Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰
Answer:
n = 1.523
Explanation:
Snell's law
1.273sin32.89 = nsin26.99
(A) The total initial momentum of the system is
(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s
(B) Momentum is conserved, so that the total momentum of the system after the collision is
35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>
where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives
<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s
(C) The kinetic energy of the system after the collision is
1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J
and before the collision, it is
1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J
so that the change in kinetic energy is
∆<em>K</em> = 25.4 J - 474 J ≈ -449 J
Answer:
He is wrong!
Explanation:
Frequency refers to how many wave lengths pass in a second and speed is how fast the wave is traveling for example the speed of light goes really fast but has a mid-level frequency.
Hope this helps! ;-)
going to the toilet, and keeping cleant. no nasty bits floating around but they are weightless !
eating. pushing a fork would push back by newton 3. can't brace themselves to cut