A box of total mass M is on a frictionless horizontal table. It is connected to a massless string that runs over a massless, fri
1 answer:
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Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C
Complete question:
Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit? Fringe C is the central maximum.
Check the image uploaded.
Answer:
The difference is 1100 nm
Explanation:
A bright fringe creates constructive interference, the wavelength is always in a multiple form.
At center fringe, the difference is (550 nm)(0) = 0
For the first maximum, the difference is (550 nm)(1) = 550 nm
For the second maximum, the difference = (550 nm)(2) = 1100 nm
Thus, for nth maxima, the difference is (550 nm)(n)
From the image uploaded, C is located on the second maximum, therefore the difference is given as (550 nm)(2) = 1100 nm
Therefore, the dot on the screen in the center of fringe E to the left slit is 1100 nm
