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Vika [28.1K]
3 years ago
14

A converging lens can produce both real and virtual images depending on the position of the object. Explain when converging lens

es can produce virtual images and describe the image produced.
A) A virtual image is produced if the object is on the focal point, the image is upright, enlarged and on the same side of the lens as the object.

B) A virtual image is produced if the object is on the focal point, the image is inverted, enlarged and on the opposite side of the lens from the object.

C) A virtual image is produced if the object is between the focal point and the lens, the image will be upright, enlarged and on the same side of the lens
as the object.

D) A virtual image is produced if the object is between the focal point and the
lens, the image will be upright, reduced and on the opposite side of the
lens from the object.

Physics
2 answers:
inessss [21]3 years ago
5 0

Answer: C

Explanation:

Svet_ta [14]3 years ago
3 0

Answer:

c

Explanation:

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The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

\dfrac{1}{2} \times 166 \times v^2 = 17457.0912

v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2)  }}

Which gives;

\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66)  } \approx 8.1

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

Time = \dfrac{Distance}{Velocity}

\mathrm{The \ time \ the\ Lone \  Ranger \  and  \ Tonto \  have,  \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s

The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

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Answer:

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Explanation:

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Answer:

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Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

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Explanation:

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