Ocean waves are hitting a beach at a rate of 3.5 Hz. What is the period of the waves?

(a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of lig

OLga [1]

Answer:

t< 75 nm

Explanation:

A soap bubble is a thin film where when the beam enters the film it has a 180º phase change due to the refractive index and the wavelength changes between

λ = λ₀ / n

In the case of constructive interference in the curve of the spherical film it is

2 nt = (m + ½) λ₀

Where t is the thickness of the film and n the refractive index that does not indicate that we use that of water n = 1.33, m is an integer. The thickness of the film for the first interference (m = 0) is

t = λ₀ / 4 n

A thickness less than this gives destructive interference.

Let's look for the thickness for the visible spectrum

Violet light λ₀ = 400 nm = 400 10⁻⁹ m

t₁ = 400 10⁻⁹ / 4 1.33

t₁ = 75.2 10-9 m

Red light λ₀ = 700 nm = 700 10⁻⁹ m

t₂ = 700 10⁻⁹ / 4 1.33

t₂ = 131.6 10⁻⁹ m

Therefore, for all wavelengths to have destructive interference, the thickness must be less than 75 10⁻⁹ m = 75 nm

b) a film like eta is very thin, it is achieved when gravity thins the pomp, but any movement or burst of air breaks it,

3 0

2 years ago

What is the minimum coefficient (ms) of static friction between the road and the car? Since the net force is equivalent to the f

kirill115 [55]

Best Answer: <span> fnet = umg </span>

3 0

2 years ago

Two vehicles approach an intersection: a truck moving eastbound at 16.0 m/s and an SUV moving southbound at 20.0 m/s. Suppose th

mario62 [17]

Answer:25.61 m/s

Explanation:

Given

truck is moving eastbound with a velocity of 16 m/s

Velocity of truck $v_t=16\hat{i}$

SUV is moving south with a velocity of 20 m/s

Velocity of SUV in vector form $v_s=-20\hat{j}$

Velocity of truck relative to the SUV

$v_{ts}=v_{t}-v_s$

$v_{ts}=16\hat{i}-(-20\hat{j})$

Magnitude of relative velocity is

$|v_{ts}|=\sqrt{16^2+20^2}$

$|v_{ts}|=25.61\ m/s$

5 0

2 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago

Which name is given to the type of friction that an objects falling through the air experience

andrey2020 [161]

It is called Air resistance

3 0

3 years ago