Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

Ainat [17]

Answer:

a

The number of fringe is z = 3 fringes

b

The ratio is $I = 0.2545I_o$

Explanation:

a

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

b

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

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

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

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

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

6 0

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This type of radiation is smallest and highest in energy: ____ ___.

madreJ [45]

Gamma rays have the highest energies and the shortest wavelengths.

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You hear that a storm is moving 15km/HR north. You have been given the storm's______

Elodia [21]

You have been given the storm's velocity.

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What is a single cell organism able to do?

Sunny_sXe [5.5K]

Survived on its own.

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3 years ago

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A space vehicle is traveling at 5425 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent

lana66690 [7]

Answer:

$V_{cE}=1489m/s$

Explanation:

Given data

Space vehicle speed=5425 km/h relative to earth

The rocket motor speed=81 km/h and mass 4m

The command has mass m

From the conservation of momentum as the system isolated

$p_{i}=p_{f}\\$

Since the motion in on direction we can drop the unit vector direction

$MV_{i}=4mV_{mE}+mV_{CE}$

Where M is the mass of space vehicle which equals to sum of the motors mass and command mass.

The velocity of the motor relative to the earth equals the velocity of the motor relative to command plus the velocity of the command relative to earth

$V_{mE}=V_{mc}+V_{cE}$

Where Vmc is the velocity of motor relative to command

This yields

$5mV_{i}=4m(V_{mc}+V_{cE})+mV_{cE}\\5V_{i}=4V_{mc}+5V_{cE}$

Substitute the given values

$V_{cE}=\frac{5V_{i}-4V_{mc}}{5}\\ V_{cE}=5425*\frac{1000}{60*60}(m/s)-\frac{4}{5}*81\frac{1000}{60*60}(m/s)\\ V_{cE}=1489m/s$

5 0

3 years ago