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3 years ago

What does Snell's law state

Physics

1 answer:

DochEvi [55]3 years ago

7 0

Answer:

The snell's law states that the ratio of sine of angle of incidence to the sine of angle of refraction is always constant for two reasons. →μ(constant) = refractive

Explanation:

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3 years ago

What are the uses of a magnet?

Doss [256]

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2 years ago

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A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

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2 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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4 years ago

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From largest to smallest they are: Universe, galaxy, solar system, star, planet, moon and asteroid.

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YOUR VERY WelCoMe!!!! :) :) :) :) :0 :)

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3 years ago

What does Snell's law state​

What does Snell's law state