Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
Multiply the masses by the respective distances:
(12 kg) (2 m) = 24 J
(25 kg) (1 m) = 25 J
so the heavier bag takes more work to lift, and (b) is the answer.
(d) is technically correct if the sacks are carrying different contents whose masses are not equal, but since we don't know what's inside each sack, assume 12 kg and 25 kg are the masses of each sack *and* their contents.
Explanation:
Given that,
Electrostatic force, ![F=3.7\times 10^{-9}\ N](https://tex.z-dn.net/?f=F%3D3.7%5Ctimes%2010%5E%7B-9%7D%5C%20N)
Distance, ![r=5\ A=5\times 10^{-10}\ m](https://tex.z-dn.net/?f=r%3D5%5C%20A%3D5%5Ctimes%2010%5E%7B-10%7D%5C%20m)
(a)
, q is the charge on the ion
![q=\sqrt{\dfrac{Fr^2}{k}}](https://tex.z-dn.net/?f=q%3D%5Csqrt%7B%5Cdfrac%7BFr%5E2%7D%7Bk%7D%7D)
![q=3.2\times 10^{-19}\ C](https://tex.z-dn.net/?f=q%3D3.2%5Ctimes%2010%5E%7B-19%7D%5C%20C)
(b) Let n is the number of electrons are missing from each ion. It can be calculated as :
![n=\dfrac{q}{e}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7Bq%7D%7Be%7D)
![n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7B3.2%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
n = 2
Hence, this is the required solution.
If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:
<h3>How is the ratio of the perpendicular parts obtained?</h3>
To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.
The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:
r = √m/e : √2m/e : √4m/2e
When resolved, the resulting ratios will be:
1: √2 : 1
Learn more about the radii of their circular paths here:
brainly.com/question/16816166
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