If you walk 4mi north and then 2mi south what is your total displacement?

Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0

2 years ago

Ite

soldier1979 [14.2K]

For the sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

<h3>What is effect of density of a medium on wavelength of a wave?</h3>

The density of a medium is directly proportional to the wavelength of a wave.

The higher the density of the medium, the longer the wavelength of a wave.

Therefore, for a sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

Learn more about density and wavelength at: brainly.com/question/9486264

#SPJ1

4 0

2 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago

PLEASE HELP ASAP!!

Stells [14]

Its B: reduce the amount of energy needed to do the work by putting the work onto something else

3 0

3 years ago

Read 2 more answers

How can the atomic number of nitrogen be determined?

Doss [256]

Answer:

So... for the element of NITROGEN, you already know that the atomic number tells you the number of electrons. That means there are 7 electrons in a nitrogen atom. Looking at the picture, you can see there are two electrons in shell one and five in shell two.

4 0

2 years ago