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1 year ago

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5. A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen. The 5th minimum occurs at 7.00° aw

ay from the central maximum. At what angle does the 18th minimum occur? A) 26.0° B) 1.94° C) 5.05° D) 0.44°

1 answer:

abruzzese [7]1 year ago

3 0

For a single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen,the angle is mathematically given as

theta=25.3

Option A is correct

<h3> What angle does the 18th minimum occur?</h3>

Generally, the equation for the the angle is mathematically given as

$\theta=n(\lambda/d)$

Therefore

$\theta 1/ \theta 2=n1(\lambda/d)/ n2(\lambda/d)$

In conclusion

theta/7=16/5

theta=10*7/5

theta=25.3

Read more about Angle

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Answer:

The angle of incidence is equal to the angle of reflection.

Explanation:

angle of incidence (i) = angle of reflection (r)

So if the angle of incidence was 45°, the angle of reflection would also be 45°.

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2 years ago

In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s

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"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) $v_{i}$ = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd equation of motion

$h = v_{y_{f}}t + \frac{1}{2}gt^{2}$

-20 = $v_{y_{f}}$ (4) + 0.5(-9.8)(4)²

$v_{y_{f}}$ (4) = 58.4

$v_{y_{f}}$ = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

$v_{f}$ = $v_{y_{f}}$ / sinθ

$v_{f}$ = 14.6 / sin 60

$v_{f}$ = 16.86 m/s

$v_{x_{f}}$ = $v_{f}$ cosθ

$v_{x_{f}}$ = 16.86 cos 60

$v_{x_{f}}$ = 8.43 m/s

To calculate the horizontal distance

d = $v_{y_{f}}$ t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

$v_{x_{f}}$ = $v_{x_{i}}$

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = ( $v_{y_{f}}$ - $v_{y_{i}}$ ) / t

9.8 = 14.6 - $v_{y_{i}}$ ) / 4

$v_{y_{i}}$ = 24.6 m/s

$v_{i}$ = $\sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }$

$v_{i}$ = $\sqrt{8.43^{2}+24.6^{2}}$

$v_{i}$ = 26 m/s

c)

cos β = $v_{x_{i}}$ / $v_{i}$

β = cos⁻¹ (8.43 / 26)

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Answer:

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Answer:

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Provided all the other parameters are constant, resistance is inversely proportional to cross sectional area

r ∝ (1/A)

And the the cross sectional Area of the wire increases with increase in thickness & decreases with thickness

So, decreasing thickness ----> Decreasing Cross sectional Area ----> Increasing resistance.

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components of the speed of the coin is given as

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2 years ago