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3 years ago

What is the critical angle of a light beam passed from a medium (n=2) to a medium (n=1.2)?

Physics

1 answer:

sertanlavr [38]3 years ago

8 0

Answer:

θ_c = 36.87°

Explanation:

Index of refraction for index medium; n_i = 2

Index of refraction for Refractive medium; n_r = 1.2

Formula to find the critical angle is given;

n_i(sin θ_c) = n_r(sin 90)

Where θ_c is critical angle.

Thus;

2 × (sin θ_c) = 1.2 × 1

(sin θ_c) = 1.2/2

(sin θ_c) = 0.6

θ_c = sin^(-1) 0.6

θ_c = 36.87°

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A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it

Veseljchak [2.6K]

Incomplete question as the unit of volume is not written correctly.So the complete question is here:

A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces 89.0 cm³?

Answer:

$d_{Density}=2.7g/cm^{3}$

Explanation:

Given data

Mass m=240g

Volume V=89.0 cm³

To find

Density d

Solution

If rock displaces 89.0 cm³ of water means volume of rock is also 89cm³

$d_{Density}=\frac{mass}{volume}\\d_{Density}=\frac{240g}{89.0cm^{3} } \\d_{Density}=2.7g/cm^{3}$

5 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .