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jolli1 [7]
2 years ago
7

Use the passage to answer the question.

Physics
1 answer:
Zielflug [23.3K]2 years ago
8 0

The reasoning in the passage is : ( C ) The spots are the result of interference, which happens with waves but not particles

<h3>Light waves </h3>

Light passed through a slit and projected onto a screen will form a pattern which is very visible and the pattern might have dark spots if there is some form of interference. Interference is present in waves and not particles.

Hence we can conclude that the reasoning in the passage is The spots are the result of interference, which happens with waves but not particles

Learn more about light waves : brainly.com/question/25847009

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Discuss potential behavioral concerns for people should they travel to Mars
Anna35 [415]

Answer:

The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...

6 0
3 years ago
Using the free-body diagram, calculate the net force
yuradex [85]

Answer:yes it is

Explanation:

5 0
3 years ago
Read 2 more answers
A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
kirill115 [55]

Answer:

the 70kg man

Explanation:

because he has more weight and is moving faster

4 0
3 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d
Butoxors [25]
Calculate the pressure due to sea water as density*depth.
That is, 
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
3 0
3 years ago
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