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IgorC [24]
3 years ago
13

In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separatio

n between adjacent bright fringes on a screen 5 m from the slits is:______a. 0.10 cm b. 0.25 cm c. 0.50 cm d. 1.0 cm e. none of the above
Physics
1 answer:
Keith_Richards [23]3 years ago
5 0

Answer:

b. 0.25cm

Explanation:

You can solve this question by using the formula for the position of the fringes:

y=\frac{m\lambda D}{d}

m: order of the fringes

lambda: wavelength 500nm

D: distance to the screen 5 m

d: separation of the slits 1mm=1*10^{-3}m

With the formula you can calculate the separation of two adjacent slits:

\Delta y=\frac{(m+1)(\lambda D)}{d}-\frac{m\lambda D }{d}=\frac{\lambda D}{d}\\\\\Delta y=\frac{(500*10^{-9}nm)(5m)}{1*10^{-3}m}=2.5*10^{-3}m=0.25cm

hence, the aswer is 0.25cm

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What is the formula that describes the magnitude of impulse on an object?
vladimir1956 [14]

Answer:

Option C.

Impulse = mass × change in velocity

Explanation:

Impulse is defined by the following the following formula:

Impulse = force (F) × time (t)

Impulse = Ft

From Newton's second law of motion,

Force = change in momentum /time

Cross multiply

Force × time = change in momentum

Recall:

Impulse = Force × time

Thus,

Impulse = change in momentum

Recall:

Momentum = mass x velocity

Momentum = mv

Chang in momentum = mass × change in velocity

Change in momentum = mΔv

Thus,

Impulse = change in momentum

Impulse = mass × change in velocity

8 0
3 years ago
Scientific research shows that birds are capable of regenerating their hair cells.
kiruha [24]

True.

It has been studied in a research study that claims birds have the ability and capability to regenerate their hair cells.

Brainliest please?

4 0
3 years ago
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When ___ attacks the surface of a metal, it becomes tarnished.​
alexandr1967 [171]

Answer:corrosion (i believe)

Explanation:

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3 years ago
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TRUE OR FALSE: In single-slit diffraction, the central band gets thinner as the width of the slit increases.
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Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi
meriva

Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Explanation:

So we are told that it is a RC circuit. We are told Q = C V [1 - e^(-t/RC)] = 12.0 V, R =  1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

Q = CV[1-e^{-t/RC} ]

The capacitor is fully charge when t approaches infinity, therefore:

Q =  \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.

6 0
3 years ago
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