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zavuch27 [327]
3 years ago
11

Which of these nebulae is the odd one out?

Physics
2 answers:
Lubov Fominskaja [6]3 years ago
5 0

Answer: The answer is D!

Explanation:

solong [7]3 years ago
3 0

Answer:

Explanation:

A

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What is the weight of a box with a mass of 150 kg on Earth?
Dmitry [639]

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

6 0
2 years ago
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Analyzed or conducted

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2 years ago
Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
Ratling [72]

(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

minima equation is sinθ  = mλ/α where, m = 4 (minima with smallest distance)

sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

tanθ = y/L

  L = 4.5 cm

From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

5 0
1 year ago
HELP ASAP ILL GIVE BRAINLIST
OlgaM077 [116]

Answer:

Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps

4 0
2 years ago
Read 2 more answers
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  f_{objective } =  43.0 \ cm

    The focal length of  A  is   f_{eye} =  10.4 \  cm

The  theoretical angular  magnification is mathematically represented as

           m = \frac{f_{objective }}{f_{eye}}  =  \frac{43.0}{10.4}

            m = \frac{f_{objective }}{f_{eye}}  =  4.175

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

3 0
3 years ago
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