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2 years ago

Due today HELP HELP HELP

Physics

1 answer:

Anna [14]2 years ago

5 0

Vas happenin!

Independent variable : amount of water each day

Dependent variable: water on the windsill

Hypotheses: Ben wants to try by adding water each day to two different places. Will that work? Will that effect the water?

Hope this helps you out

*smiles*

-Zayn Malik

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3 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 634-nm

eimsori [14]

Answer:

minimum thickness of the coating = 122.868 nm

Explanation:

Given data

lens index of refraction = 1.29

wavelength = 634 nm

glass index of refraction = 1.53

to find out

minimum thickness of the coating

solution

we have given non reflective coating

we know that minimum thickness of the coating formula

minimum thickness of the coating = Wavelength / 4n

here n is coating index of refraction

so put here both value to get thickness

minimum thickness of the coating = Wavelength / 4n

minimum thickness of the coating = 634 / 4 ( 1.29 )

so minimum thickness of the coating = 122.868 nm

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3 years ago

A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate

frosja888 [35]

Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

Explanation:

The total charge is distributed over the two objects:

$Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\$

The plate and the rod must have $Q_{total}/2\\$ . So the charge transferred from the plate to the rod is:

$Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\$

Number of electrons:

$N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons$

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When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
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where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
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