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elena-14-01-66 [18.8K]

3 years ago

5

A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surroun

ded by air and if its index of refraction is 1.33, what is the thinnest thickness the soap film can be? (c = 3.00 × 108 m/s)

1 answer:

denis23 [38]3 years ago

4 0

Answer:

$1.07004\times 10^{-7}\ m$

Explanation:

$n_s$ = Refractive index of bubble = 1.33

f = Frequency of light = $5.27\times 10^{14}\ Hz$

c = Speed of light = $3\times 10^8\ m/s$

The wavelength of light is given by

$\lambda=\dfrac{2n_st}{m-\dfrac{1}{2}}$

Wavelength is also given by

$\lambda=\dfrac{c}{f}$

m = 1 for minimum thickness

$\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m$

The minimum thickness is $1.07004\times 10^{-7}\ m$

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Can any1 tell if my answer is right

jok3333 [9.3K]

Answer:

The correct answer would be C. 5.0kg

Explanation:

The mass of an object never changes unless parts of the object are taken away. In other words, although the gravitational force is different on the moon then on the earth the mass of the object would remain the same.

6 0

2 years ago

G = 10 N/kg or 10 m/s2

Irina18 [472]

Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

a) The initial potential energy is:

$U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)$

$U_{g} = 40\,J$

b) The efficiency of the bounce is:

$\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%$

$\eta = 70\,\%$

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

$K = U_{g}$

$U_{g} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

$v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }$

$v = 20\,\frac{m}{s}$

5 0

3 years ago

1) A person riding their bike on a nice cloudy day travels 50 meters in 20 seconds.

Rasek [7]

Answer:

5.59

Explanation:

50 meters in 10 seconds is 11.18, which is an easy way to remember. Just divided by 2

5 0

3 years ago

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du

Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}$ .

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is $700\,{\text{kg}}$ .

The velocity of the car before collision is $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The velocity of the car after collision is $4.5\text{ m}/\text{s}$ .

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

$\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}$ …… (I)

Here, ${p_i}$ is the initial momentum of the car, $m$ is the mass of the car and ${v_i}$ is the initial velocity of the car.

Substitute $700\,{\text{kg}}$ for $m$ and $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (I).

$\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}$

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

$\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}$ …… (II)

Here, ${p_f}$ is the final momentum and ${v_f}$ is the final velocity.

Substitute $700\,{\text{kg}}$ for $m$ and - $4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_f}$ in equation (II).

$\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

$\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}$ …… (III)

Here, $\Delta p$ is the change in the momentum of the car.

Substitute $- 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${p_f}$ and $20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (III).

$\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}$

Thus, the change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}$ .

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

7 0

3 years ago

The current supplied by a battery in a portable device is typically about 0.151 A. Find the number of electrons passing through

butalik [34]

Answer:

n = 1.7*10²² electrons.

Explanation:

As the current, by definition, is the rate of change of charge, assuming that the current was flowing at a steady rate of .151 A during the 5 hours, we can find the total charge that passed perpendicular to the cross-section of the circuit, as follows:

$I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s$

⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C

As this charge is carried by electrons, we can express this value as the product of the elementary charge e (charge of a single electron) times the number of electrons flowing during that time, as follows:

Δq = n*e

Solving for e:

$n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22 electrons.$

6 0

3 years ago

Read 2 more answers