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2 years ago

5

-How can a camera help someone see something he or she wouldn't normally be able to see?

1 answer:

joja [24]2 years ago

4 0

Answer:

》How can a camera help someone see something he or she wouldn't normally be able to see?

- Although the human eye is able to observe fast events as they happen, it is not able to focus on a single point of time. We cannot freeze motion with our eyes. With a camera, however, so long as there is enough light, we can freeze motion. ... The camera can capture 'the moment', while your eye cannot.

》In what way is laser different from light produced by other sources?

-The major difference between laser light and light generated by white light sources (such as a light bulb) is that laser light is monochromatic, directional and coherent. ... Light from a conventional sources, such as a light bulb diverges, spreading in all directions.

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lozanna [386]

I am not so sure about this it is too difficult

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3 years ago

Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomical o

inysia [295]

Answer:

Telescope

Explanation:

Telescope is usually defined as an optical instrument that is commonly used to observe the objects in a magnified way that are located at a large distance from earth. These telescopes are comprised of lenses and curved mirrors that are needed to be arranged in a proper way in order to have a prominent look. It is commonly used by the astronomers.

This was first constructed by Hans Lippershey in the year 1608.

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3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

A cannon is shot downwards with an initial speed of 10m/s at an angle of 20 degrees South of East off a 60m cliff. Find the init

Sliva [168]

Answer:hmm

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3 years ago

The wasted energy in a charger is:

kirza4 [7]

Thermal is the wasted energy in a charger

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3 years ago