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insens350 [35]
3 years ago
6

What type of specialized cell in the eye is used for detecting low levels of light?

Physics
2 answers:
Hunter-Best [27]3 years ago
8 0

Answer:

There are about six to seven million cones in a human eye and are most concentrated towards the macula. Cones are less sensitive to light than the rod cells in the retina (which support vision at low light levels), but allow the perception of color.

pashok25 [27]3 years ago
8 0

Answer:

Rods

Explanation:

There are about six to seven million cones in a human eye and are most concentrated towards the macula. Cones are less sensitive to light than the rod cells in the retina (which support vision at low light levels), but allow the perception of color.

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3 0
3 years ago
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What is the velocity of a wave that has a wavelength of 20 meters and frequency of 0.5<br> Hz?
k0ka [10]

Answer:

v = 10 m/s

Explanation:

recall that velocity is related to wavelength and frequency by the formula

v = fλ

where v = velocity, f = frequency and λ= wavelength

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Which wind blows 30 latitude in both hemisphere almost to the equator
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2 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

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3 years ago
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