A conservative force acts on a particle. We can conclude:

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t $_{min$ = λ/2n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20

we substitute

t $_{min$ = 750 / 2(1.20)

t $_{min$ = 750 / 2.4

t $_{min$ = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t $_{min$ = λ/4n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50

we substitute

t $_{min$ = 750 / 4(1.50)

t $_{min$ = 750 / 6

t $_{min$ = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0

2 years ago

When hot and cold air meet, the hot air rises to the top. What causes the hot air to rise???

vekshin1

Hot air rises because when you heat air (or any other gas for that matter), it expands. When the air expands, it becomes less dense than the airaround it. The less dense hot air then floats in the more dense cold air much like wood floats on water because wood is less dense than water.

3 0

3 years ago

An object has a kinetic energy of 225 j and a momentum of magnitude 28.3 kg · m/s. (a) find the speed of the object. m/s (b) fin

Irina18 [472]

A. hope this helps ;0

5 0

3 years ago

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.

8 0

3 years ago

A 6.0 kg object undergoes an acceleration of 2.0 m/s2 .

Alenkinab [10]

A. Fnet=ma
6*2=12N of force acting on the object in the direction it is accelerating

B. Fnet=ma
4*2=8N of force action on the object in the direction it is accelerating

6 0

3 years ago