Answer:
Explanation:
In case of oil slick a thin layer of oil is formed on water . This thin layer creates a rainbow of colour . The phenomenon is due to interference of light waves , one reflected from the upper surface of oil and the other reflected from the lower surface of the oil.
For formation of bright colour
2 μ t = ( 2n + 1 ) λ / 2
μ is refractive index of oil , t is thickness of oil layer λ is wave length of light falling on the layer .
given μ = 1.2 , λ = 750 x 10⁻⁹ ,
2 x 1.2 t = ( 2n + 1 ) 750 x 10⁻⁹ / 2
For minimum thickness n = 0
2.4 t = 375 x 10⁻⁹
t = 156.25 n m
B ) If the refractive index of layer of medium below oil is less than that of oil , the condition of formation of colour changes
The new condition is
2 μ t = n λ
2 x 1.5 t = 750 nm , n = 1 for minimum wavelength .
t = 250 nm
C ) Light mostly transmitted means dark spot is formed at that point .
For that to be observed from water side , the condition is
2 μ t = ( 2n + 1 ) λ / 2
λ = 4μ t / ( 2n + 1 )
For maximum wavelength n = 0
λ = 4μ t
= 4 x 1.5 x 200 nm
= 1200 nm .
It is wasted, most likely as light, in this case, or it is lost during the transport of electricity.
Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
medium
Explanation:
<em>A sound </em><em>medium</em><em> is defined as channel through which sound can travel or be transmitted. </em>
Sound medium could be in the form gases, liquids, solids or plasmas. Space is made up of vacuum and therefore, has no medium within it. Hence, space cannot transmit sound in any form or allows sound to travel through it.
Answer:
at t=46/22, x=24 699/1210 ≈ 24.56m
Explanation:
The general equation for location is:
x(t) = x₀ + v₀·t + 1/2 a·t²
Where:
x(t) is the location at time t. Let's say this is the height above the base of the cliff.
x₀ is the starting position. At the base of the cliff we'll take x₀=0 and at the top x₀=46.0
v₀ is the initial velocity. For the ball it is 0, for the stone it is 22.0.
a is the standard gravity. In this example it is pointed downwards at -9.8 m/s².
Now that we have this formula, we have to write it two times, once for the ball and once for the stone, and then figure out for which t they are equal, which is the point of collision.
Ball: x(t) = 46.0 + 0 - 1/2*9.8 t²
Stone: x(t) = 0 + 22·t - 1/2*9.8 t²
Since both objects are subject to the same gravity, the 1/2 a·t² term cancels out on both side, and what we're left with is actually quite a simple equation:
46 = 22·t
so t = 46/22 ≈ 2.09
Put this t back into either original (i.e., with the quadratic term) equation and get:
x(46/22) = 46 - 1/2 * 9.806 * (46/22)² ≈ 24.56 m