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Ede4ka [16]
3 years ago
11

Where should an object be placed in front of a concave mirror’s principal axis to form an image that is real, inverted, larger t

han the object, and farther from the mirror than the object is?
a. at the focal point on the principal axis
b. at the center of curvature on the principal axis
c. between the center of curvature and focal point on the principal axis
d. between the focal point and the vertex on the principal axis
Physics
2 answers:
malfutka [58]3 years ago
8 0
<em> I believe the answer is letter C.</em>
mart [117]3 years ago
4 0

Answer:

option C

Explanation:

In a concave mirror for the image to be real, inverted  and enlarged the object must be place between the center of curvature and focal point on the principal axis. To obtain an enlarged image object must not in proximity of the mirror, it should be far. The image formed will be on the same side as object( real ). but image will be inverted.

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Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

In order to determine the gravitational potential energy of the brick, we must know the mass (m) of the brick, its acceleration due to gravity (g) since it is acting under the influence of gravitational force and the distance between the top of the hill and the ground level. (The height).

Potential energy of a body is calculated as mass × acceleration due to gravity × height.

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you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh
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Answer:

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Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

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sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

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d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

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