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3 years ago

8

I need plz some on help me

1 answer:

Alex3 years ago

4 0

I think the answer is repulsive.... Sorry if I get this wrong? :)

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KATRIN_1 [288]

C Occupational Safety and Health Administration

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1 year ago

If are spaced closely together on the map,there is a drastic temperature change over the distance

AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

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3 years ago

Read 2 more answers

Describe an example where materials are prepared as fluids so that they can be moved easily

den301095 [7]

Answer:

In general solids are easier to transport than liquids, but the above metal example is a valid one and the only other one that comes to mind is that of concrete. It is mixed as a liquid and transported as such, but then sprayed or laid down to dry and form a solid surface or filler.

Explanation:

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3 years ago

A good train took 7 hours to complete its journey. For the first 3 hours, it travelled an average speed of 186km/h. What was the

Sliva [168]

I really hope this helps

4 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

3 years ago