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Scilla [17]
3 years ago
11

In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation

between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Physics
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

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A ramp , a doorstop and a car Jack are made with what simple machine ?
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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

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Answer with Explanation:

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Current,I=19 A

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\theta=15^{\circ}

15 degree above the horizontal  in the northward direction.

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