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3 years ago

11

When light falls on objects, it interacts with them in different ways as shown in the images. What type of interaction is seen i

n each image?

1 answer:

agasfer [191]3 years ago

7 0

Answers with explanation

1)

Transmission

If all the light passes through a medium without any absorption then transmittance is 100%.

2)

Refraction

Refraction is the bending of a wave when it enters a medium where its speed is different and rays are again refracted when they leave that medium,

3)

Reflection

Reflected rays don't pass through the medium instead rays bounces off an object at an angle.

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

The initial speed is $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

=> $KE_i = \frac{1}{2} * m * (2.05 *10^{4})^2$

=> $KE_i = 2.101 *10^{8} \ \ m \ \ J$

And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

Generally from the law energy conservation we have that

$T__{E}} = KE_p$

So

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

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3 years ago

Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle o

Kazeer [188]

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, $\theta$ = 15.8°

First bright fringe means , m = 1

So,

$d\times sin\ 15.8^0=1\times \597\ nm$

$d\times 0.2723=1\times \597\ nm$

$d=2192.43481\ nm$

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

<u>Distance between slits ≅ 2.2 µm</u>

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3 years ago

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A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass

bogdanovich [222]

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun Ф= \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }

b) minimun Ф = $\frac{\pi }{2}$ - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

y = y₀ + v₀ t - ½ g t²

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

0 = y₀ - ½ g t²

t = $\sqrt{2y_{o}/g}$

The bucket-spring system has a simple harmonic motion, which is described by

x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

w = $\sqrt{k/m}$

In the initial state

w = \sqrt{k/2}

When the mass changes

w ’= \sqrt{k/3}

the displacement in each case is

x = A cos (wt)

for the new case

x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

cos (w´t + Ф) = 1

w’t + Ф = 0

Ф = -w ’t

Ф = - $\sqrt{k/3}$ $\sqrt{2y_{o}/g}$

\sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }

b) the function is minimun if

cos (w’t + fi) = 0

w’t + Ф = π / 2

Ф = π / 2 - w ’t

Ф = $\frac{\pi }{2}$ - \sqrt{\frac{2}{3} \frac{2 y_{o} }{g} }

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3 years ago

Please help on this one

PolarNik [594]

If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

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3 years ago

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A soap bubble usually pops because some part of it becomes too thin due to evaporation or drainage of fluid. The change in thick

Lapatulllka [165]

Answer:

To understand why the change in thickness also changes the color of light the bubble reflects, we definitely know that

mλ = dsinθ

when d sin θ = mλ, we expect constructive interference

d = distance between the slits

θ = the angle relative to the incident direction

m = order of the interference

Hence, a change in the thickness will also cause a change in the wavelength, and as wavelength changes, various colors are generated.

The light's path difference is therefore generated as a result of the thickness variation of the soap bubble.

Explanation:

To understand why the change in thickness also changes the color of light the bubble reflects, we definitely know that

mλ = dsinθ

when d sin θ = mλ, we expect constructive interference

d = distance between the slits

θ = the angle relative to the incident direction

m = order of the interference

Hence, a change in the thickness will also cause a change in the wavelength, and as wavelength changes, various colors are generated.

The light's path difference is therefore generated as a result of the thickness variation of the soap bubble.

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3 years ago