Snell's law: n1Sinα=n2Sin β where α=Incidence angle, β=angle of refraction, n1 and n2 are the indices of refraction for water and air respectively.
Therefore,
Sinα=n2/n1 Sinβ For refracted ray to be along the surface of water, β=90° and thus Sinβ = 1
Sinα=n2/n1= 1/1.33 = 0.7519 => α=sin^-1 (0.7519) = 48.75°
When light moves from a medium of higher index of refraction (such as water) to medium of lesser index of refraction (such as air), the refracted ray is bend such that α is bigger than β. This is internal refraction. At some value of α, β approaches 90°. This incidence angle is called critical incidence angle. Therefore, the current scenario is shows critical angle of incidence.
Answer:
Not enough information given. Are there answer choices? Or more details given in the question?
Explanation:
Answer:
True
Explanation:
If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell
According to Gauss law
∅ = EA =-Q/∈₀
Where ∅ is the electric flux through the gaussian surface and E is the electric field strength
If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero
Answer:
Explanation:
Let the height of the ladder be L
Also:
- Let
- Let
When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.
So, we have:
--- Pythagoras Theorem
When the base is 9ft from the wall, this means that:
Substitute 9 for x and 10 for L in
Make 
the subject
Make y the subject
<em>Hence, the true distance at that point is approximately 4.36ft</em>
Yes it can because it had lots of force
