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2 years ago

A box exerts 10,000 Pa of pressure on the ground. If the box weighs 1000 N, how much area is in contact with the ground?

Physics

1 answer:

lara [203]2 years ago

4 0

Pressure = (total force) / (Area)

10,000 Pa = (1,000 N) / (Area)

Multiply each side by (Area) :

(10,000 Pa) x (Area) = 1,000 N

Divide each side by (10,000 Pa) :

Area = (1,000 N) / (10,000 Pa)

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The number of fringe is z = 3 fringes

The ratio is $I = 0.2545I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

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