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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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A couch is pushed with a horizontal force of 80 N and moves the couch a
Lapatulllka [165]

Answer:

400 J

Explanation:

Work = force × distance

W = (80 N) (5 m)

W = 400 J

5 0
1 year ago
Read 2 more answers
The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee
enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

  • As the collision is perfectly elastic, kinetic energy must be conserved.
  • The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

        v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}

  • As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

  • As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)    

  • The maximum energy stored in the in the spring is given by the following expression:

       U =\frac{1}{2} *k * A^{2}

  • where A = maximum compression of the spring.
  • This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
  • When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
  • Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
  • As total kinetic energy must be conserved, the following condition must be met:

       KE_{10} = KE_{1f}  + KE_{2f}

  • So, it is clear that KE₂f  < KE₁₀
  • Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

  • As explained above, if total kinetic energy must be conserved:

        KE_{10} = KE_{1f}  + KE_{2f}

  • So as kinetic energy is always positive, KEf₂ < KE₁₀.
4 0
1 year ago
Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They
denis-greek [22]

Answer:

  v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

         Em₀ = U = m g h

Final point. Lower

        Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        m g h = ½ m v²

         v² = 2gh

         v = √ 2gh

Let's calculate

         v = √ (2 9.8 150)

         v = 54.2 m / s

3 0
1 year ago
Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of
Aleks [24]

We can solve the problem by using Kepler's third law, which states:

\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s

r=384000 km=3.84 \cdot 10^8 m

We can re-arrange the equation and find the Earth's mass:

M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg

4 0
2 years ago
in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at
Tema [17]
<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

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3 0
1 year ago
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