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vlada-n [284]
3 years ago
8

A rubber ducky is placed 20 cm from a thin convex lens with a focal length of 15 cm. Which statement correctly describes the nat

ure and position of the image formed of the rubber ducky? A) Real and upright on the side of the lens opposite the rubber ducky. B) Real and inverted on the side of the lens opposite the rubber ducky. C) Imaginary and upright on the side of the lens opposite the rubber ducky. D) Imaginary and inverted on the same side of the lens as the rubber ducky.
Physics
2 answers:
blondinia [14]3 years ago
6 0
I believe the answer is B, a real and inverted image is formed on the side of the lens opposite the rubber ducky. The focal length is 15 cm and therefore the center of curvature (2F) will be 30 cm. When the object is placed between F and 2F (in this case 20 cm) in front of a convex lens, an inverted, real image is formed on the other side of the lens.
andre [41]3 years ago
6 0

Answer:

B

Explanation:

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Density can be kg/m^3 or g/cm3
In g/cm3 density =mass /volume =111g/23cm3
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At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
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As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

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\\ \sf\bull\dashrightarrow Momentum=Mass\times Velocity

\\ \sf\bull\dashrightarrow Momentum=2200(55)

\\ \sf\bull\dashrightarrow Momentum=121000kgm/s

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Answer: The height of the fluid rise is 0.01m

Explanation:

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h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

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