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stiv31 [10]
4 years ago
8

Why can cameras see objects too faint to be seen by the eye? A.They have larger detectors. B.They can record light for a longer

period of time. C.Their detectors are more sensitive to light than the eye. D.They have more pixels than the eye. E.They have larger lenses.
Physics
1 answer:
r-ruslan [8.4K]4 years ago
8 0

Answer:

B.They can record light for a longer period of time.

Explanation:

The human eye and a camera are very similar in their design and functioning. Each has its own merits and demerits. The one area where camera beats the eye is the ability to see the unseen. There are fainter stars in the night sky which will be invisible to naked eye but you point the camera towards the same area and take a long exposure shot. Voila! You will see a lot of stars.

The main reason is exposure time. It is a measure of how long the shutter of the camera is exposed to the light. in case of our eye, the exposure time will be the time for which our retina was exposed to incident light without being 'refreshed'. Since the retina keeps getting refreshed and we can't keep our eyes open for more than a few seconds/minutes we can't see the faint stars or the colors in the celestial objects. As the color development takes some time. Camera has this feature where you can increase the exposure time to even hours thus making them better at capturing fainter stars/objects.

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We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to
makkiz [27]

electromagnetic spectrum is consisting of many frequency range which is from gamma rays to radio waves

they are of various wavelength and different energy levels

minimum wavelength will occurs at Gamma rays

and maximum wavelength at Radio waves

the list of increasing order of wavelength is as following

Gamma rays < X rays < Ultraviolet < Visible Light < Infrared Waves < Radio Waves

so least to maximum order is

1. Gamma rays

2. X rays

3 Ultraviolet

4 Visible light

5 Infrared waves

6 Radio waves

5 0
3 years ago
12. When it is snowing, you should ___________________. A. drive slowly B. stay farther behind the vehicle ahead C. turn on your
maw [93]

it is D all the above.


4 0
3 years ago
If you start skating down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill, your
baherus [9]
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom.  The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.






6 0
3 years ago
Pilots often take advantage of the ____,which are highly-speed winds between 7km and 16 km above earths surface.
lukranit [14]
I believe it is called or referred to as the "Jet Stream". During World War II, allied pilots encountered high speed winds in the upper air. They named those winds after the fastest planes they came up against: fighters equipped with jet engines! Jet stream winds in winter time can reach up to 300 MPH as well!
6 0
3 years ago
Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of
Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

A=2\times10^{-6}m^2

* The young modulus of the steel is,

Y=2.1\times10^{11}\text{ Pa}

Solution:

The young modulus of the steel in terms of the force and extension is,

Y=\frac{F\times l}{A\times dl}

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0
1 year ago
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