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tensa zangetsu [6.8K]
3 years ago
7

Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac

ent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance ????=55.9 cm from the wall. The reflected beam strikes the wall at a height ????=38.9 cm above the surface of the mirror. Find the angle of incidence thetai at which the laser beam strikes the mirror.
Physics
1 answer:
fgiga [73]3 years ago
6 0

Answer:

θr = 55.2 °

Explanation:

The law of reflection states that the angle of incidence and the angles of reflection is the same.

Let's use trigonometry to find the angle between the reflected ray, where the horizontal distance x = 55.9 cm and the vertical distance is y = 38.9 cm

    tan θ = y / x

    tan θ = 38.9 / 55.9

    θ = tan⁻¹ (0.6959)

    θ = 34.8°

This angle is measured with respect to the x-axis (horizontal), but in general the angles in optics are measured from the y-axis so that the angle is

    θ = 90 - 34.8

    θr = 55.2 °

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jeka94

Answer:

What does that even mean?

Explanation:

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3 years ago
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A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w
frosja888 [35]

Answer:

(a) n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) n_{2} = 1.349

(c) v_{1} = 2.04\times 10^{8}\ m/s

(d) v_{2} = 2.22\times 10^{8}\ m/s

Solution:

As per the question:

Refractive index of medium 1, n_{1} = 1.47

Angle of refraction for medium 1, \theta_{1} = 59^{\circ}

Angle of refraction for medium 2, \theta_{1} = 69^{\circ}

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

n_{1}sin\theta_{1} = n_{2}sin\theta_{2}

where

n_{2} = Refractive Index of medium 2

Now,

n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}

n_{2} = 1.349

(c) To calculate the velocity of light in medium 1:

We know that:

Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}

Thus for medium 1

n_{1} = \frac{c}{v_{1}

v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s

(d) To calculate the velocity of light in medium 2:

For medium 2:

n_{2} = \frac{c}{v_{2}

v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s

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A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin
lianna [129]

Answer:

<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>

Explanation:

<u>Net Force </u>

The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.

The acceleration is proportional to the net force and inversely proportional to the mass of the object.

If the object has zero net force, it won't get accelerated and its velocity will remain constant.

The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.

Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.

The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.

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3 years ago
Ph11_UnitPacket2019
frozen [14]

Let's see

Use snells law

\\ \rm\Rrightarrow \dfrac{n_1}{n_2}=\dfrac{sini}{sinr}

\\ \rm\Rrightarrow \mu=\dfrac{sin30}{sin19.9}

\\ \rm\Rrightarrow \mu=0.5/0.34

\\ \rm\Rrightarrow \mu=1.47

It may be glass

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3 years ago
A model plane has a mass of 0.75 kg and is flying 12 m above the ground
Grace [21]

Answer:

Option C. 210 J.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Velocity (v) = 18 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Total Mechanical energy (ME) =?

Next, we shall determine the potential energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Height (h) = 12 m

Acceleration due to gravity (g) = 9.8 m/s²

Potential energy (PE) =?

PE = mgh

PE = 0.75 × 9.8 × 12

PE = 88.2 J

Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:

Mass (m) = 0.75 Kg

Velocity (v) = 18 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.75 × 18²

KE = ½ × 0.75 × 324

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Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:

Potential energy (PE) = 88.2 J

Kinetic energy (KE) = 121.5 J

Total Mechanical energy (ME) =?

ME = PE + KE

ME = 88.2 + 121.5

ME = 209.7 J

ME ≈ 210 J

Therefore, the total mechanical energy of the plane is 210 J.

8 0
3 years ago
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