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3 years ago

Which is the destructive interference formula for diffraction grating problems? dsin = dcos = dcos = n dsin = n.

1 answer:

5 0

The destructive interference formula for diffraction grating problems is $d sin \theta = (n+\frac{1}{2})\lambda$ .

<h3>What is the definition of destructive interference?</h3>

Destructive interference happens when the maxima of two waves are 180° out of phase a positive displacement of one wave is canceled exactly by a negative displacement of the other wave.

The formula for brighter patches resulting from constructive interference and darker patches resulting from destructive interference in a diffraction grating is:

$\rm d sin \theta = n \lambda$

The grating spacing is denoted by d, the angle of light is denoted by a the fringe order is denoted by n, and the wavelength is denoted by $\rm \lambda$ .

The destructive interference formula is now based on the fact that destructive interference occurs between the fringes.

Hence the destructive interference formula for diffraction grating problems is $d sin \theta = (n+\frac{1}{2})\lambda$ .

To learn more about destructive interference refer to the link;

brainly.com/question/16098226

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Read 2 more answers

A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

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3 years ago

At what water temperature will additional heat energy need to be added before the temperature will change again?

Katyanochek1 [597]

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)

Hope this helps! (If correct, please rank as brainliest answer) :)

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3 years ago