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Andrei [34K]
3 years ago
11

Walking along a 6-meter beam without falling helps to develop __________.

Physics
2 answers:
Gala2k [10]3 years ago
7 0

Walking along a 6-meter beam without falling helps to develop C. BALANCE.


IrinaVladis [17]3 years ago
4 0
C is correct answer.


Walking along a 6 meter beam without falling helps to develop balance.


Hope it helped.

-Charlie
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Which group of measurements is the most precise? A) 2 g, 3 g, 4 g B) 2 g, 2.5 g, 3 g C) 2.0 g, 3.0 g, 4.0 g, 5.0 g D) 2.0 g, 3.0
siniylev [52]
The answer to your question is A
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Question 3 of 10
Mazyrski [523]

Answer:

B . energy cannot be created or destroyed

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a 280 nm thin film with index of refraction 1.6 floats on waterwhat is the largest wavelength of reflected light for which const
aksik [14]

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

3 0
3 years ago
Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern
marusya05 [52]

Answer:

I = 0.636*Imax

Explanation:

(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:

I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)   (1)

I: intensity of light

Imax: maximum intensity of light

d: separation between slits = 0.200mm = 0.200 *10^-3 m

L: distance from the screen = 613cm = 0.613 m

y: distance to the central peak of the interference pattern

λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m

You replace the values of all variables in the equation (1):

I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}

Hence, the fraction of the maximum intensity is I = 0.636*Imax

6 0
3 years ago
Please i need help! ill give brainliest toooo!
Igoryamba

Answer:

b

Explanation:

7 0
3 years ago
Read 2 more answers
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