Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?

Answers

Answer 1

Answer:

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt = λ / 2

t = λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

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The cart is initially at rest. ForceF is applied to the cart for time ? t, after which the car has speed v. Suppose the same force is applied for the same time to a second cart with twice the mass. Friction is negligible. Afterward, the second cart's speed will be
a) 2v
b) 1/2v
c) v
d) 4v
e) 1/4v

Answers

Answer:

$v'=(1)/(2)v$

Explanation:

Given that,

Initial speed of the cart, u = 0

Let F force is applied to the cart for time $\Delta t$ after which the car has speed v. The force on an object is given by :

F = ma

m is the mass of the cart

We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,

$F=(mv')/(t)$

$v'=(F t)/(2m)$

$v'=(1)/(2)v$

So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).

High-energy particles are observed in laboratories by photographing the tracks they leave in certain detectors; the length of the track depends on the speed of the particle and its lifetime. A particle moving at 0.993c leaves a track 1.15 mm long. What is the proper lifetime of the particle

Answers

Answer:

Lifetime = 4.928 x 10^-32 s

Explanation:

(1 / v2 – 1 / c2) x2 = T2

T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225

T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6

T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s

Final answer:

To find the proper lifetime of the particle, we can use the time dilation equation and the Lorentz factor. Plugging in the given values, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Explanation:

To find the proper lifetime of the particle, we can use the time dilation equation, which states that the proper time (time experienced in the frame of reference of the particle) is equal to the time observed in the laboratory frame of reference divided by the Lorentz factor. The Lorentz factor can be calculated using the equation γ = 1/√(1 - (v/c)^2), where v is the velocity of the particle and c is the speed of light. Given that the particle is moving at 0.993c, the Lorentz factor is approximately 22.82.

Next, we can use the equation Δx = βγcτ, where Δx is the length of the track, β is the velocity of the particle in units of the speed of light (v/c), γ is the Lorentz factor, c is the speed of light, and τ is the proper lifetime of the particle. Plugging in the given values, we have 1.15 mm = 0.993 * 22.82 * c * τ. Solving for τ, we find that the proper lifetime of the particle is approximately 5.42 × 10^-9 seconds.

Learn more about Proper lifetime of high-energy particles here:

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John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda says the object's mass is proportional to its density and to its volume. Which one, if either, is correct?A. They are both wrong
B. John is correct, but Linda is wrong
C. John is wrong, but Linda is correct
D. They are both correct.
E. John must be wrong, because Linda always wins these arguments.

Answers

Answer:

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

The most interpersonal constructive passion response to relational conflict is..

Answers

Answer:

loyalty

Explanation:

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = (7.4)/(cos(70))$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)