41.5 is the answer that i got. hope this helps!
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
To know more about Fringes refer to: brainly.com/question/15649748
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Answer:
The maximum potential energy of the system is 0.2 J
Explanation:
Hi there!
When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.
The equation of elastic potential energy (EPE) is the following:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretching distance.
The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.
Then:
EPE = 1/2 · 40 N/m · (0.1 m)²
EPE = 0.2 J
The maximum potential energy of the system is 0.2 J
Answer:
c is the one that makes the most sense
Explanation:
