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max2010maxim [7]
3 years ago
6

In a double-slit experiment, the slit separation is 2.0 mm, and two wavelengths, 750 nm and 900 nm, illuminate the slits. A scre

en is placed 2.0 m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other?
Physics
1 answer:
Zanzabum3 years ago
5 0

Answer:

4.5 x 10⁻6m

Explanation:

<em>From the formula given,</em>

<em>ʎ/d = x/L</em>

<em>The wavelength of 750 nm will be: 750 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x/ 2.0 </em>

<em> Now</em>

<em>50 x 10⁻⁹ / 2.0 x 10⁻³ 3 =  x 2.0  = x = 7.5 x 10⁻⁴m</em>

<em>The wavelength of 900 nm  will be:  900 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0 </em>

<em> 00 x 10 ⁻⁹ / 2.0 x 10⁻³ = x /2.0 = x = 9.0 x 10⁻⁴</em>

<em>Then</em>

<em>9.0  x 10⁻⁴ / 7.5 x 10⁻⁴ = 1.2</em>

<em>to get to the first integer, 1.3 must be 1.2  is the multiplied  5 times t (which will be a fringe (bright)) which is 6. </em>

<em>Therefore, what this mean is there will be 6 fringes of wavelength 750 nm and 5 fringes of wavelength 900 nm .</em>

<em>From the central brightest fringe.  the mixing fringe will be  4.5 x 10⁻6m  or 4500 nm .</em>

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Answer:

The answer is A/B, they're the same answer anyways.

Explanation:

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Identify three specific situations in which machines make work easier.
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3 years ago
An eagle carrying a trout flies above a lake along a horizontal path. The eagle drops the trout from a height of 6.1 m. The fish
Snezhnost [94]

Answer:

7.1 m/s

Explanation:

First, find the time it takes for the fish to reach the water.

Given in the y direction:

Δy = 6.1 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²

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Δx = 7.9 m

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4 0
3 years ago
Read 2 more answers
A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai
patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

<u>t = 4.79 s </u>

The equation for velocity is as follows:

v = v0  + a* t

Where:

v = velocity

v0 =  initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>

b) The time interval was calculated above, using the equation of  the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

<u>t = 1.55 s</u> ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = <u>6.34 s</u>

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

<u>The car lands 24 m from the base of the cliff.</u>

PHEW!, it was a very complete problem :)

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2 years ago
A meteor is
Leya [2.2K]

Answer:

A meteor is  B) an icy body with a long tail extending from it.

Explanation:

Meteors are very small dust particles that, when penetrating into the Earth's atmosphere, burn quickly by rubbing with the gases of the same. Some meteors, those with larger dimensions and appreciable weights, are brighter and can describe longer trajectories, showing longer.

In other words, the meteoroids, celestial bodies can vary in size between 100 micrometers up to 50 meters, they collide with the atmosphere of our planet and if the particles are of a small size, upon impact they enter combustion creating a flash, is what we know as meteor or shooting star. Therefore, the meteor is a luminous phenomenon that leaves behind a persistent trail.

So, <u><em>a meteor is  B) an icy body with a long tail extending from it.</em></u>

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2 years ago
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