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3 years ago

According to Coulomb’s Law, the force between two charged objects is related to _____.

Physics

1 answer:

Gekata [30.6K]3 years ago

7 0

The square<span> </span>of the distance between<span> them</span>

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A slingshot is used to shoot a small ball upward into the air; when the pouch is released, the sling propels the ball upward. as

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Explanation:

kinetic to gravitational

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3 years ago

List several examples of applied force, normal force, and friction that you’ve observed in your life.

grandymaker [24]

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

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3 years ago

A shopper walks westward 2.8 meters in the first 5 minutes and then eastward 9.2 meters in the next 10 minutes. What is the shop

Alecsey [184]

Answer:

12m

Explanation:

Given parameters:

Distance walked westward = 2.8m

Time of travel = 5min

Distance walked eastward = 9.2m

Time of travel = 10min

Unknown:

The total shopper's travel distance = ?

Solution:

Total distance traveled is the sum of the length of path covered by a body. It is a scalar quantity.

Total distance = distance walked westward + distance walked eastward

Total distance = 2.8m + 9.2m = 12m

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3 years ago

What is the main reason plants and animals have sex?

Klio2033 [76]

Evolution of sexual production

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3 years ago

Read 2 more answers

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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3 years ago