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Anuta_ua [19.1K]

2 years ago

7

A jeweler is determining the optical properties of an unknown blue gemstone. She uses an angle of incidence of 62°, and measures

an angle of refraction of 30°. Assuming the index of refraction of air is 1.0, what is the index of refraction of the gem?

1 answer:

Juliette [100K]2 years ago

3 0

Answer:

n = 1.76

Explanation:

According to the rule of ( n1 sin theta1 = n2 sin theta2 )

we know both angles so we insert them to the law and apply n1 = 1

so 1/2 = n2 sin 62 and we get the final answer

You might be interested in

A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block

Luden [163]

Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, $\theta=20^{\circ}$

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

$W=F\ cos\theta\times d$

$W=30\times cos(20)\times 3$

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

7 0

2 years ago

At a location where the acceleration due to gravity is 9.807 m/s2, the atmospheric pressure is 9.891 × 104 Pa. A barometer at th

Otrada [13]

Answer:

$8616.7468 \ kg/m^3$

Explanation:

Pressure is measured is $p=\rho gh$ here p is pressure $\rho$ is density and h is height

We have given pressure $p=9.891\times 10^4\ Pa$ acceleration due to gravity $g=9.9870\ m/sec^2$ height =1.163 m

$\rho =\frac{p}{gh}=\frac{9.891\times 10^4}{9.870\times 1.163}=8616.7468 \ kg/m^3$

5 0

2 years ago

Read 2 more answers

. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

faust18 [17]

Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

$h = gt^2/2$

$10 = 10t^2/2$

$t^2 = 2$

$t = \sqrt{2} = 1.414 s$

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

$\theta = \alpha t^2/2$

$\theta = 15 * 2/2 = 15 rad$

As each revolution is 2π, the total number of revolution that she could make is: 15 / 2π = 2.39 rev

3 0

2 years ago

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

Sauron [17]

They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)

4 0

2 years ago

Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

8 0

2 years ago