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
= λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t = 750 / 2(1.20)
t = 750 / 2.4
t = 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 = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t = 750 / 4(1.50)
t = 750 / 6
t = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Answer:
A
Explanation:
They drove 30km north. The displacement adds up to 25km therefore making the distance greater
Hope this helps!
Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
Answer:
The ball would hit the floor approximately 
after leaving the table.
The ball would travel approximately 
horizontally after leaving the table.
(Assumption: 
.)
Explanation:
Let 
denote the change to the height of the ball. Let 
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let 
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:
.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was 
.
The height of the table was 
. Therefore, after hitting the floor, the ball would be 
below where it was before leaving the table. Hence, 
.
The equation becomes:
.
Solve for :
.
In other words, it would take approximately for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at 
) until the ball hits the floor.
The ball was in the air for approximately 
and would have travelled approximately 
horizontally during the flight.
