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torisob [31]
3 years ago
6

Yung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a sc

reen 1.10 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.2 mm from the center of the central bright fringe. What is the separation of the two slits?
Physics
1 answer:
Lady bird [3.3K]3 years ago
4 0

Answer:

1.082 mm

Explanation:

From the question, we can see that we were given The following

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1.1 m

We know that Y(20) = 10.2 mm = 10.2*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1.1 * 502*10^-9)/10.2*10^-3

d = 1.1*10^-5 / 10.2*10^-3

d = 1.082 mm

Therefore, we can say that the distance of separation between the two slits is 1.082 mm

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Yuliya22 [10]

Answer:

Explanation:

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v₁ = (m₁-m₂)/(m₁+m₂) u₁ + 2m₁m₂/(m₁+m₂) u₂

=( .06-.09 / .06+.09 ) 5.5 + (2 x .06 x .09 / .06+.09 ) 3.4

=( -.03 / .15) x 5.5 + (2 x .0054 / .15) x 3.4

= -1.1 +.2448

= - 0 .8552 m/s

Its direction will be - opposite direction

the formula for velocity of .09kg  is

v₂ = (m₂-m₁)/(m₁+m₂) u₂ + 2m₁m₂/(m₁+m₂) u₁

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4 0
3 years ago
Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo
kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

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here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

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3 years ago
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Answer:

a)  Total mass form, density and axis of rotation location are  True

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Explanation:

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3 years ago
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Answer:b) atoms

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8 0
2 years ago
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