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2 years ago

10

Beyond what point must an object be squeezed for it to be a black hole? A : Schwarzschild Radius. B : Crackpoint. C : Event Hori

zon. D : Hawking Boundary.

2 answers:

7nadin3 [17]2 years ago

8 0

Answer:

A:Schwarzschild radius

djverab [1.8K]2 years ago

7 0

Answer:

A.

Explanation:

An object that is below its Schwarrzfield radius is a black hole. For our Sun its about 3 km.

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Read 2 more answers

If the index of refraction for a certain glass is 1.5, and the angle of refraction is 15 degrees for a ray of light

LiRa [457]

Answer:

$\theta_i = 23^o$

Explanation:

The refractive index of a medium is given by the following formula:

$\eta = \frac{Sin\theta_i}{Sin\theta_r}$

where,

η = refractive index = 1.5

$\theta_i$ = angle of incidence = ?

$\theta_r$ = angle of refraction = 15°

Therefore,

$1.5 = \frac{Sin\theta_i}{Sin15^o}\\\\Sin\theta_i=(1.5)Sin15^o\\\\\theta_i = Sin^{-1}(0.388)\\\\$

$\theta_i = 23^o$

4 0

3 years ago

A red laser (740 nm) shines on a double slit (slit separation = 0.165 mm). What is the angle of the fourth order maximum (that i

SVEN [57.7K]

Answer:

The angle of the fourth order maximum is 1.027 degrees.

Explanation:

It is given that,

Wavelength of red laser, $\lambda=740\ nm=740\times 10^{-9}\ m$

Slit separation, $d=0.165\ mm= 0.000165\ m$

We need to find the angle of the fourth order maximum. For maximum, the equation is given by :

$d\ sin\theta=n\lambda$

$sin\theta=\dfrac{n\lambda}{d}$

$sin\theta=\dfrac{4\times 740\times 10^{-9}}{0.000165}$

$\theta=1.027^{\circ}$

So, the angle of the fourth order maximum is 1.027 degrees. Hence, this is the required solution.

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3 years ago