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marysya [2.9K]
2 years ago
9

A diffraction grating is illuminated with yellow light at normal incidence. The pattern seen on a screen behind the grating cons

ists of three yellow spots, one at zero degrees (straight through) and one each at ±45°. You now add red light of equal intensity, coming in the same direction as the yellow light. The new pattern consists of1. red spots at 0 and +/- 45
2. yellow spots at 0 and +/- 45
3. Orange spots at 0 and +/- 45
4. an orange spot at 0, yellow spots at +/- 45, and red spots slightly further out
5. an orange spot at 0, yellow spots at +/- 45, and red spots slightly closer in
Physics
1 answer:
dedylja [7]2 years ago
4 0

Answer:

4. an orange spot at 0, yellow spots at +/- 45, and red spots slightly further out

Explanation:

At central place ( zero degree ) central maxima of both the color will be formed resulting into formation of orange spot there. At +/- 45 degree , yellow spot will be formed as before . For red light , wave length is larger so fringe width will be greater . Hence fringe will be formed farther away from the region of central maxima. It means red spot will be observed farther away or out.

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Init. A
Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

           v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

           Δx = 20 - 15 +20

           Δx = 25 m

therefore we calculate

         v = 25/75

         v = 1/3 m / s = 0.333 m / s

in the direction of the truck

7 0
2 years ago
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work
SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string T=5 N

2\pi r=2

r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 90^{\circ}

W=\int\vec{F}\cdot \vec{r}

W=\int Fdr\cos 90

W=0

4 0
2 years ago
Kenny and Candy decided to sit on a see-saw while visiting a local play park. Candy, of mass
pochemuha

Answer:

(i) 208 cm from the pivot

(ii) Move further from the pivot

Explanation:

(i) Sum of the moments about the pivot of the seesaw is zero.

∑τ = Iα

(50 kg) (10 N/kg) (2.5 m) + (60 kg) (10 N/kg) x = 0

1250 Nm + 600 N x = 0

x = -2.08 m

Kenny should sit 208 cm on the other side of the pivot.

(ii) To increase the torque, Kenny should move away from the pivot.

4 0
2 years ago
Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

7 0
1 year ago
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
aleksandr82 [10.1K]

Answer:

(a) W=-19.25J

(b) W=-52.8J

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

W=nRTln(\frac{V_1}{V_2} )

Whereas the moles are computed from the given mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Thus, the work is:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Regards.

4 0
2 years ago
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