Ask question

Ask question

All categories

Ray Of Light [21]

3 years ago

10

sports photographers often use large aperture, long focal length lenses. what limitations do these lenses impose on the photogra

phs?

1 answer:

Brut [27]3 years ago

6 0

Answer:

The depth of focus achievable with those lenses is small.

Explanation:

A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.

Hope this helps!

You might be interested in

What is the speed of an object that travels 60 metres in 4 seconds

Zina [86]

Answer:

$s = \frac{d}{t} = \frac{60}{4} \\ \boxed{speed = 15m. {sec}^{ - 1} }$

4 0

3 years ago

Read 2 more answers

Centripetal is a word used in science to mean

bogdanovich [222]

: the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation a string on the end of which a stone is whirled about exerts centripetal<span> force on the stone — compare centrifugal force.</span>

7 0

3 years ago

Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model

AysviL [449]

Answer:

Explanation:

If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.

n = c / v

If light travels faster in warmer air, it will have a lower refractive index

nh < nc

Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:

n1 * sin(θ1) = n2 * sin(θ2)

sin(θ2) = sin(θ1) * n1/n2

If blue light from the sky passing through the hot air will cross to the cold air, then

n1 = nh

n2 = nc

Then:

n1 < n2

So:

n1/n2 < 1

The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.

6 0

4 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

A 9.6 cm diameter circular loop of wire is in a 1.10 T magnetic field perpendicular to the plane of the loop. The loop is remove

Natali5045456 [20]

Answer:

Thus induced emf is 0.0531 V

Solution:

As per the question:

Diameter of the loop, $d = 9.6\ cm = 0.096\ m$

Thus the radius of the loop, R = 0.048 m

Time in which the loop is removed, t = 0.15 s

Magnetic field, B = 1.10 T

Now,

The average induced emf, e is given by Lenz Law:

$e = - \frac{\Delta \phi_{B}}{\Delta t}$

$e = - \frac{\Delta \phi_{B}}{\Delta t}$

where

$\phi_{B}$ = magnetic flux = $A\Delta B$

where

A = cross sectional area

Also, we know that:

$e = - \frac{A\Delta B}{\Delta t}$

$e = - \frac{\pi r^{2}\times (0 - 1.10)}{0.15}$

$e = - \frac{\pi \times 0.048^{2}\times (0 - 1.10)}{0.15}$

e = 0.0531 V

The sketch is shown in the figure, where I indicates the direction of the induced current.

3 0

3 years ago

Read 2 more answers