Question from my vector statics class (deals with vectors from physics)

Physics

2 answers:

Minchanka [31]2 years ago

5 0

free body diagram at pulley C:

along AC at 55deg, 2*T (for tension) is pulling up

along CB at 25deg, T is pulling up

combining, 2T*sin55 + T*sin25 = 1800

solving, T= 873.4N

Ooops...me bad - mis-read the Q. plz report n delete this!

jeka57 [31]2 years ago

4 0

Haven't done one like this in awhile but I see no one is answering so I gave it a try. I think it's right but let me know if you see something fishy...

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen

Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}$

$sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m$