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aniked [119]
1 year ago
9

The angle of incidence for a ray of light passing through the centre of curvature of a concave mirror is.

Physics
1 answer:
evablogger [386]1 year ago
8 0

The angle of incidence for a ray of light passing through the center of curvature of a concave mirror is 0°.

The angle of incidence is the angle between the surface's normal and the incident ray. For a concave mirror, the normal of the surface is along the center of the curvature, and a ray of light passed through a center of curvature passes through the normal of the surface.

The ray of light retreats its path making a zero angle of reflection. The law of reflection state that the angle of incidence is equal to the angle of reflection; therefore, the angle of incidence of a concave surface passed through the center of curvature is zero degrees.

Learn more about the  angle of incidence here:

brainly.com/question/3432273

#SPJ4

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Answer:

The thrown rock will strike the ground 2.42s earlier than the dropped rock.

Explanation:

<u>Known Data</u>

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We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}, then clearing for t_{D}.

t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s

<u>Time of the Thrown Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}, then, 0=-4.9t_{T}^{2}-29t_{T}+300, as it is a second-grade polynomial, we find that its positive root is t_{T}=5.4s

Finally, we can find how much earlier does the thrown rock strike the ground, so t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s

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