elctrons have a negative charge

Explanation:

6 0
3 years ago
Please help have due very soon?thank you
Andru [333]
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The equation shows neutralization of an acid and a base to produce a salt and water.
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7 0
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Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind
IrinaVladis [17]
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3 0
3 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

4 0
2 years ago
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